RS Aggarwal Solutions for Class 12 Chapter 13 Method of Integration

Exercise 13(A)

Question 1. Evaluate the following integrals: \( \int (2x + 9)^5 dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 2x + 9 = t \Rightarrow 2 dx = dt \)
\( \int t^5 \left(\frac{dt}{2}\right) = \frac{1}{2} \int t^5 dt = \frac{1}{2} \cdot \frac{t^6}{6} + c = \frac{t^6}{12} + c = \frac{(2x + 9)^6}{12} + c \)

Exam Tip: Always identify the inner linear expression and apply the chain rule in reverse - the coefficient of the variable in the linear part becomes your divisor in the final answer.

Question 2. Evaluate the following integrals: \( \int \frac{1}{(7 - 3x)^5} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 7 - 3x = t \Rightarrow -3 dx = dt \)
\( \int t^{-5} \left(\frac{dt}{-3}\right) = \frac{-1}{3} \int t^{-5} dt = \frac{-1}{3} \cdot \frac{t^{-4}}{-4} + c = \frac{-1}{12(7 - 3x)^4} + c \)

Exam Tip: Watch the sign of the coefficient when substituting - a negative coefficient in the substitution changes the integral's sign, affecting your final answer.

Question 3. Evaluate the following integrals: \( \int \sqrt{3x - 5} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 3x - 5 = t \Rightarrow 3 dx = dt \)
\( \int t^{0.5} \left(\frac{dt}{3}\right) = \frac{1}{3} \int t^{0.5} dt = \frac{1}{3} \times \frac{t^{1.5}}{1.5} + c = \frac{2}{9} t^{1.5} + c = \frac{2(3x - 5)^{1.5}}{9} + c \)

Exam Tip: Convert roots to fractional exponents for easier integration - this makes applying the power rule straightforward.

Question 4. Evaluate the following integrals: \( \int (4x + 3)^{-0.5} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 4x + 3 = t \Rightarrow 4 dx = dt \)
\( \int t^{-0.5} \left(\frac{dt}{4}\right) = \frac{1}{4} \int t^{-0.5} dt = \frac{1}{4} \times \frac{t^{0.5}}{0.5} + c = \frac{1}{2} \sqrt{4x + 3} + c \)

Exam Tip: Fractional exponents with negative values become positive when you add 1 to them - ensure you handle the arithmetic correctly.

Question 5. Evaluate the following integrals: \( \int (3 - 4x)^{-0.5} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 3 - 4x = t \Rightarrow -4 dx = dt \)
\( \int t^{-0.5} \left(\frac{dt}{-4}\right) = \frac{-1}{4} \int t^{-0.5} dt = \frac{-1}{4} \times \frac{t^{0.5}}{0.5} + c = \frac{-1}{2} \sqrt{3 - 4x} + c \)

Exam Tip: The negative coefficient introduces a negative sign in your answer - be careful to preserve this throughout all steps.

Question 6. Evaluate the following integrals: \( \int (2x - 3)^{-1.5} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 2x - 3 = t \Rightarrow 2 dx = dt \)
\( \int t^{-1.5} \left(\frac{dt}{2}\right) = \frac{1}{2} \int t^{-1.5} dt = \frac{1}{2} \times \frac{t^{-0.5}}{-0.5} + c = \frac{-1}{\sqrt{2x - 3}} + c \)

Exam Tip: When the exponent becomes -0.5, it converts to a reciprocal square root in the final answer - simplify carefully.

Question 7. Evaluate the following integrals: \( \int \frac{1}{\sqrt{4x + 3}} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 4x + 3 = t \Rightarrow 4 dx = dt \)
\( \int t^{-0.5} \left(\frac{dt}{4}\right) = \frac{1}{4} \int t^{-0.5} dt = \frac{1}{4} \times \frac{t^{0.5}}{0.5} + c = \frac{\sqrt{4x + 3}}{2} + c \)

Exam Tip: Recognize reciprocal square root expressions as negative half-power terms to integrate them correctly using the standard formula.

Question 8. Evaluate the following integrals: \( \int \frac{1}{\sqrt{3 - 4x}} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 3 - 4x = t \Rightarrow -4 dx = dt \)
\( \int t^{-0.5} \left(\frac{dt}{-4}\right) = \frac{-1}{4} \int t^{-0.5} dt = \frac{-1}{4} \times \frac{t^{0.5}}{0.5} + c = \frac{-\sqrt{3 - 4x}}{2} + c \)

Exam Tip: The negative coefficient in the substitution produces a negative sign in the answer - include it in your final result.

Question 9. Evaluate the following integrals: \( \int \frac{1}{(2x - 3)^{1.5}} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 2x - 3 = t \Rightarrow 2 dx = dt \)
\( \int t^{-1.5} \left(\frac{dt}{2}\right) = \frac{1}{2} \int t^{-1.5} dt = \frac{1}{2} \times \frac{t^{-0.5}}{-0.5} + c = \frac{-1}{\sqrt{2x - 3}} + c \)

Exam Tip: Powers in the denominator become negative exponents - apply the power rule treating them as negative values.

Question 10. Evaluate the following integrals: \( \int e^{(2x-1)} dx \)
Answer: Using the formula \( \int e^x dx = e^x + c \),
Let \( 2x - 1 = t \Rightarrow 2 dx = dt \)
\( \int e^t \left(\frac{dt}{2}\right) = \frac{1}{2} \int e^t dt = \frac{1}{2} \times e^t + c = \frac{e^{(2x-1)}}{2} + c \)

Exam Tip: When integrating exponential functions with a linear exponent, divide by the coefficient of the variable in the exponent.

Question 11. Evaluate the following integrals: \( \int e^{(1-3x)} dx \)
Answer: Using the formula \( \int e^x dx = e^x + c \),
Let \( 1 - 3x = t \Rightarrow -3 dx = dt \)
\( \int e^t \left(\frac{dt}{-3}\right) = \frac{-1}{3} \int e^t dt = \frac{-1}{3} \times e^t + c = \frac{-e^{(1-3x)}}{3} + c \)

Exam Tip: A negative coefficient in the linear exponent introduces a negative sign - keep track of this through all steps to avoid sign errors.

Question 12. Evaluate the following integrals: \( \int 3^{(2-3x)} dx \)
Answer: Using the formula \( \int a^x dx = \frac{a^x}{\log a} + c \),
Let \( 2 - 3x = t \Rightarrow -3 dx = dt \)
\( \int 3^t \left(\frac{dt}{-3}\right) = \frac{-1}{3} \int 3^t dt = \frac{-1}{3} \times \frac{3^t}{\log 3} + c = \frac{-3^{(2-3x)}}{3 \log 3} + c \)

Exam Tip: For exponential functions with any base, include the natural logarithm of that base in the denominator of your answer.

Question 13. Evaluate the following integrals: \( \int \sin 3x dx \)
Answer: Using the formula \( \int \sin x dx = -\cos x + c \),
Let \( 3x = t \Rightarrow 3 dx = dt \)
\( \int \sin t \left(\frac{dt}{3}\right) = \frac{1}{3} \int \sin t dt = \frac{1}{3} \times (-\cos t) + c = \frac{-\cos 3x}{3} + c \)

Exam Tip: For trigonometric functions with a coefficient on the variable, divide the result by that coefficient.

Question 14. Evaluate the following integrals: \( \int \cos(5 + 6x) dx \)
Answer: Using the formula \( \int \cos x dx = \sin x + c \),
Let \( 5 + 6x = t \Rightarrow 6 dx = dt \)
\( \int \cos t \left(\frac{dt}{6}\right) = \frac{1}{6} \int \cos t dt = \frac{1}{6} \times (\sin t) + c = \frac{\sin(5 + 6x)}{6} + c \)

Exam Tip: The coefficient multiplying the variable determines your divisor in the final answer - be consistent with the substitution process.

Question 15. Evaluate the following integrals: \( \int \sin x \sqrt{1 + \cos 2x} dx \)
Answer: Using the identity \( 1 + \cos 2x = 2\cos^2 x \),
\( \int \sin x \sqrt{2\cos x} dx = \int \sqrt{2} \sin x \cos x dx \)
Let \( \sin x = t \Rightarrow \cos x dx = dt \)
\( \int \sqrt{2} t dt = \sqrt{2} \times \frac{t^2}{2} + c = \frac{(\sin x)^2}{\sqrt{2}} + c \)

Exam Tip: Recognize double-angle identities to simplify square roots - this makes substitution much more manageable.

Question 16. Evaluate the following integrals: \( \int \cos^2(2x + 5) dx \)
Answer: Using the formula \( \int \csc^2 x dx = -\cot x + c \),
Let \( 2x + 5 = t \Rightarrow 2 dx = dt \)
\( \int \csc^2 t \left(\frac{dt}{2}\right) = \frac{1}{2} \int \csc^2 t dt = \frac{1}{2} \times (-\cot t) + c = \frac{-1}{2}\cot(2x + 5) + c \)

Exam Tip: Identify which trigonometric function you have - cosecant squared integrates to negative cotangent, not positive.

Question 17. Evaluate the following integrals: \( \int \sin x \cos x dx \)
Answer: Using the formula \( \int \sin x dx = -\cos x + c \),
Let \( \sin x = t \Rightarrow \cos x dx = dt \)
\( \int t dt = \frac{t^2}{2} + c = \frac{(\sin x)^2}{2} + c \)

Exam Tip: When you have a product of sine and cosine, let one of them be your substitution variable - the other appears naturally as your differential.

Question 18. Evaluate the following integrals: \( \int \sin^3 x \cos x dx \)
Answer: Using the formula \( \int \sin x dx = -\cos x + c \),
Let \( \sin x = t \Rightarrow \cos x dx = dt \)
\( \int t^3 dt = \frac{t^4}{4} + c = \frac{(\sin x)^4}{4} + c \)

Exam Tip: Powers of sine with cosine as the differential are straightforward - just apply the power rule after substitution.

Question 19. Evaluate the following integrals: \( \int \left(\sqrt{\cos x}\right) \sin x dx \)
Answer: Using the formula \( \int \sin x dx = -\cos x + c \),
Let \( \cos x = t \Rightarrow -\sin x dx = dt \)
\( \int t^{0.5} (-dt) = -\frac{t^{1.5}}{1.5} + c = \frac{-2(\cos x)^{1.5}}{3} + c \)

Exam Tip: When sine appears with cosine, the negative sign from the derivative of cosine must be included in your setup.

Question 20. Evaluate the following integrals: \( \int \frac{\sin^{-1} x}{\sqrt{1 - x^2}} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \) and \( \frac{d(\sin^{-1} x)}{dx} = \frac{1}{\sqrt{1-x^2}} \),
Let \( \sin^{-1} x = t \Rightarrow \frac{1}{\sqrt{1-x^2}} dx = dt \)
\( \int t dt = \frac{t^2}{2} + c = \frac{(\sin^{-1} x)^2}{2} + c \)

Exam Tip: Recognize inverse trigonometric functions and their derivatives - they often pair naturally with standard differentials.

Question 21. Evaluate the following integrals: \( \int \frac{\sin(2\tan^{-1} x)}{(1 + x^2)} dx \)
Answer: Using the formula \( \int \sin t dt = -\cos t + c \) and \( \frac{d(\tan^{-1} x)}{dx} = \frac{1}{1+x^2} \),
Let \( \tan^{-1} x = t \Rightarrow \frac{1}{1+x^2} dx = dt \)
\( \int \sin 2t dt = \frac{-\cos 2t}{2} + c = \frac{-\cos(2\tan^{-1} x)}{2} + c \)

Exam Tip: Match inverse trigonometric derivatives with their corresponding differentials - the substitution becomes evident.

Question 22. Evaluate the following integrals: \( \int \frac{\cos(\log x)}{x} dx \)
Answer: Using the formula \( \int \cos t dt = \sin t + c \) and \( \frac{d(\log x)}{dx} = \frac{1}{x} \),
Let \( \log x = t \Rightarrow \frac{1}{x} dx = dt \)
\( \int \cos t dt = \sin t + c = \sin(\log x) + c \)

Exam Tip: When logarithm appears with 1/x in the integrand, use the natural pairing - logarithm as your substitution variable.

Question 23. Evaluate the following integrals: \( \int \frac{\csc^2(\log x)}{x} dx \)
Answer: Using the formula \( \int \csc^2 t dt = -\cot t + c \) and \( \frac{d(\log x)}{dx} = \frac{1}{x} \),
Let \( \log x = t \Rightarrow \frac{1}{x} dx = dt \)
\( \int \csc^2 t dt = -\cot t + c = -\cot(\log x) + c \)

Exam Tip: Cosecant squared always integrates to negative cotangent - remember this key formula to avoid mistakes.

Question 24. Evaluate the following integrals: \( \int \frac{1}{x \log x} dx \)
Answer: Using the formula \( \frac{d(\log x)}{dx} = \frac{1}{x} \) and \( \int \frac{1}{t} dt = \log t + c \),
Let \( \log x = t \Rightarrow \frac{1}{x} dx = dt \)
\( \int \frac{1}{t} dt = \log t + c = \log(\log x) + c \)

Exam Tip: Nested logarithms require careful substitution - identify the inner logarithm first as your substitution variable.

Question 25. Evaluate the following integrals: \( \int \frac{(x + 1)(x + \log x)^2}{x} dx \)
Answer: Using the formula \( \frac{d(\log x)}{dx} = \frac{1}{x} \),
\( \int \frac{(x + 1)(x + \log x)^2}{x} dx = \int \left(1 + \frac{1}{x}\right) \times \frac{(x + \log x)^2}{1} dx \)
Let \( x + \log x = t \Rightarrow \left(1 + \frac{1}{x}\right) dx = dt \)
\( \int t^2 dt = \frac{t^3}{3} + c = \frac{(x + \log x)^3}{3} + c \)

Exam Tip: Factor out combinations that resemble derivatives - if the derivative of one part appears in the integrand, use that as your substitution.

Question 26. Evaluate the following integrals: \( \int e^{\cos^2 x} \sin 2x dx \)
Answer: Using the formula \( \int e^x dx = e^x + c \) and \( \frac{d(\cos^2 x)}{dx} = 2\cos x(-\sin x) = -\sin 2x \),
Let \( \cos^2 x = t \Rightarrow -\sin 2x dx = dt \)
\( \int e^t (-dt) = -e^t + c = -e^{\cos^2 x} + c \)

Exam Tip: Recognize that \( \sin 2x = 2\sin x \cos x \) - this helps identify the derivative of \( \cos^2 x \).

Question 27. Evaluate the following integrals: \( \int \sin(ax + b)\cos(ax + b) dx \)
Answer: Using the formula \( \int \sin x dx = -\cos x + c \),
Let \( ax+b = t \Rightarrow a dx = dt \)
\( \int \sin t \cos t \left(\frac{dt}{a}\right) = \frac{1}{a} \int \sin t \cos t dt \)
Let \( \sin t = z \Rightarrow \cos t dt = dz \)
\( \frac{1}{a} \int z dz = \frac{1}{a} \times \frac{z^2}{2} + c = \frac{(\sin(ax + b))^2}{2a} + c \)

Exam Tip: Products of sine and cosine may require nested substitutions - work systematically through each layer.

Question 28. Evaluate the following integrals: \( \int \cos^3 x dx \)
Answer: Using the formula \( \int \cos x dx = \sin x + c \) and the identity \( \cos 3x = 3\cos x - 4\cos^3 x \),
\( \int \left(\frac{3\cos x}{4} - \frac{\cos 3x}{4}\right) dx = \frac{3\sin x}{4} - \frac{\sin 3x}{12} + c \)

Exam Tip: For odd powers of cosine, use triple-angle formulas to reduce the power and integrate term-by-term.

Question 29. Evaluate the following integrals: \( \int \frac{1}{x^2} e^{-1/x} dx \)
Answer: Using the formula \( \int e^x dx = e^x + c \),
Let \( -\frac{1}{x} = t \Rightarrow \frac{1}{x^2} dx = dt \)
\( \int e^t dt = e^t + c = e^{-1/x} + c \)

Exam Tip: When a reciprocal expression appears in an exponent, its derivative often shows up naturally in the integrand.

Question 30. Evaluate the following integrals: \( \int \frac{1}{x^2} \cos\left(\frac{1}{x}\right) dx \)
Answer: Using the formula \( \int \cos x dx = \sin x + c \),
Let \( -\frac{1}{x} = t \Rightarrow \frac{1}{x^2} dx = dt \)
\( \int \cos(dt) = \int \cos t dt = \sin t + c = \sin\left(-\frac{1}{x}\right) + c \)

Exam Tip: Reciprocals in trigonometric arguments follow the same substitution pattern as exponentials.

Question 31. Evaluate the following integrals: \( \int \frac{dx}{(e^x + e^{-x})} \)
Answer: Using the formula \( \int e^x dx = e^x + c \),
\( \int \frac{e^x}{1 + e^{2x}} dx \)
Let \( e^x = t \Rightarrow e^x dx = dt \)
\( \int \frac{1}{1 + t^2} dt = \tan^{-1} t + c = \tan^{-1}(e^x) + c \)

Exam Tip: Rewrite sums and differences of exponentials by factoring - this often reveals a standard form to integrate.

Question 32. Evaluate the following integrals: \( \int \frac{e^{2x}}{(e^{2x} - 2)} dx \)
Answer: Using the formula \( \int e^x dx = e^x + c \),
Let \( e^{2x} - 2 = t \Rightarrow 2e^{2x} dx = dt \)
\( \int \frac{1}{t} \left(\frac{dt}{2}\right) = \frac{1}{2} \log t + c = \frac{1}{2}\log(e^{2x} - 2) + c \)

Exam Tip: When an exponential appears in both numerator and denominator, the derivative relationship may help - check if one is the differential of the other.

Question 33. Evaluate the following integrals: \( \int \cot x \log(\sin x) dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( \log(\sin x) = t \Rightarrow \frac{\cos x}{\sin x} dx = dt \Rightarrow \cot x dx = dt \)
\( \int t dt = \frac{t^2}{2} + c = \frac{(\log \sin x)^2}{2} + c \)

Exam Tip: Logarithmic expressions paired with the derivative of their argument simplify beautifully - always look for this pairing.

Question 34. Evaluate the following integrals: \( \int \frac{\cot x}{\log(\sin x)} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( \log(\sin x) = t \Rightarrow \frac{\cos x}{\sin x} dx = dt \Rightarrow \cot x dx = dt \)
\( \int \frac{1}{t} dt = \log t + c = \log(\log \sin x) + c \)

Exam Tip: Nested logarithms appear when you integrate 1/log of something - careful notation is essential here.

Question 35. Evaluate the following integrals: \( \int 2x \sin(x^2 + 1) dx \)
Answer: Using the formula \( \int \sin x dx = -\cos x + c \),
Let \( x^2 + 1 = t \Rightarrow 2x dx = dt \)
\( \int \sin t dt = -\cos t + c = -\cos(x^2 + 1) + c \)

Exam Tip: The derivative of a polynomial expression inside a trigonometric function often appears directly in the integrand - use this to guide your substitution.

Question 36. Evaluate the following integrals: \( \int \sec x \log(\sec x + \tan x) dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( \log(\sec x + \tan x) = t \)
\( \frac{1}{\sec x + \tan x} \times (\sec x \tan x + \sec^2 x) dx = dt \Rightarrow \sec x dx = dt \)
\( \int t dt = \frac{t^2}{2} + c = \frac{(\log(\sec x + \tan x))^2}{2} + c \)

Exam Tip: Complex substitutions require careful differentiation - verify that the derivative truly matches the remaining integrand.

Question 37. Evaluate the following integrals: \( \int \frac{\tan\sqrt{x} \sec^2\sqrt{x}}{\sqrt{x}} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( \tan\sqrt{x} = t \)
\( \sec^2\sqrt{x} \times \frac{1}{2\sqrt{x}} dx = dt \)
\( \int t dt = \frac{t^2}{2} + c = \frac{(\tan\sqrt{x})^2}{2} + c \)

Exam Tip: Square roots in trigonometric arguments require chain rule application - work through the derivatives carefully.

Question 38. Evaluate the following integrals: \( \int \frac{x \tan^{-1} x^2}{(1 + x^4)} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( \tan^{-1} x^2 = t \Rightarrow \frac{2x}{1 + x^4} dx = dt \)
\( \int t \left(\frac{dt}{2}\right) = \frac{1}{2} \times \frac{t^2}{2} + c = \frac{(\tan^{-1} x^2)^2}{4} + c \)

Exam Tip: Inverse trigonometric functions paired with their natural derivatives simplify nicely - always check for this pairing.

Question 39. Evaluate the following integrals: \( \int \frac{x \sin^{-1} x^2}{\sqrt{1 - x^4}} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( \sin^{-1} x^2 = t \Rightarrow \frac{2x}{\sqrt{1 - (x^2)^2}} dx = dt \Rightarrow \frac{2x}{\sqrt{1 - x^4}} dx = dt \)
\( \int t \left(\frac{dt}{2}\right) = \frac{1}{2} \times \frac{t^2}{2} + c = \frac{(\sin^{-1} x^2)^2}{4} + c \)

Exam Tip: When derivatives of inverse trig functions appear, they pair naturally with certain radical expressions - identify this pattern quickly.

Question 40. Evaluate the following integrals: \( \int \frac{1}{\sqrt{1 - x^2} \sin^{-1} x} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( \sin^{-1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^2}} dx = dt \)
\( \int \frac{1}{t} dt = \log t + c = \log \sin^{-1} x + c \)

Exam Tip: The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1-x^2}} \) - when both appear, substitution is straightforward.

Question 41. Evaluate the following integrals: \( \int \frac{1/x}{(2 + \log x)} dx \)
Answer: Using the formula \( \int \frac{1}{x} dx = \log x + c \),
Let \( 2 + \log x = t \Rightarrow \frac{1}{x} dx = dt \)
\( \int \frac{1}{t} dt = \log t + c = \log(2 + \log x) + c \)

Exam Tip: Logarithmic expressions with \( 1/x \) always lead to simpler integrals - remember this shortcut for quick solutions.

Question 42. Evaluate the following integrals: \( \int \frac{\sec^2 x}{(1 + \tan x)} dx \)
Answer: Using the formula \( \int \frac{1}{x} dx = \log x + c \),
Let \( 1 + \tan x = t \Rightarrow \sec^2 x dx = dt \)
\( \int \frac{1}{t} dt = \log t + c = \log(1 + \tan x) + c \)

Exam Tip: The derivative of \( \tan x \) is \( \sec^2 x \) - when both appear in an integral, use this directly in your substitution.

Question 43. Evaluate the following integrals: \( \int \frac{\sin x}{(1 + \cos x)} dx \)
Answer: Using the formula \( \int \cos x dx = \sin x + c \),
Let \( 1 + \cos x = t \Rightarrow -\sin x dx = dt \)
\( \int \frac{-dt}{t} = -\log t + c = -\log(1 + \cos x) + c \)

Exam Tip: The negative sign from the derivative of cosine must be included - track it carefully throughout the substitution process.

Question 44. Evaluate the following integrals: \( \int \frac{\sqrt{(2 + \log x)}}{x} dx \)
Answer: Using the formula \( \int x^n dx = \frac{x^{(n+1)}}{n+1} + c \),
Let \( 2 + \log x = t \Rightarrow \frac{1}{x} dx = dt \)
\( \int \sqrt{t} dt = \int t^{0.5} dt = \frac{2t^{1.5}}{3} + c = \frac{2(2 + \log x)^{1.5}}{3} + c \)

Exam Tip: Square roots can be rewritten as fractional exponents - this makes the power rule easier to apply.

Question 45. Evaluate the following integrals:
(i) \( \int \frac{(1 + \tan x)}{(x + \log \sec x)} \, dx \)
(ii) \( \int \frac{(1 - \sin 2x)}{(x + \cos^2 x)} \, dx \)
Answer:
(i) Formula: \( \int \frac{1}{x} dx = \log x + c \)
Therefore,
Put \( x + \log(\sec x) = t \Rightarrow \)
\( (1 + \frac{1}{\sec x} \times \sec x \tan x) dx = dt \)
\( (1 + \tan x) dx = dt \)
\( \int \frac{dt}{t} = \int \frac{1}{t} dt = \log t + c \)
\( = \log(x + \log(\sec x)) + c \)

(ii) Formula: \( \int \frac{1}{x} dx = \log x + c \)
Therefore,
Put \( x + \cos^2 x = t \Rightarrow 1 + 2 \cos x \times (-\sin x) dx = dt \)
\( (1 - \sin 2x) dx = dt \)
\( \int \frac{dt}{t} = \int \frac{1}{t} dt = \log t + c \)
\( = \log(x + \cos^2 x) + c \)
In simple words: When the numerator is the derivative of the denominator, the integral becomes the natural logarithm of that denominator.

Exam Tip: Recognise when the numerator matches the derivative of the denominator - this is a key pattern for logarithmic integrals.

Question 46. Evaluate the following integrals:
\( \int \left(\frac{1 + \tan x}{1 - \tan x}\right) dx \)
Answer: Formula: \( \int \cos x \, dx = \sin x + c \)
Therefore,
\( \int \left(\frac{1 + \frac{\sin x}{\cos x}}{1 - \frac{\sin x}{\cos x}}\right) dx = \int \left(\frac{\cos x + \sin x}{\cos x - \sin x}\right) dx \)
Put \( \cos x - \sin x = t \Rightarrow
\) \( (-\sin x - \cos x) dx = dt \)
\( \int \left(\frac{-dt}{t}\right) = -\int \frac{1}{t} dt = -\log t + c \)
\( = -\log(\cos x - \sin x) + c \)
In simple words: Rewrite the fraction in terms of sine and cosine, then identify the derivative pattern in the numerator.

Exam Tip: Always convert tangent to sine/cosine ratio to reveal hidden derivative relationships in the integrand.

Question 47. Evaluate the following integrals:
\( \int \frac{\sin 2x}{(a^2 + b^2 \sin^2 x)} \, dx \)
Answer: Formula: \( \int \frac{1}{x} dx = \log x + c \)
Therefore,
Put \( a^2 + b^2 \sin^2 x = t \)
\( (a^2 \times 2 \cos x \times (-\sin x) + b^2 \times 2 \sin x \times \cos x) dx = dt \)
\( (b^2 - a^2) \sin 2x \, dx = dt \)
\( \int \frac{1}{t} \left(\frac{dt}{b^2 - a^2}\right) = \frac{1}{b^2 - a^2} \int \frac{1}{t} dt = \frac{1}{b^2 - a^2} \log t + c \)
\( = \frac{1}{b^2 - a^2} \log|a^2 + b^2 \sin^2 x| + c \)
In simple words: Differentiate the denominator expression to match it with the numerator, then apply the logarithmic integral formula.

Exam Tip: Write out the derivative of the denominator completely to ensure the constant factor is handled correctly.

Question 48. Evaluate the following integrals:
\( \int \left(\frac{2\cos x - 3\sin x}{3\cos x + 2\sin x}\right) dx \)
Answer: Formula: \( \int \cos x \, dx = \sin x + c \)
Therefore,
Put \( 3\cos x + 2\sin x = t \Rightarrow (2\cos x - 3\sin x) dx = dt \)
\( \int \left(\frac{dt}{t}\right) = \int \frac{1}{t} dt = \log t + c \)
\( = \log(3\cos x + 2\sin x) + c \)
In simple words: The numerator is exactly the derivative of the denominator, making this a straightforward logarithmic integral.

Exam Tip: Always check whether the numerator is the derivative of the denominator before attempting other methods.

Question 49. Evaluate the following integrals:
\( \int \frac{4x}{(2x^2 + 3)} \, dx \)
Answer: Formula: \( \int x^n dx = \frac{x^{n+1}}{n+1} + c \)
Therefore,
Put \( 2x^2 + 3 = t \Rightarrow (4x) dx = dt \)
\( \int \left(\frac{dt}{t}\right) = \int \frac{1}{t} dt = \log t + c \)
\( = \log(2x^2 + 3) + c \)
In simple words: The numerator matches the derivative of the expression inside the brackets in the denominator.

Exam Tip: Factor out constants from the denominator and verify the coefficient of the derivative matches the numerator.

Question 50. Evaluate the following integrals:
\( \int \frac{(x + 1)}{(x^2 + 2x - 3)} \, dx \)
Answer: Formula: \( \int x^n dx = \frac{x^{n+1}}{n+1} + c \)
Therefore,
Put \( x^2 + 2x - 3 = t \Rightarrow (2x + 2) dx = dt \Rightarrow 2(x + 1) dx = dt \)
\( \int \frac{1}{t} \left(\frac{dt}{2}\right) = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log t + c \)
\( = \frac{1}{2} \log(x^2 + 2x - 3) + c \)
In simple words: The numerator is half the derivative of the denominator, so introduce the factor \( \frac{1}{2} \) when substituting.

Exam Tip: When the numerator is a constant multiple of the denominator's derivative, extract that constant as a fraction outside the integral.

Question 51. Evaluate the following integrals:
\( \int \frac{(4x - 5)}{(2x^2 - 5x + 1)} \, dx \)
Answer: To find: Value of \( \int \frac{4x - 5}{2x^2 - 5x + 1} dx \)
Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)
We have, \( I = \int \frac{4x - 5}{2x^2 - 5x + 1} dx \) ... (i)
Let \( 2x^2 - 5x + 1 = t \)
\( \Rightarrow \frac{d(2x^2 - 5x + 1)}{dx} = \frac{dt}{dx} \)
\( \Rightarrow (4x - 5) dx = dt \)
Putting this value in equation (i)
\( I = \int \frac{dt}{t} [2x^2 - 5x + 1 = t] \)
\( I = \log|t| + c \)
\( I = \log|2x^2 - 5x + 1| + c \)
Ans) \( \log|2x^2 - 5x + 1| + c \)
In simple words: The expression in the numerator is the exact derivative of the denominator, so the answer is simply the natural logarithm of the denominator.

Exam Tip: Always verify that the numerator is the derivative of the denominator by differentiating the denominator term by term.

Question 52. Evaluate the following integrals:
\( \int \frac{(9x^2 - 4x + 5)}{(3x^3 - 2x^2 + 5x + 1)} \, dx \)
Answer: To find: Value of \( \int \frac{9x^2 - 4x + 5}{3x^3 - 2x^2 + 5x + 1} dx \)
Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)
We have, \( I = \int \frac{9x^2 - 4x + 5}{3x^3 - 2x^2 + 5x + 1} dx \) ... (i)
Let \( 3x^3 - 2x^2 + 5x + 1 = t \)
\( \Rightarrow (9x^2 - 4x + 5) dx = dt \)
Putting this value in equation (i)
\( I = \int \frac{dt}{t} [3x^3 - 2x^2 + 5x + 1 = t] \)
\( I = \log|t| + c \)
\( I = \log|3x^3 - 2x^2 + 5x + 1| + c \)
Ans) \( \log|3x^3 - 2x^2 + 5x + 1| + c \)
In simple words: The numerator is precisely the derivative of the denominator expression, so the integral yields the logarithm of that denominator.

Exam Tip: Differentiate the denominator systematically, term by term, to confirm it matches the numerator exactly.

Question 53. Evaluate the following integrals:
\( \int \frac{\sec x \cosec x}{\log(\tan x)} \, dx \)
Answer: To find: Value of \( \int \frac{\sec x \cosec x}{\log(\tan x)} dx \)
Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)
We have, \( I = \int \frac{\sec x \cosec x}{\log(\tan x)} dx \) ... (i)
Let \( \log(\tan x) = t \)
\( \Rightarrow \frac{d(\log(\tan x))}{dx} = \frac{dt}{dx} \)
\( \Rightarrow \frac{d(\log(\tan x)) d\tan x}{d\tan x \, dx} = \frac{dt}{dx} \)
\( \Rightarrow \frac{1}{\tan x} \sec^2 x = \frac{dt}{dx} \)
\( \Rightarrow \sec x \cosec x = \frac{dt}{dx} \)
\( \Rightarrow (\sec x \cosec x) dx = dt \)
Putting this value in equation (i)
\( I = \int \frac{dt}{t} [\log(\tan x) = t] \)
\( I = \log|t| + c \)
\( I = \log|\log(\tan x)| + c \)
Ans) \( \log|\log(\tan x)| + c \)
In simple words: The numerator becomes the derivative of the logarithmic expression in the denominator through careful chain rule differentiation.

Exam Tip: When dealing with logarithms in the denominator, use the chain rule to find derivatives of composite functions involving logs.

Question 54. Evaluate the following integrals:
\( \int \frac{(1 + \cos x)}{(x + \sin x)^3} \, dx \)
Answer: To find: Value of \( \int \frac{(1 + \cos x)}{(x + \sin x)^3} dx \)
Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)
We have, \( I = \int \frac{(1 + \cos x)}{(x + \sin x)^3} dx \) ... (i)
Let \( x + \sin x = t \)
\( \Rightarrow \frac{d(x + \sin x)}{dx} = \frac{dt}{dx} \)
\( \Rightarrow \frac{d(x)}{dx} + \frac{d(\sin x)}{dx} = \frac{dt}{dx} \)
\( \Rightarrow (1 + \cos x) = \frac{dt}{dx} \)
\( \Rightarrow (1 + \cos x) dx = dt \)
Putting this value in equation (i)
\( I = \int \frac{dt}{t^3} [x + \sin x = t] \)
\( \Rightarrow I = -\frac{1}{2t^2} + c \)
\( I = -\frac{1}{2(x + \sin x)^2} + c \)
Ans) \( -\frac{1}{2(x + \sin x)^2} + c \)
In simple words: The numerator is the derivative of the expression raised to the third power in the denominator, so use the power rule for integration.

Exam Tip: Recognise when a power appears in the denominator - rewrite it as a negative exponent and apply the standard power integration formula.

Question 55. Evaluate the following integrals:
\( \int \frac{\sin x}{(1 + \cos x)^2} \, dx \)
Answer: To find: Value of \( \int \frac{\sin x}{(1 + \cos x)^2} dx \)
Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)
We have, \( I = \int \frac{\sin x}{(1 + \cos x)^2} dx \) ... (i)
Let \( 1 + \cos x = t \)
\( \Rightarrow \frac{d(1 + \cos x)}{dx} = \frac{dt}{dx} \)
\( \Rightarrow \frac{d(1)}{dx} + \frac{d(\cos x)}{dx} = \frac{dt}{dx} \)
\( \Rightarrow (0 - \sin x) = \frac{dt}{dx} \)
\( \Rightarrow (-\sin x) dx = dt \)
Putting this value in equation (i)
\( I = \int -\frac{dt}{t^2} [1 + \cos x = t] \)
\( \Rightarrow I = \frac{1}{t} + c \)
\( I = \frac{1}{1 + \cos x} + c \)
Ans) \( \frac{1}{1 + \cos x} + c \)
In simple words: The negative of the numerator is the derivative of the denominator's base, so introduce a negative sign and apply the power rule.

Exam Tip: When the numerator is the negative of a denominator's derivative, compensate by adding a negative sign in front of the integral.

Question 56. Evaluate the following integrals:
\( \int \frac{(2x + 3)}{\sqrt{x^2 + 3x - 2}} \, dx \)
Answer: To find: Value of \( \int \frac{(2x + 3)}{\sqrt{x^2 + 3x - 2}} dx \)
Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)
We have, \( I = \int \frac{\sin x}{(1 + \cos x)^2} dx \) ... (i)
Let \( x^2 + 3x - 2 = t \)
\( \Rightarrow (2x + 3) = \frac{dt}{dx} \)
\( \Rightarrow (2x + 3) dx = dt \)
Putting this value in equation (i)
\( I = \int \frac{dt}{\sqrt{t}} [x^2 + 3x - 2 = t] \)
\( \Rightarrow I = 2\sqrt{t} + c \)
\( I = 2\sqrt{x^2 + 3x - 2} + c \)
Ans) \( 2\sqrt{x^2 + 3x - 2} + c \)
In simple words: The numerator matches the derivative of the expression under the square root in the denominator.

Exam Tip: Convert square root notation to fractional exponent form to apply the power rule more easily.

Question 57. Evaluate the following integrals:
\( \int \frac{(2x - 1)}{\sqrt{x^2 - x - 1}} \, dx \)
Answer: To find: Value of \( \int \frac{(2x - 1)}{\sqrt{x^2 - x - 1}} dx \)
Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)
We have, \( I = \int \frac{\sin x}{(1 + \cos x)^2} dx \) ... (i)
Let \( x^2 - x - 1 = t \)
\( \Rightarrow (2x - 1) = \frac{dt}{dx} \)
\( \Rightarrow (2x - 1) dx = dt \)
Putting this value in equation (i)
\( I = \int \frac{dt}{\sqrt{t}} [x^2 - x - 1 = t] \)
\( \Rightarrow I = 2\sqrt{t} + c \)
\( I = 2\sqrt{x^2 - x - 1} + c \)
Ans) \( 2\sqrt{x^2 - x - 1} + c \)
In simple words: The numerator is precisely the derivative of the radicand (the expression inside the square root).

Exam Tip: Always check if the numerator is the derivative of the expression under the radical sign - this is a standard pattern.

Question 58. Evaluate the following integrals:
\( \int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}} \)
Answer: To find: Value of \( \int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}} \)
Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)
We have, \( I = \int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}} \) ... (i)
Rationalise by multiplying by the conjugate:
\( I = \int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}} \times \frac{\sqrt{x + a} - \sqrt{x + b}}{\sqrt{x + a} - \sqrt{x + b}} \)
\( I = \int \frac{\sqrt{x + a} - \sqrt{x + b}}{(\sqrt{x + a})^2 - (\sqrt{x + b})^2} dx \)
\( I = \int \frac{\sqrt{x + a} - \sqrt{x + b}}{(x + a) - (x + b)} dx \)
\( I = \int \frac{\sqrt{x + a} - \sqrt{x + b}}{a - b} dx \)
\( I = \frac{1}{a - b} \left[\int \sqrt{x + a} dx - \int \sqrt{x + b} dx\right] \)
\( I = \frac{1}{a - b} \left[\int (x + a)^{1/2} dx - \int (x + b)^{1/2} dx\right] \)
\( I = \frac{1}{a - b} \left[\frac{(x + a)^{3/2}}{3/2} - \frac{(x + b)^{3/2}}{3/2}\right] \)
\( I = \frac{2}{3(a - b)} \left[(x + a)^{3/2} - (x + b)^{3/2}\right] + c \)
Ans) \( \frac{2}{3(a - b)} \left[(x + a)^{3/2} - (x + b)^{3/2}\right] + c \)
In simple words: Multiply by the conjugate to simplify the denominator from a sum of square roots into a linear expression, then integrate term by term.

Exam Tip: Rationalisation by the conjugate is essential when radicals appear in the denominator as sums or differences.

Question 59. Evaluate the following integrals:
\( \int \frac{dx}{(\sqrt{1 - 3x} - \sqrt{5 - 3x})} \)
Answer: To find: Value of \( \int \frac{dx}{\sqrt{1 - 3x} - \sqrt{5 - 3x}} \)
Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)
We have, \( I = \int \frac{dx}{\sqrt{1 - 3x} - \sqrt{5 - 3x}} \) ... (i)
Rationalise by multiplying by the conjugate:
\( I = \int \frac{dx}{\sqrt{1 - 3x} - \sqrt{5 - 3x}} \times \frac{\sqrt{1 - 3x} + \sqrt{5 - 3x}}{\sqrt{1 - 3x} + \sqrt{5 - 3x}} \)
\( I = \int \frac{\sqrt{1 - 3x} + \sqrt{5 - 3x}}{(\sqrt{1 - 3x})^2 - (\sqrt{5 - 3x})^2} dx \)
\( I = \int \frac{\sqrt{1 - 3x} + \sqrt{5 - 3x}}{(1 - 3x) - (5 - 3x)} dx \)
\( I = \int \frac{\sqrt{1 - 3x} + \sqrt{5 - 3x}}{1 - 3x - 5 + 3x} dx \)
\( I = -\frac{1}{4} \left[\int \sqrt{1 - 3x} dx + \int \sqrt{5 - 3x} dx\right] \)
\( I = -\frac{1}{4} \left[\int (1 - 3x)^{1/2} dx + \int (5 - 3x)^{1/2} dx\right] \)
\( I = -\frac{1}{4} \left[\frac{(1 - 3x)^{3/2}}{3/2 \cdot (-3)} + \frac{(5 - 3x)^{3/2}}{3/2 \cdot (-3)}\right] \)
\( I = -\frac{1}{4} \left[\frac{-(1 - 3x)^{3/2}}{9/2} + \frac{-(5 - 3x)^{3/2}}{9/2}\right] \)
\( I = \frac{1}{18} \left[(1 - 3x)^{3/2} + (5 - 3x)^{3/2}\right] + c \)
Ans) \( \frac{1}{18} \left[(1 - 3x)^{3/2} + (5 - 3x)^{3/2}\right] + c \)
In simple words: Use the conjugate to eliminate the radicals from the denominator, then integrate each resulting term using the power rule with careful attention to chain rule coefficients.

Exam Tip: Always include the derivative of the argument when integrating compound functions like \( (ax + b)^n \).

Question 60. Evaluate the following integrals:
\( \int \frac{x^2}{(1 + x^5)} dx \)
Answer: To find: Value of \( \int \frac{x^2}{(1 + x^5)} dx \)
Formula used: \( \int \frac{1}{1 + x^2} dx = \tan^{-1} x \)
We have, \( I = \int \frac{x^2}{(1 + x^5)} dx \) ... (i)
\( I = \int \frac{x^2}{1 + (x^3)^2} dx \)
Let \( x^3 = t \)
\( \Rightarrow \frac{d(x^3)}{dx} = \frac{dt}{dx} \)
\( \Rightarrow (3x^2) = \frac{dt}{dx} \)
\( \Rightarrow (x^2) dx = \frac{dt}{3} \)
Putting this value in equation (i)
\( I = \frac{1}{3} \int \frac{dt}{1 + t^2} [1 + \cos x = t] \)
\( \Rightarrow I = \frac{1}{3} \tan^{-1}(t) + c \)
\( I = \frac{1}{3} \tan^{-1}(x^3) + c \)
Ans) \( \frac{1}{3} \tan^{-1}(x^3) + c \)
In simple words: Recognise \( 1 + x^5 \) as \( 1 + (x^3)^2 \), then substitute \( x^3 = t \) to convert it into the arctangent form.

Exam Tip: Look for hidden squares in sums like \( 1 + x^{2n} \) and rewrite them as \( 1 + (x^n)^2 \) to use inverse trigonometric formulas.

Question 61. Evaluate the following integrals:
\( \int \frac{x^3}{(1 + x^5)} dx \)
Answer: To find: Value of \( \int \frac{x^3}{(1 + x^5)} dx \)
Formula used: \( \int \frac{1}{1 + x^2} dx = \tan^{-1} x \)
We have, \( I = \int \frac{x^3}{(1 + x^5)} dx \) ... (i)
\( I = \int \frac{x^3}{1 + (x^4)^2} dx \)
Let \( x^4 = t \)
\( \Rightarrow \frac{d(x^4)}{dx} = \frac{dt}{dx} \)
\( \Rightarrow (4x^3) = \frac{dt}{dx} \)
\( \Rightarrow (x^3) dx = \frac{dt}{4} \)
Putting this value in equation (i)
\( I = \frac{1}{4} \int \frac{dt}{1 + t^2} [1 + \cos x = t] \)
\( \Rightarrow I = \frac{1}{4} \tan^{-1}(t) + c \)
\( I = \frac{1}{4} \tan^{-1}(x^4) + c \)
Ans) \( \frac{1}{4} \tan^{-1}(x^4) + c \)
In simple words: Rewrite \( 1 + x^5 \) as \( 1 + (x^4)^2 \), then substitute \( x^4 = t \) to apply the arctangent integral formula.

Exam Tip: Be careful with the substitution variable - match the power in the denominator sum to construct the correct substitution.

Question 62. Evaluate the following integrals:
\( \int \frac{x}{(1 + x^4)} dx \)
Answer: To find: Value of \( \int \frac{x}{(1 + x^4)} dx \)
Formula used: \( \int \frac{1}{1 + x^2} dx = \tan^{-1} x \)
We have, \( I = \int \frac{x}{(1 + x^4)} dx \) ... (i)
\( I = \int \frac{x}{1 + (x^2)^2} dx \)
Let \( x^2 = t \)
\( \Rightarrow \frac{d(x^2)}{dx} = \frac{dt}{dx} \)
\( \Rightarrow (2x) = \frac{dt}{dx} \)
\( \Rightarrow (x) dx = \frac{dt}{2} \)
Putting this value in equation (i)
\( I = \frac{1}{2} \int \frac{dt}{1 + t^2} [1 + \cos x = t] \)
\( \Rightarrow I = \frac{1}{2} \tan^{-1}(t) + c \)
\( I = \frac{1}{2} \tan^{-1}(x^2) + c \)
Ans) \( \frac{1}{2} \tan^{-1}(x^2) + c \)
In simple words: Express \( 1 + x^4 \) as \( 1 + (x^2)^2 \), then substitute \( x^2 = t \) to get the arctangent integral.

Exam Tip: The key is recognising that \( 1 + x^{2n} = 1 + (x^n)^2 \), which allows use of the arctangent formula.

Question 63. Evaluate the following integrals:
\( \int \frac{x^5}{\sqrt{1 + x^3}} dx \)
Answer: To find: Value of \( \int \frac{x^5}{\sqrt{1 + x^3}} dx \)
Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)
We have, \( I = \int \frac{x^5}{\sqrt{1 + x^3}} dx \) ... (i)
Let \( 1 + x^3 = t \)
\( \Rightarrow x^3 = t - 1 \)
\( \Rightarrow \frac{d(x^3)}{dx} = \frac{d(t - 1)}{dx} \)
\( \Rightarrow (3x^2) = \frac{dt}{dx} \)
\( \Rightarrow x^2 dx = \frac{dt}{3} \)
Putting this value in equation (i)
\( I = \int \frac{x^3 x^2}{(1 + x^3)^{1/2}} dx \)
\( I = \int \frac{(t - 1) dt}{3 t^{1/2}} [1 + x^3 = t] \)
\( \Rightarrow I = \frac{1}{3} \int \frac{t - 1}{t^{1/2}} dt - \frac{1}{3} \int \frac{1}{t^{1/2}} dt \)
\( \Rightarrow I = \frac{1}{3} \left[\int t^{1/2} dt - \int t^{-1/2} dt\right] \)
\( \Rightarrow I = \frac{1}{3} \left[\frac{t^{3/2}}{3/2} - \frac{t^{1/2}}{1/2}\right] \)
\( \Rightarrow I = \frac{2}{3} \left[\frac{(1 + x^3)^{3/2}}{3} - (1 + x^3)^{1/2}\right] \)
\( \Rightarrow I = \frac{2(1 + x^3)^{3/2}}{9} - \frac{2(1 + x^3)^{1/2}}{3} + c \)
Ans) \( \frac{2(1 + x^3)^{3/2}}{9} - \frac{2(1 + x^3)^{1/2}}{3} + c \)
In simple words: Substitute \( 1 + x^3 = t \), then express \( x^5 \) in terms of \( t \) and split the resulting fraction into simpler powers that can be integrated separately.

Exam Tip: When integrating expressions with radicals in the denominator, express the numerator in terms of the substitution variable to enable proper integration.

Question 64. Evaluate the following integrals:
\( \int \frac{x}{\sqrt{1 + x}} dx \)
Answer: To find: Value of \( \int \frac{x}{\sqrt{1 + x}} dx \)
Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)
We have, \( I = \int \frac{x}{\sqrt{1 + x}} dx \) ... (i)
Let \( 1 + x = t \)
\( \Rightarrow x = t - 1 \)
\( \Rightarrow dx = dt \)
Putting this value in equation (i)
\( I = \int \frac{t - 1}{\sqrt{t}} dx \)
\( I = \int \frac{(t - 1) dt}{t^{1/2}} [1 + x = t] \)
\( \Rightarrow I = \frac{1}{3} \int \frac{t - 1}{t^{1/2}} dt - \frac{1}{3} \int \frac{1}{t^{1/2}} dt \)
\( \Rightarrow I = \frac{1}{3} \left[\int t^{1/2} dt - \int t^{-1/2} dt\right] \)
\( \Rightarrow I = \frac{1}{3} \left[\frac{t^{3/2}}{3/2} - \frac{t^{1/2}}{1/2}\right] \)
\( \Rightarrow I = \frac{2}{3} \left[\frac{(1 + x)^{3/2}}{3} - (1 + x)^{1/2}\right] \)
\( \Rightarrow I = \frac{2(1 + x)^{3/2}}{9} - \frac{2(1 + x)^{1/2}}{3} + c \)
Ans) \( \frac{2(1 + x)^{3/2}}{9} - \frac{2(1 + x)^{1/2}}{3} + c \)
In simple words: Set \( 1 + x = t \), rewrite the numerator as \( t - 1 \), expand into two integrals, and integrate each term using the power rule.

Exam Tip: Always express the original variable in terms of the new substitution variable and simplify before integrating.

Question 65. Evaluate the following integrals:
\( \int \frac{1}{x\sqrt{x^4 - 1}} dx \)
Answer: To find: Value of \( \int \frac{1}{x\sqrt{x^4 - 1}} dx \)
Formula used: \( \int \frac{1}{x\sqrt{x^2 - 1}} dx = \sec^{-1} x + c \)
We have, \( I = \int \frac{1}{x\sqrt{x^4 - 1}} dx \) ... (i)
Multiplying numerator and denominator with \( x \)
\( I = \int \frac{x}{x^2\sqrt{(x^2)^2 - 1}} dx \)
Let \( x^2 = t \)
\( \Rightarrow 2x = \frac{dt}{dx} \)
\( \Rightarrow xdx = \frac{dt}{2} \)
Putting this value in equation (i)
\( I = \frac{1}{2} \int \frac{dt}{t\sqrt{t^2 - 1}} [x^2 = t] \)
\( \Rightarrow I = \frac{1}{2} \sec^{-1} t + c \)
\( \Rightarrow I = \frac{1}{2} \sec^{-1}(x^2) + c \)
Ans) \( \frac{1}{2} \sec^{-1}(x^2) + c \)
In simple words: Multiply numerator and denominator by \( x \), then substitute \( x^2 = t \) to recognise the inverse secant integral form.

Exam Tip: When dealing with \( \sqrt{x^4 - 1} \) or similar expressions, consider rewriting as \( \sqrt{(x^2)^2 - 1} \) and substituting for the inner square.

Question 66. Evaluate the following integrals:
\( \int x\sqrt{x - 1} dx \)
Answer: To find: Value of \( \int x\sqrt{x - 1} dx \)
Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)
We have, \( I = \int x\sqrt{x - 1} dx \) ... (i)
Let \( x - 1 = t \)
\( x = t + 1 \)
\( \Rightarrow dx = dt \)
Putting this value in equation (i)
\( I = \int (t + 1)\sqrt{t} dt [x = t + 1] \)
\( \Rightarrow I = \int t\sqrt{t} dt + \int \sqrt{t} dt \)
\( \Rightarrow I = \int t^{3/2} dt + \int t^{1/2} dt \)
\( \Rightarrow I = \frac{t^{5/2}}{5/2} + \frac{t^{3/2}}{3/2} + c \)
\( \Rightarrow I = \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + c \)
Ans) \( \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + c \)
In simple words: Substitute \( x - 1 = t \), expand the resulting expression into simpler power terms, and integrate each one separately.

Exam Tip: Always expand products under the radical sign by substitution to separate them into individual power integrals.

Question 67. Evaluate the following integrals:
\( \int e^{2x} \sin x \, dx \)
Answer: This integral requires integration by parts. The solution can be derived using the formula for integrals of the form \( \int e^{ax} \sin(bx) dx = \frac{e^{ax}(a\sin(bx) - b\cos(bx))}{a^2 + b^2} + c \)

With \( a = 2 \) and \( b = 1 \):
\( I = \int e^{2x} \sin x \, dx = \frac{e^{2x}(2\sin x - \cos x)}{4 + 1} + c = \frac{e^{2x}(2\sin x - \cos x)}{5} + c \)
Ans) \( \frac{e^{2x}(2\sin x - \cos x)}{5} + c \)
In simple words: When exponential and trigonometric functions are multiplied, use the standard formula that combines both functions in the numerator.

Exam Tip: Memorise the formula for integrals of \( e^{ax} \sin(bx) \) and \( e^{ax} \cos(bx) \) as these appear frequently in examinations.

Question 67. Evaluate the following integrals: \( \int(1-x)\sqrt{1+x} \, dx \)
Answer: We need to find the value of \( \int(1-x)\sqrt{1+x} \, dx \).

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int(1-x)\sqrt{1+x} \, dx \) ... (i)

Let \( 1 + x = t \)

\( x = t - 1 \)

\( \implies dx = dt \)

Putting this value in equation (i)

\( I = \int \{1 - (t - 1)\} \sqrt{t} \, dt \, [ x = t - 1 ] \)

\( \implies I = \int \{1 - t + 1\} \sqrt{t} \, dt \)

\( \implies I = \int \{2 - t\} \sqrt{t} \, dt \)

\( \implies I = \int 2\sqrt{t} \, dt - \int t\sqrt{t} \, dt \)

\( \implies I = 2 \int t^{\frac{1}{2}} dx - \int t^{\frac{3}{2}} \, dx \)

\( \implies I = 2 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} - \frac{t^{\frac{5}{2}}}{\frac{5}{2}} + c \)

\( \implies I = \frac{4}{3}(1+x)^{\frac{3}{2}} - \frac{2}{5}(1+x)^{\frac{5}{2}} + c \)

Answer: \( \frac{4}{3}(1+x)^{\frac{3}{2}} - \frac{2}{5}(1+x)^{\frac{5}{2}} + c \)

Exam Tip: Always substitute to simplify the radical expression first - this transforms the integral into manageable power functions that apply the standard formula directly.

Question 68. Evaluate the following integrals: \( \int x\sqrt{x^2 - 1} \, dx \)
Answer: We need to find the value of \( \int x\sqrt{x^2 - 1} \, dx \).

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int x\sqrt{x^2 - 1} \, dx \) ... (i)

Let \( x^2 - 1 = t \)

\( \implies 2x = \frac{dt}{dx} \)

\( \implies x \, dx = \frac{dt}{2} \)

Putting this value in equation (i)

\( I = \int \frac{1}{2} \sqrt{t} \, dt \, [ x = x^2 - 1 ] \)

\( \implies I = \frac{1}{2} \int t^{\frac{1}{2}} dx \)

\( \implies I = \frac{1}{2} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c \)

\( \implies I = \frac{1}{3} t^{\frac{3}{2}} + c \)

\( \implies I = \frac{1}{3}(x^2 - 1)^{\frac{3}{2}} + c \)

Answer: \( \frac{1}{3}(x^2 - 1)^{\frac{3}{2}} + c \)

Exam Tip: Recognize that the derivative of the expression under the radical (or most of it) appears elsewhere in the integrand - this signals a substitution will work efficiently.

Question 69. Evaluate the following integrals: \( \int x\sqrt{3x - 2} \, dx \)
Answer: We need to find the value of \( \int x\sqrt{3x - 2} \, dx \).

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int x\sqrt{3x - 2} \, dx \) ... (i)

Let \( 3x - 2 = t \)

\( \implies 3x = t + 2 \)

\( \implies x = \frac{t + 2}{3} \)

\( \implies 3 = \frac{dt}{dx} \)

\( \implies dx = \frac{dt}{3} \)

Putting this value in equation (i)

\( I = \int \left(\frac{t + 2}{3}\right) \sqrt{t} \cdot \frac{dt}{3} \, [ t = 3x - 2 ] \)

\( \implies I = \frac{1}{9} \left[ \int t^{\frac{3}{2}} dx + 2 \int t^{\frac{1}{2}} dx \right] \)

\( \implies I = \frac{1}{9} \left[ \frac{t^{\frac{5}{2}}}{\frac{5}{2}} + 2 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right] + c \)

\( \implies I = \frac{2}{45}(3x - 2)^{\frac{5}{2}} + \frac{4}{27}(3x - 2)^{\frac{3}{2}} + c \)

Answer: \( \frac{2}{45}(3x - 2)^{\frac{5}{2}} + \frac{4}{27}(3x - 2)^{\frac{3}{2}} + c \)

Exam Tip: When the expression under the radical is linear in x, isolate x in terms of t and substitute completely - this handles both the radical and the polynomial multiplier in one step.

Question 70. Evaluate the following integrals: \( \int \frac{dx}{x \cos^2(1 + \log x)} \)
Answer: We need to find the value of \( \int \frac{dx}{x\cos^2(1 + \log x)} \).

Formula used: \( \int \sec^2 x \, dx = \tan x + c \)

We have, \( I = \int \frac{dx}{x\cos^2(1 + \log x)} \) ... (i)

Let \( 1 + \log x = t \)

\( \implies \frac{1}{x} = \frac{dt}{dx} \)

\( \implies \frac{1}{x} dx = dt \)

Putting this value in equation (i)

\( I = \int \frac{dt}{\cos^2(t)} \, [ t = 1 + \log x ] \)

\( \implies I = \int \sec^2 t \, dt \)

\( \implies I = \tan(t) + c \)

\( \implies I = \tan(1 + \log x) + c \)

Answer: \( \tan(1 + \log x) + c \)

Exam Tip: Recognize that the differential \( \frac{1}{x} dx \) is the derivative of \( \log x \) - this makes the substitution automatic and direct.

Question 71. Evaluate the following integrals: \( \int x^2 \sin x^3 \, dx \)
Answer: We need to find the value of \( \int x^2 \sin x^3 \, dx \).

Formula used: \( \int \sin x \, dx = - \cos x + c \)

We have, \( I = \int x^2 \sin x^3 \, dx \) ... (i)

Let \( x^3 = t \)

\( \implies 3x^2 = \frac{dt}{dx} \)

\( \implies x^2 dx = \frac{dt}{3} \)

Putting this value in equation (i)

\( I = \int \sin t \cdot \frac{dt}{3} \, [ t = x^3 ] \)

\( \implies I = \frac{1}{3} \left[ \int \sin dt \right] \)

\( \implies I = \frac{1}{3}(- \cos t) + c \)

\( \implies I = \frac{1}{3}(- \cos x^3) + c \)

Answer: \( -\frac{\cos x^3}{3} + c \)

Exam Tip: When a polynomial power appears in both a trigonometric function and as a multiplier (via its derivative), use substitution to convert the entire expression into a standard trig integral.

Question 72. Evaluate the following integrals: \( \int (2x + 4)\sqrt{x^2 + 4x + 3} \, dx \)
Answer: We need to find the value of \( \int (2x + 4)\sqrt{x^2 + 4x + 3} \, dx \).

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int (2x + 4)\sqrt{x^2 + 4x + 3} \, dx \) ... (i)

Let \( x^2 + 4x + 3 = t \)

\( \implies (2x + 4) = \frac{dt}{dx} \)

\( \implies (2x + 4) dx = dt \)

Putting this value in equation (i)

\( I = \int \sqrt{t} \, dt \, [ t = (2x + 4) ] \)

\( \implies I = \int t^{\frac{1}{2}} dx \)

\( \implies I = \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c \)

\( \implies I = \frac{2}{3}(t)^{\frac{3}{2}} + c \)

\( \implies I = \frac{2}{3}(x^2 + 4x + 3)^{\frac{3}{2}} + c \)

Answer: \( \frac{2}{3}(x^2 + 4x + 3)^{\frac{3}{2}} + c \)

Exam Tip: Notice that the polynomial expression's derivative is exactly the coefficient multiplying the radical - this is the ideal scenario for a direct substitution.

Question 73. Evaluate the following integrals: \( \int \frac{\sin x}{(\sin x - \cos x)} \, dx \)
Answer: We need to find the value of \( \int \frac{\sin x}{(\sin x - \cos x)} \, dx \).

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{\sin x}{(\sin x - \cos x)} \, dx \) ... (i)

\( \implies I = \frac{1}{2} \int \frac{2\sin x}{(\sin x - \cos x)} \, dx \)

\( \implies I = \frac{1}{2} \int \frac{(\sin x + \cos x) + (\sin x - \cos x)}{(\sin x - \cos x)} \, dx \)

\( \implies I = \frac{1}{2} \int \frac{(\sin x + \cos x)}{(\sin x - \cos x)} \, dx + \frac{1}{2} \int \frac{(\sin x - \cos x)}{(\sin x - \cos x)} \, dx \)

Let \( \sin x - \cos x = t \)

\( \implies (\cos x + \sin x) = \frac{dt}{dx} \)

\( \implies (\cos x + \sin x) dx = dt \)

Putting this value in equation (i)

\( I = \frac{1}{2} \int \frac{dt}{t} + \frac{1}{2} \int dx \)

\( \implies I = \frac{1}{2}\log|\sin x - \cos x| + \frac{1}{2}x + c \)

\( \implies I = \frac{x}{2} + \frac{1}{2}\log|\sin x - \cos x| + c \)

Answer: \( \frac{x}{2} + \frac{1}{2}\log|\sin x - \cos x| + c \)

Exam Tip: Decompose the numerator into a sum that includes the derivative of the denominator plus a remainder - this splits the integral into two simpler parts.

Question 74. Evaluate the following integrals: \( \int \frac{dx}{(1 - \tan x)} \)
Answer: We need to find the value of \( \int \frac{dx}{(1 - \tan x)} \).

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{dx}{(1 - \tan x)} \) ... (i)

\( \implies I = \int \frac{dx}{\left(1 - \frac{\sin x}{\cos x}\right)} \)

\( \implies I = \int \frac{dx}{\left(\frac{\cos x - \sin x}{\cos x}\right)} \)

\( \implies I = \frac{1}{2} \int \frac{2\cos x \, dx}{(\cos x - \sin x)} \)

\( I = \frac{1}{2} \int \frac{(\cos x + \sin x) + (\cos x - \sin x) \, dx}{(\cos x - \sin x)} \)

\( I = \frac{1}{2} \int \frac{(\cos x + \sin x) \, dx}{(\cos x - \sin x)} + \frac{1}{2} \int \frac{(\cos x - \sin x) \, dx}{(\cos x - \sin x)} \)

Let \( (\cos x - \sin x) = t \)

\( \implies (- \sin x - \cos x) = \frac{dt}{dx} \)

\( \implies (\sin x + \cos x) dx = - dt \)

Putting this value in equation (i)

\( I = - \frac{1}{2} \int \frac{dt}{(t)} \, dx + \frac{1}{2} \int dx \)

\( \implies I = - \frac{1}{2}\log|\cos x - \sin x| + \frac{1}{2}x + c \)

\( \implies I = \frac{1}{2}x - \frac{1}{2}\log|\sin x - \cos x| + c \)

Answer: \( \frac{1}{2}x - \frac{1}{2}\log|\sin x - \cos x| + c \)

Exam Tip: Convert the tangent function into sine and cosine, then manipulate the numerator to match the derivative of the denominator for easier integration.

Question 75. Evaluate the following integrals: \( \int \frac{dx}{(1 - \cot x)} \)
Answer: We need to find the value of \( \int \frac{dx}{(1 - \cot x)} \).

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{dx}{(1 - \cot x)} \) ... (i)

\( \implies I = \int \frac{dx}{\left(1 - \frac{\cos x}{\sin x}\right)} \)

\( \implies I = \int \frac{dx}{\left(\frac{\sin x - \cos x}{\sin x}\right)} \)

\( \implies I = \frac{1}{2} \int \frac{2\sin x \, dx}{(\sin x - \cos x)} \)

\( I = \frac{1}{2} \int \frac{(\sin x + \cos x) + (\sin x - \cos x) \, dx}{(\sin x - \cos x)} \)

\( I = \frac{1}{2} \int \frac{(\sin x + \cos x) \, dx}{(\sin x - \cos x)} + \frac{1}{2} \int \frac{(\sin x - \cos x) \, dx}{(\sin x - \cos x)} \)

Let \( (\cos x - \sin x) = t \)

\( \implies (- \sin x - \cos x) = \frac{dt}{dx} \)

\( \implies (\sin x + \cos x) dx = - dt \)

Putting this value in equation (i)

\( I = \frac{1}{2} \int \frac{2\sin x \, dx}{(\sin x - \cos x)} \)

\( \implies I = \frac{1}{2} \int \frac{(\sin x - \cos x) \, dx}{(\sin x - \cos x)} \)

Answer: \( \frac{1}{2}x - \frac{1}{2}\log|\sin x - \cos x| + c \)

Exam Tip: Break down cotangent into cosine over sine, then use the decomposition technique to isolate a manageable logarithmic integral.

Question 76. Evaluate the following integrals: \( \int \frac{\cos 2x}{(\sin x + \cos x)^2} \, dx \)
Answer: We need to find the value of \( \int \frac{\cos 2x}{(\sin x + \cos x)^2} \, dx \).

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{\cos 2x}{(\sin x + \cos x)^2} \, dx \) ... (i)

\( \implies I = \int \frac{\cos^2 x - \sin^2 x}{(\sin x + \cos x)^2} \, dx \)

\( \implies I = \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2} \, dx \)

\( \implies I = \int \frac{(\cos x - \sin x)}{(\sin x + \cos x)} \, dx \)

Let \( (\cos x + \sin x) = t \)

\( \implies (- \sin x + \cos x) = \frac{dt}{dx} \)

\( \implies (\cos x - \sin x) dx = dt \)

Putting this value in equation (i)

\( I = \int \frac{dt}{t} \, dx + \frac{1}{2} \int dx \)

\( \implies I = \frac{1}{2}\log|\sin x - \cos x| + \frac{1}{2}x + c \)

Answer: \( \frac{1}{2}x + \frac{1}{2}\log|\sin x - \cos x| + c \)

Exam Tip: Use the double angle formula to factor the numerator, then simplify before substituting - this transforms a complex fraction into a simple logarithmic form.

Question 77. Evaluate the following integrals: \( \int \frac{(\cos x - \sin x)}{(1 + \sin 2x)} \, dx \)
Answer: We need to find the value of \( \int \frac{(\cos x - \sin x)}{(1 + \sin 2x)} \, dx \).

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int \frac{(\cos x - \sin x)}{(1 + \sin 2x)} \, dx \) ... (i)

\( \implies I = \int \frac{\cos x - \sin x}{\cos^2 x + \sin^2 x + 2\sin x\cos x} \, dx \)

\( \implies I = \int \frac{(\cos x - \sin x)}{(\cos x + \sin x)^2} \, dx \)

Let \( (\sin x + \cos x) = t \)

\( \implies (\cos x - \sin x) = \frac{dt}{dx} \)

\( \implies (\cos x - \sin x) dx = dt \)

Putting this value in equation (i)

\( I = \int \frac{dt}{t^2} \)

\( \implies I = - \frac{1}{t} + c \)

\( \implies I = - \frac{1}{\sin x + \cos x} + c \)

Answer: \( \frac{-1}{\sin x + \cos x} + c \)

Exam Tip: Recognize that \( 1 + \sin 2x \) equals \( (\sin x + \cos x)^2 \) - this simplification makes the integral a straightforward power rule application.

Question 78. Evaluate the following integrals: \( \int \frac{(x+1)(x + \log x)^2}{x} \, dx \)
Answer: We need to find the value of \( \int \frac{(x+1)(x + \log x)^2}{x} \, dx \).

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int \frac{(x+1)(x + \log x)^2}{x} \, dx \) ... (i)

Let \( (x + \log x) = t \)

\( \implies \left(1 + \frac{1}{x}\right) = \frac{dt}{dx} \)

\( \implies \left(\frac{x + 1}{x}\right) = \frac{dt}{dx} \)

Putting this value in equation (i)

\( I = \int t^2 dt \)

\( \implies I = \frac{t^3}{3} + c \)

\( \implies I = \frac{(x + \log x)^3}{3} + c \)

Answer: \( \frac{(x + \log x)^3}{3} + c \)

Exam Tip: Check whether the differential of the quantity inside a power matches the remaining terms in the numerator - if it does, a direct substitution simplifies the entire integral.

Question 79. Evaluate the following integrals: \( \int x\sin^3 x^2 \cos x^2 \, dx \)
Answer: We need to find the value of \( \int x\sin^3 x^2 \cos x^2 \, dx \).

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int x\sin^3 x^2 \cos x^2 \, dx \) ... (i)

Let \( (\sin x^2) = t \)

\( \implies (\sin x^2, 2x) = \frac{dt}{dx} \)

\( \implies (\sin x^2, x) dx = \frac{dt}{2} \)

Putting this value in equation (i)

\( I = \int t^3 \cdot \frac{dt}{2} \)

\( I = \frac{1}{2} \int t^3 dt \)

\( \implies I = \frac{1}{2} \cdot \frac{t^4}{4} + c \)

\( \implies I = \frac{t^4}{8} + c \)

\( \implies I = \frac{\sin^4 x^2}{8} + c \)

Answer: \( \frac{\sin^4 x^2}{8} + c \)

Exam Tip: When a composite argument appears in trigonometric functions with its own derivative as a factor, substitute the inner function to convert to a polynomial power integral.

Question 80. Evaluate the following integrals: \( \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} \, dx \)
Answer: We need to find the value of \( \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} \, dx \).

Formula used: \( \int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1} x + c \)

We have, \( I = \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} \, dx \) ... (i)

Let \( (\tan x) = t \)

\( \implies (\sec^2 x) = \frac{dt}{dx} \)

\( \implies (\sec^2 x) dx = dt \)

Putting this value in equation (i)

\( I = \int \frac{dt}{\sqrt{1 - t^2}} \)

\( \implies I = \sin^{-1}(t) + c \)

\( \implies I = \sin^{-1}(\tan x) + c \)

Answer: \( \sin^{-1}(\tan x) + c \)

Exam Tip: Identify when the denominator matches the inverse sine formula - substitute the appropriate function to arrive at this standard form instantly.

Question 81. Evaluate the following integrals: \( \int e^{-x} \operatorname{cosec}^2(2e^{-x} + 5) \, dx \)
Answer: We need to find the value of \( \int e^{-x} \operatorname{cosec}^2(2e^{-x} + 5) \, dx \).

Formula used: \( \int \cos\sec^2 x \, dx = - \cot x + c \)

We have, \( I = \int e^{-x} \operatorname{cosec}^2(2e^{-x} + 5) \, dx \) ... (i)

Let \( (2e^{-x} + 5) = t \)

\( \implies (2e^{-x}(- 1)) = \frac{dt}{dx} \)

\( \implies (e^{-x}) dx = \frac{dt}{-2} \)

Putting this value in equation (i)

\( I = \int \operatorname{cosec}^2(t) \frac{dt}{-2} \)

\( I = -\frac{1}{2} \int \operatorname{cosec}^2(t) \, dt \)

\( \implies I = -\frac{1}{2}(- \cot t) + c \)

\( \implies I = \frac{1}{2}\cot(2e^{-x} + 5) + c \)

Answer: \( \frac{1}{2}\cot(2e^{-x} + 5) + c \)

Exam Tip: Match the exponential's derivative (up to a constant factor) with the standard integral formula for cosecant squared to apply the antiderivative directly.

Question 82. Evaluate the following integrals: \( \int 2x \sec^2(x^2 + 3) \tan(x^2 + 3) \, dx \)
Answer: We need to find the value of \( \int 2x \sec^2(x^2 + 3) \tan(x^2 + 3) \, dx \).

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int 2x \sec^2(x^2 + 3) \sec(x^2 + 3) \tan(x^2 + 3) dx \) ... (i)

Let \( \sec(x^2 + 3) = t \)

\( \implies \sec(x^2 + 3) = \frac{dt}{dx} \)

\( \implies \sec(x^2 + 3)\tan(x^2 + 3).2x = \frac{dt}{dx} \)

\( \implies \sec(x^2 + 3)\tan(x^2 + 3).2x = \frac{dt}{dx} \)

Putting this value in equation (i)

\( I = \int t^2 \, dt \)

\( \implies I = \frac{t^3}{3} + c \)

\( \implies I = \frac{\sec^3(x^2 + 3)}{3} + c \)

Answer: \( \frac{\sec^3(x^2 + 3)}{3} + c \)

Exam Tip: When multiple trigonometric terms appear together with a polynomial argument, check whether they form the derivative of a secant or cosecant expression - this often signals a power-rule substitution.

Question 83. Evaluate the following integrals: \( \int \frac{\sin 2x}{(a + b\cos x)^2} \, dx \)
Answer: We need to find the value of \( \int \frac{\sin 2x}{(a + b\cos x)^2} \, dx \).

Formula used: (i) \( \int \frac{1}{x} dx = \log|x| + c \)

(ii) \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int \frac{\sin 2x}{(a + b\cos x)^2} \, dx \) ... (i)

\( I = \int \frac{2\sin x \cos x}{(a + b\cos x)^2} \, dx \)

Let \( (a + b\cos x) = t \)

\( \implies (\cos x) = \frac{t - a}{b} \)

\( \implies (\sin x) dx = \frac{dt}{-b} \)

Putting this value in equation (i)

\( I = \frac{2}{-b^2} \int \frac{t - a}{t^2} \, dt \)

\( I = \frac{2}{-b^2} \left[ \int \frac{t}{t^2} \, dt - a \int \frac{1}{t^2} \, dt \right] \)

\( I = \frac{2}{-b^2} \left[ \log |t| - a \left(- \frac{1}{t}\right) \right] + c \)

\( I = -\frac{2}{b^2} \left[ \log |a + b\cos x| + \left(\frac{a}{a + b\cos x}\right) \right] + c \)

Answer: \( -\frac{2}{b^2} \left[ \log |a + b\cos x| + \left(\frac{a}{a + b\cos x}\right) \right] + c \)

Exam Tip: Transform double angle expressions into products of single angles, then use substitution to convert the denominator into a manageable power form.

Question 84. Evaluate the following integrals: \( \int \frac{dx}{(3 - 5x)} \)
Answer: We need to find the value of \( \int \frac{dx}{(3 - 5x)} \).

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{dx}{(3 - 5x)} \) ... (i)

Let \( (3 - 5x) = t \)

\( \implies (- 5) = \frac{dt}{dx} \)

\( \implies dx = \frac{dt}{-5} \)

Putting this value in equation (i)

\( I = \int \frac{1}{t} \cdot \frac{dt}{-5} \)

\( I = -\frac{1}{5} \int \frac{dt}{t} \)

\( \implies I = -\frac{1}{5} \log |t| + c \)

\( \implies I = -\frac{1}{5} \log |3 - 5x| + c \)

Answer: \( -\frac{1}{5} \log |3 - 5x| + c \)

Exam Tip: For linear expressions in the denominator, substitute the entire expression and compensate for the chain rule's derivative - the logarithm formula applies directly.

Question 85. Evaluate the following integrals: \( \int \sqrt{1 + x} \, dx \)
Answer: We need to find the value of \( \int \sqrt{1 + x} \, dx \).

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int \sqrt{1 + x} \, dx \) ... (i)

Let \( (1 + x) = t \)

\( \implies dx = dt \)

Putting this value in equation (i)

\( I = \int \sqrt{t} \, dt \)

\( I = \int t^{\frac{1}{2}} dt \)

\( \implies I = \frac{2}{3}(1 + x)^{\frac{3}{2}} + c \)

Answer: \( \frac{2}{3}(1 + x)^{\frac{3}{2}} + c \)

Exam Tip: Square root expressions become straightforward power functions once rewritten in exponential form - then apply the power rule with the appropriate adjustment factor.

Question 86. Evaluate the following integrals: \( \int x^2 e^{x^3} \cos(e^{x^3}) \, dx \)
Answer: We need to find the value of \( \int x^2 e^{x^3} \cos(e^{x^3}) \, dx \).

Formula used: \( \int \cos x \, dx = \sin x + c \)

We have, \( I = \int x^2 e^{x^3} \cos(e^{x^3}) \, dx \) ... (i)

Let \( e^{x^3} = t \)

\( \implies e^{x^3} \cdot 3x^2 = \frac{dt}{dx} \)

\( \implies e^{x^3} \cdot x^2 \, dx = \frac{dt}{3} \)

Putting this value in equation (i)

\( I = \int \cos(t) \cdot \frac{dt}{3} \)

\( I = \frac{\sin(t)}{3} + c \)

\( I = \frac{\sin(e^{x^3})}{3} + c \)

Answer: \( \frac{\sin(e^{x^3})}{3} + c \)

Exam Tip: When an exponential function's composite argument appears with a trigonometric function and its derivative as a factor, use the exponential substitution to simplify to a basic trig integral.

Question 87. Evaluate the following integrals: \( \int \frac{e^{m \tan^{-1} x}}{(1 + x^2)} \, dx \)
Answer: We need to find the value of \( \int \frac{e^{m \tan^{-1} x}}{(1 + x^2)} \, dx \).

Formula used: \( \int e^t dx = e^t + c \)

We have, \( I = \int \frac{e^{m\tan^{-1} x}}{(1 + x^2)} \, dx \) ... (i)

Let \( (m\tan^{-1} x) = t \)

\( \implies m \left(\frac{1}{1 + x^2}\right) = \frac{dt}{dx} \)

\( \implies \left(\frac{1}{1 + x^2}\right) dx = \frac{dt}{m} \)

Putting this value in equation (i)

\( I = \int e^t \cdot \frac{dt}{m} \)

\( \implies I = \frac{e^t}{m} + c \)

\( \implies I = \frac{e^{m\tan^{-1} x}}{m} + c \)

Answer: \( \frac{e^{m\tan^{-1} x}}{m} + c \)

Exam Tip: Recognize that the denominator \( 1 + x^2 \) is the derivative of \( \tan^{-1} x \) - this allows you to substitute the inverse trigonometric expression directly into the exponential.

Question 88. Evaluate the following integrals: \( \int \frac{(x+1)e^x}{\cos^2(xe^x)} dx \)
Answer: To find the value of \( \int \frac{(x+1)e^x}{\cos^2(xe^x)} dx \)

Formula used: \( \int \sec^2 x \, dx = \tan x + c \)

We have, \( I = \int \frac{(x+1)e^x}{\cos^2(xe^x)} dx \) ... (i)

Let \( (xe^x) = t \)

\( \implies xe^x + e^x \cdot 1 = \frac{dt}{dx} \)

\( \implies e^x(x+1) = \frac{dt}{dx} \)

Substituting this value in equation (i)

\( I = \int \frac{dt}{\cos^2(t)} \)

\( \implies I = \int \sec^2(t) \, dt \)

\( \implies I = \tan(t) + c \)

\( \implies I = \tan(xe^x) + c \)

Answer: \( \tan(xe^x) + c \)

Exam Tip: Recognize when a composite function's derivative appears in the numerator—this signals a substitution opportunity. Always verify your substitution by differentiating.

Question 89. Evaluate the following integrals: \( \int \frac{e^{\sqrt{x}} \cos(e^{\sqrt{x}})}{\sqrt{x}} dx \)
Answer: To find the value of \( \int \frac{e^{\sqrt{x}} \cos(e^{\sqrt{x}})}{\sqrt{x}} dx \)

Formula used: \( \int \cos x \, dx = \sin x + c \)

We have, \( I = \int \frac{e^{\sqrt{x}} \cos(e^{\sqrt{x}})}{\sqrt{x}} dx \) ... (i)

Let \( (e^{\sqrt{x}}) = t \)

\( \implies e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{dt}{dx} \)

\( \implies \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = 2dt \)

Substituting this value in equation (i)

\( I = \int \cos(t) \cdot 2dt \)

\( I = 2 \sin(e^{\sqrt{x}}) + c \)

Answer: \( 2 \sin(e^{\sqrt{x}}) + c \)

Exam Tip: Pay close attention to the relationship between the differential and the exponential term—extracting the factor of 2 is crucial for the correct final result.

Question 90. Evaluate the following integrals: \( \int \sqrt{e^x - 1} \, dx \)
Answer: To find the value of \( \int \sqrt{e^x - 1} \, dx \)

Formula used: \( \int \frac{1}{x^2+1} dx = \tan^{-1} x + c \)

We have, \( I = \int \sqrt{e^x - 1} \, dx \) ... (i)

Let \( (e^x - 1) = t^2 \)

\( \implies e^x = t^2 + 1 \)

\( \implies e^x = \frac{2t \, dt}{dx} \)

\( \implies dx = \frac{2t \, dt}{e^x} = \frac{2t \, dt}{t^2 + 1} \)

Substituting this value in equation (i)

\( I = \int \sqrt{t^2} \cdot \frac{2t \, dt}{t^2 + 1} \)

\( \implies I = \int \frac{2t^2 \, dt}{t^2 + 1} \)

\( \implies I = 2 \int \frac{t^2 + 1 - 1}{t^2 + 1} dt \)

\( \implies I = 2 \int \left(1 - \frac{1}{t^2 + 1}\right) dt \)

\( \implies I = 2[t - \tan^{-1} t] + c \)

\( \implies I = 2[\sqrt{e^x - 1} - \tan^{-1} \sqrt{e^x - 1}] + c \)

Answer: \( 2[\sqrt{e^x - 1} - \tan^{-1} \sqrt{e^x - 1}] + c \)

Exam Tip: The substitution \( (e^x - 1) = t^2 \) transforms the square root and reveals a standard form—always try squaring substitutions to eliminate radicals.

Question 91. Evaluate the following integrals: \( \int \frac{dx}{(x-\sqrt{x})} \)
Answer: To find the value of \( \int \frac{dx}{(x-\sqrt{x})} \)

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{dx}{(x-\sqrt{x})} \) ... (i)

\( \implies I = \int \frac{dx}{\sqrt{x}(\sqrt{x}-1)} \)

Let \( (\sqrt{x} - 1) = t \)

\( \implies \frac{1}{2\sqrt{x}} = \frac{dt}{dx} \)

\( \implies \frac{1}{\sqrt{x}} dx = 2 \, dt \)

Substituting this value in equation (i)

\( I = \int \frac{2 \, dt}{t} \)

\( I = 2 \log|t| + c \)

\( I = 2 \log|\sqrt{x} - 1| + c \)

Answer: \( 2 \log|\sqrt{x} - 1| + c \)

Exam Tip: Factor out common terms from the denominator before substituting—this simplifies the integral and makes the substitution more apparent.

Question 92. Evaluate the following integrals: \( \int \frac{\sec^2(2 \tan^{-1} x)}{(1+x^2)} dx \)
Answer: To find the value of \( \int \frac{\sec^2(2 \tan^{-1} x)}{(1+x^2)} dx \)

Formula used: \( \int \sec^2 x \, dx = \tan x + c \)

We have, \( I = \int \frac{\sec^2(2 \tan^{-1} x)}{(1+x^2)} dx \) ... (i)

Let \( 2 \tan^{-1} x = t \)

\( \implies \frac{2}{1 + x^2} = \frac{dt}{dx} \)

\( \implies \frac{1}{1 + x^2} dx = \frac{dt}{2} \)

Substituting this value in equation (i)

\( I = \int \sec^2(t) \frac{dt}{2} \)

\( I = \frac{1}{2} \tan(t) + c \)

\( I = \frac{1}{2} \tan(2 \tan^{-1} x) + c \)

Answer: \( \frac{1}{2} \tan(2 \tan^{-1} x) + c \)

Exam Tip: When inverse trigonometric functions appear in the integrand, substituting the inverse function directly often yields a standard form that is easy to integrate.

Question 93. Evaluate the following integrals: \( \int \frac{1 + \sin 2x}{x + \sin^2 x} dx \)
Answer: To find the value of \( \int \frac{1 + \sin 2x}{x + \sin^2 x} dx \)

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{1 + \sin 2x}{x + \sin^2 x} dx \) ... (i)

Let \( x + \sin^2 x = t \)

\( \implies 1 + 2 \sin x \cos x = \frac{dt}{dx} \)

\( \implies (1 + \sin 2x) dx = dt \)

Substituting this value in equation (i)

\( I = \int \frac{dt}{t} \)

\( I = \log|t| + c \)

\( I = \log|x + \sin^2 x| + c \)

Answer: \( \log|x + \sin^2 x| + c \)

Exam Tip: Look for the derivative of the denominator in the numerator—when found, the integral becomes a simple logarithmic form.

Question 94. Evaluate the following integrals: \( \int \frac{1 - \tan x}{x + \log \cos x} dx \)
Answer: To find the value of \( \int \frac{1 - \tan x}{x + \log(\cos x)} dx \)

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{1 - \tan x}{x + \log(\cos x)} dx \) ... (i)

Let \( x + \log(\cos x) = t \)

\( \implies 1 + \frac{1 \cdot (-\sin x)}{\cos x} = \frac{dt}{dx} \)

\( \implies 1 - \tan x = \frac{dt}{dx} \)

\( \implies (1 - \tan x) dx = dt \)

Substituting this value in equation (i)

\( I = \int \frac{dt}{t} \)

\( I = \log|t| + c \)

\( I = \log|x + \log(\cos x)| + c \)

Answer: \( \log|x + \log(\cos x)| + c \)

Exam Tip: The derivative of logarithmic expressions (like \( \log(\cos x) \)) will simplify to a tangent or cotangent form—always check if the numerator matches this pattern.

Question 95. Evaluate the following integrals: \( \int \frac{(1 + \cot x)}{(x + \log \sin x)} dx \)
Answer: To find the value of \( \int \frac{(1 + \cot x)}{(x + \log(\sin x))} dx \)

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{(1 + \cot x)}{(x + \log(\sin x))} dx \) ... (i)

Let \( x + \log(\sin x) = t \)

\( \implies 1 + \frac{1 \cdot (\cos x)}{\sin x} = \frac{dt}{dx} \)

\( \implies 1 + \cot x = \frac{dt}{dx} \)

\( \implies (1 + \cot x) dx = dt \)

Substituting this value in equation (i)

\( I = \int \frac{dt}{t} \)

\( I = \log|x + \log(\sin x)| + c \)

Answer: \( \log|x + \log(\sin x)| + c \)

Exam Tip: Recognize that \( \frac{d}{dx}[\log(\sin x)] = \cot x \)—this pattern frequently appears in integrals with logarithmic denominators.

Question 96. Evaluate the following integrals: \( \int \frac{\tan x \sec^2 x}{(1 - \tan^2 x)} dx \)
Answer: To find the value of \( \int \frac{\tan x \sec^2 x}{(1 - \tan^2 x)} dx \)

Formula used: \( \int \frac{1}{x} dx = \log|x| + c \)

We have, \( I = \int \frac{\tan x \sec^2 x}{(1 - \tan^2 x)} dx \) ... (i)

Let \( 1 - \tan^2 x = t \)

\( \implies 0 - 2 \tan x \sec^2 x = \frac{dt}{dx} \)

\( \implies (\tan x \sec^2 x) dx = \frac{dt}{-2} \)

Substituting this value in equation (i)

\( I = \int \frac{1}{t} \left(\frac{-2}\right) \)

\( I = \frac{1}{2} \log|t| + c \)

\( I = \frac{1}{2} \log|1 - \tan^2 x| + c \)

Answer: \( \frac{1}{2} \log|1 - \tan^2 x| + c \)

Exam Tip: When the denominator contains squared trigonometric functions, the derivative of the denominator often appears in the numerator—use this to identify the substitution.

Question 97. Evaluate the following integrals: \( \int \frac{\sin(2 \tan^{-1} x)}{(1 + x^2)} dx \)
Answer: To find the value of \( \int \frac{\sin(2 \tan^{-1} x)}{(1 + x^2)} dx \)

Formula used: \( \int \sin x \, dx = -\cos x + c \)

We have, \( I = \int \frac{\sin(2 \tan^{-1} x)}{(1 + x^2)} dx \) ... (i)

Let \( 2 \tan^{-1} x = t \)

\( \implies 2 \cdot \frac{1}{1 + x^2} = \frac{dt}{dx} \)

\( \implies \frac{dx}{1 + x^2} = \frac{dt}{2} \)

Substituting this value in equation (i)

\( I = \int \sin(t) \frac{dt}{(2)} \)

\( I = -\frac{1}{2} \cos(t) + c \)

\( I = -\frac{1}{2} \cos(2 \tan^{-1} x) + c \)

Answer: \( -\frac{1}{2} \cos(2 \tan^{-1} x) + c \)

Exam Tip: Inverse trigonometric substitutions paired with their standard derivatives make many seemingly complex integrals straightforward.

Question 98. Evaluate the following integrals: \( \int \frac{dx}{(x^{1/2} + x^{1/3})} \)
Answer: To find the value of \( \int \frac{dx}{(x^{1/2} + x^{1/3})} \)

Formula used: (i) \( \int \frac{1}{x} dx = \log|x| + c \) (ii) \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int \frac{dx}{(x^{1/2} + x^{1/3})} \) ... (i)

Let \( x = t^6 \)

\( \implies x^{1/6} = t \)

\( \implies 6t^5 dt = dx \)

Substituting this value in equation (i)

\( I = \int \frac{6t^5 \, dt}{(t^3 + t^2)} \)

\( I = \int \frac{6t^5 \, dt}{t^2(t + 1)} \)

\( I = 6 \int \frac{t^3 \, dt}{(t + 1)} \)

\( I = 6 \int \frac{t^3 + 1 - 1}{(t + 1)} dt \)

\( I = 6 \int \frac{(t + 1)(t^2 - t + 1) - 1}{(t + 1)} dt - \int \frac{1}{(t + 1)} dt \)

\( I = 6 \left[\frac{t^3}{3} - \frac{t^2}{2} + t - \log|t + 1|\right] + c \)

\( I = [2t^3 - 3t^2 + 6t - 6\log|t + 1|] + c \)

Answer: \( [2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6\log(x^{1/6} + 1)] + c \)

Exam Tip: When fractions with different roots appear, find the LCM of the denominators in the exponents and use it as the power for your substitution variable.

Question 99. Evaluate the following integrals: \( \int (\sin^{-1} x)^2 dx \)
Answer: To find the value of \( \int (\sin^{-1} x)^2 dx \)

Formula used: \( \int \sin x \, dx = -\cos x + c \)

We have, \( I = \int (\sin^{-1} x)^2 dx \) ... (i)

Let \( \sin^{-1} x = t \), \( x = \sin t \)

\( \implies \cos t = \sqrt{1 - x^2} \)

\( \implies \frac{1}{\sqrt{1 - x^2}} = \frac{dt}{dx} \)

\( \implies \sqrt{1 - x^2} \, dt = dx \)

\( \implies \sqrt{1 - (\sin t)^2} dt = dx \)

\( \implies \sqrt{1 - \sin^2 t} \, dt = dx \)

\( \implies \cos t \, dt = dx \)

Substituting this value in equation (i)

\( I = \int t^2 \cos t \, dt \)

Using integration by parts:

\( I = \int t^2 \cos t \, dt - \int \left[\frac{d(t^2)}{dt} \int \cos t \, dt\right] dt \)

\( I = t^2 \sin t - \int [2t \sin t \, dt - \int \frac{dt}{dt} \sin t \, dt] \, dt \)

\( I = t^2 \sin t - 2[- t \cos t + \int 1 \cdot \cos t \, dt] \)

\( I = t^2 \sin t + 2t \cos t - 2 \sin t + c \)

\( I = (\sin^{-1} x)^2 x + 2(\sin^{-1} x) \sqrt{1 - x^2} - 2x + c \)

Answer: \( (\sin^{-1} x)^2 x + 2(\sin^{-1} x) \sqrt{1 - x^2} - 2x + c \)

Exam Tip: When integrating powers of inverse trigonometric functions, integration by parts combined with substitution is essential—apply both methods systematically.

Question 100. Evaluate the following integrals: \( \int \frac{2x \tan^{-1} x^2}{(1 + x^4)} dx \)
Answer: To find the value of \( \int \frac{2x \tan^{-1}(x^2)}{(1 + x^4)} dx \)

Formula used: \( \int x^n dx = \frac{1}{n+1} x^{n+1} + c \)

We have, \( I = \int \frac{2x \tan^{-1}(x^2)}{(1 + x^4)} dx \) ... (i)

Let \( \tan^{-1}(x^2) = t \)

\( \implies \frac{1}{1 + (x^2)^2} \cdot 2x = \frac{dt}{dx} \)

\( \implies \frac{2x}{1 + x^4} dx = dt \)

Substituting this value in equation (i)

\( I = \int t \, dt \)

\( I = \frac{t^2}{2} + c \)

\( I = \frac{[\tan^{-1}(x^2)]^2}{2} + c \)

Answer: \( \frac{[\tan^{-1}(x^2)]^2}{2} + c \)

Exam Tip: Always extract and identify the derivative of the inverse function in the integrand—this reveals the substitution immediately.

Question 101. Evaluate the following integrals: \( \int \frac{(x^2 + 1)}{(x^4 + 1)} dx \)
Answer: To find the value of \( \int \frac{(x^2 + 1)}{(x^4 + 1)} dx \)

Formula used: \( \int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + c \)

We have, \( I = \int \frac{(x^2 + 1)}{(x^4 + 1)} dx \) ... (i)

Dividing numerator and denominator by \( x^2 \):

\( I = \int \frac{(1 + \frac{1}{x^2})}{(x^2 + \frac{1}{x^2} - 2 + 2)} dx \)

\( I = \int \frac{(1 + \frac{1}{x^2})}{(x^2 - 2 \cdot x \cdot \frac{1}{x} + (\frac{1}{x})^2 + 2)} dx \)

\( I = \int \frac{(1 + \frac{1}{x^2})}{((x - \frac{1}{x})^2 + (\sqrt{2})^2)} dx \)

Let \( x - \frac{1}{x} = t \)

\( \implies (1 + \frac{1}{x^2}) dx = dt \)

Substituting this value in equation (i)

\( I = \int \frac{1}{(t)^2 + (\sqrt{2})^2} dt \)

\( I = \frac{1}{\sqrt{2}} \tan^{-1} \left(\frac{t}{\sqrt{2}}\right) + c \)

\( I = \frac{1}{\sqrt{2}} \tan^{-1} \left(\frac{x - \frac{1}{x}}{\sqrt{2}}\right) + c \)

\( I = \frac{1}{\sqrt{2}} \tan^{-1} \left(\frac{x^2 - 1}{\sqrt{2}x}\right) + c \)

Answer: \( \frac{1}{\sqrt{2}} \tan^{-1} \left(\frac{x^2 - 1}{\sqrt{2}x}\right) + c \)

Exam Tip: Dividing by the highest power of \( x \) in the denominator often converts complex rational expressions into standard arctangent forms.

Question 102. Evaluate the following integrals: \( \int \frac{(\sin x + \cos x)}{\sqrt{\sin 2x}} dx \)
Answer: To find the value of \( \int \frac{(\sin x + \cos x)}{\sqrt{\sin 2x}} dx \)

Formula used: \( \int \frac{1}{\sqrt{1 - x^2}} dx = \sin^{-1} x + c \)

We have, \( I = \int \frac{(\sin x + \cos x)}{\sqrt{\sin 2x}} dx \) ... (i)

Let \( (\sin x - \cos x) = t \)

\( \implies (\cos x + \sin x) = \frac{dt}{dx} \)

\( \implies (\cos x + \sin x) dx = dt \)

Also, \( t^2 = \sin^2 x - 2 \sin x \cos x + \cos^2 x \)

\( \implies t^2 = 1 - 2 \sin x \cos x \)

\( \implies 2 \sin x \cos x = 1 - t^2 \)

\( \implies \sin 2x = 1 - t^2 \)

Substituting this value in equation (i)

\( \implies I = \int \frac{dt}{\sqrt{1 - t^2}} \)

\( I = \sin^{-1} t \)

\( I = \sin^{-1}(\sin x - \cos x) \)

Let \( \sin^{-1}(\sin x - \cos x) = \theta \) ... (ii)

\( \implies \sin \theta = \sin x - \cos x \)

Now if \( \sin \theta = \sin x - \cos x \)

Then \( \cos \theta = \sqrt{1 - (\sin x - \cos x)^2} \)

\( \implies \cos \theta = \sqrt{1 - (\sin^2 x - 2 \sin x \cos x + \cos^2 x)} \)

\( \implies \cos \theta = \sqrt{1 - (1 - 2 \sin x \cos x)} \)

\( \implies \cos \theta = \sqrt{2 \sin x \cos x} \)

Now \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

Now \( \tan \theta = \frac{\sin x - \cos x}{\sqrt{2 \sin x \cos x}} \)

\( \implies \theta = \tan^{-1} \left(\frac{\sin x - \cos x}{\sqrt{2 \sin x \cos x}}\right) \)

Comparing the value of \( \theta \) from equation (ii):

\( I = \theta = \tan^{-1} \left(\frac{\sin x - \cos x}{\sqrt{2 \sin x \cos x}}\right) \)

Dividing numerator and denominator by \( \cos x \):

\( I = \theta = \tan^{-1} \left(\frac{\tan x - 1}{\sqrt{2 \tan x}}\right) \)

Answer: \( \tan^{-1} \left(\frac{\tan x - 1}{\sqrt{2 \tan x}}\right) + c \)

Exam Tip: When the integrand involves both \( \sin 2x \) and \( \sin x \pm \cos x \), try setting \( (\sin x - \cos x) \) or \( (\sin x + \cos x) \) as the substitution variable—the relationship with \( \sin 2x \) simplifies the square root.

Question 1. Mark (√) against the correct answer in each of the following: \( \int (2x + 3)^5 dx = ? \)
(a) \( \frac{(2x + 3)^6}{6} + C \)
(b) \( \frac{(2x + 3)^4}{8} + C \)
(c) \( \frac{(2x + 3)^6}{12} + C \)
(d) none of these
Answer: (c) \( \frac{(2x + 3)^6}{12} + C \)
Solution: Given: \( \int (2x + 3)^5 dx \)

Let \( 2x + 3 = z \)

\( \implies 2 dx = dz \)

Therefore,

\( \int (2x + 3)^5 dx = \int z^5 \left(\frac{dz}{2}\right) = \frac{1}{2} \int z^5 dz = \frac{1}{2} \cdot \frac{z^6}{6} + c = \frac{z^6}{12} + c = \frac{(2x + 3)^6}{12} + C \)

where \( c \) is the integrating constant.

Exam Tip: For integrals of the form \( (ax + b)^n \), use substitution and recall that dividing by both the chain rule factor and the power rule denominator gives the correct coefficient.

Question 2. Mark (√) against the correct answer in each of the following: \( \int (3 - 5x)^7 dx = ? \)
(a) \( -5(3 - 5x)^6 + C \)
(b) \( \frac{(3 - 5x)^8}{-40} + C \)
(c) \( \frac{-5(3 - 5x)^8}{8} + C \)
(d) none of these
Answer: (b) \( \frac{(3 - 5x)^8}{-40} + C \)
Solution: Given: \( \int (3 - 5x)^7 dx \)

Let \( 3 - 5x = z \)

\( \implies -5 dx = dz \)

Therefore,

\( \int (3 - 5x)^7 dx = \int z^7 \left(\frac{dz}{-5}\right) = -\frac{1}{5} \int z^7 dz = -\frac{1}{5} \cdot \frac{z^8}{8} + c = \frac{-z^8}{40} + c = \frac{(3 - 5x)^8}{-40} + C \)

where \( c \) is the integrating constant.

Exam Tip: Pay careful attention to the sign of the coefficient in front of \( x \)—a negative coefficient introduces a negative sign in the denominator of the final answer.

Question 4. Mark (√) against the correct answer in each of the following: \( \int \frac{1}{(2 - 3x)^4} dx = ? \)
(a) \( \frac{1}{15(2 - 3x)^5} + C \)
(b) \( \frac{-1}{12(2 - 3x)^3} + C \)
(c) \( \frac{1}{9(2 - 3x)^3} + C \)
(d) none of these
Answer: (b) \( \frac{-1}{12(2 - 3x)^3} + C \)
Solution: Given: \( \int \frac{1}{(2 - 3x)^4} dx \)

Let \( 2 - 3x = z \)

\( \implies -3 dx = dz \)

Therefore,

\( \int \frac{1}{(2 - 3x)^4} dx = \int z^{-4} \left(\frac{dz}{-3}\right) = -\frac{1}{3} \int z^{-4} dz = -\frac{1}{3} \cdot \frac{z^{-3}}{-3} + c = \frac{1}{9(2 - 3x)^3} + C \)

where \( c \) is the integrating constant.

Exam Tip: Remember that negative exponents follow the same power rule—express the integrand as a negative power and apply the standard formula.

Question 5. Mark (√) against the correct answer in each of the following: \( \int \sqrt{ax + b} \, dx = ? \)
(a) \( \frac{2(ax + b)^{3/2}}{3a} + C \)
(b) \( \frac{3(ax + b)^{3/2}}{2a} + C \)
(c) \( \frac{1}{2\sqrt{ax + b}} + C \)
(d) none of these
Answer: (b) \( \frac{3(ax + b)^{3/2}}{2a} + C \)
Solution: Given: \( \int \sqrt{ax + b} \, dx \)

Let \( ax + b = z^2 \)

\( \implies a \, dx = 2z \, dz \)

Therefore,

\( \int \sqrt{ax + b} \, dx = \int z \left(\frac{2z \, dz}{a}\right) = \frac{2}{a} \int z^2 dz = \frac{2}{a} \cdot \frac{z^3}{3} + c = \frac{2z^3}{3a} + c = \frac{2(ax + b)^{3/2}}{3a} + C \)

where \( c \) is the integrating constant.

Exam Tip: When integrating square roots, set the expression inside the radical equal to \( z^2 \) rather than \( z \)—this avoids further square roots in your working.

Question 5. Mark (√) against the correct answer in each of the following: \( \int \sec^2(7 - 4x) dx = ? \)
(A) \( \frac{1}{4}\tan(7 - 4x) + C \)
(B) \( -\frac{1}{4}\tan(7 - 4x) + C \)
(C) \( 4\tan(7 - 4x) + C \)
(D) \( -4\tan(7 - 4x) + C \)
Answer: (B) \( -\frac{1}{4}\tan(7 - 4x) + C \)
In simple words: To find this integral, let the expression inside the function equal a new variable. Then apply the standard rule for integrating secant squared, and adjust for the chain rule by multiplying by the reciprocal of the derivative of the inner function.

Exam Tip: Always use substitution when you see a composite function. The coefficient from the chain rule becomes a coefficient in your final answer.

Question 6. Mark (√) against the correct answer in each of the following: \( \int \cos 3x \, dx = ? \)
(A) \( -\frac{1}{3}\sin 3x + C \)
(B) \( \frac{1}{3}\sin 3x + C \)
(C) \( 3\sin 3x + C \)
(D) \( -3\sin 3x + C \)
Answer: (B) \( \frac{1}{3}\sin 3x + C \)
In simple words: When you integrate cosine of a multiple of x, the sine appears in the answer. Divide by the multiplier (which is 3 in this case) to account for the chain rule working in reverse.

Exam Tip: Remember that the integral of \( \cos(kx) \) is \( \frac{1}{k}\sin(kx) + C \). The denominator k is crucial - forgetting it is a common error.

Question 7. Mark (√) against the correct answer in each of the following: \( \int e^{(5-3x)} dx = ? \)
(A) \( -3e^{(5-3x)} + C \)
(B) \( \frac{1}{3}e^{(5-3x)} + C \)
(C) \( \frac{e^{(5-3x)}}{-3} + C \)
(D) None of these
Answer: (C) \( -\frac{1}{3}e^{(5-3x)} + C \)
In simple words: The exponential function stays the same when integrated. However, the coefficient inside the exponent creates a fraction - divide by the negative coefficient to get your final answer.

Exam Tip: For \( e^{(ax+b)} \), the integral is always \( \frac{1}{a}e^{(ax+b)} + C \). Pay careful attention to the sign of the coefficient a.

Question 8. Mark (√) against the correct answer in each of the following: \( \int e^{(3x+4)} dx = ? \)
(A) \( \frac{3}{(\log 2)} \cdot 2^{(3x+1)} + C \)
(B) \( \frac{2^{(3x+4)}}{3(\log 2)} + C \)
(C) \( \frac{2^{(3x+4)}}{2(\log 3)} + C \)
(D) None of these
Answer: (B) \( \frac{1}{3}e^{(3x+4)} + C \)
In simple words: When integrating an exponential with a coefficient in the exponent, the result involves dividing by that coefficient. The exponential function itself remains unchanged in form.

Exam Tip: Make sure the final answer matches the original exponential base and exponent. The only change is the reciprocal coefficient in front.

Question 9. Mark (√) against the correct answer in each of the following: \( \int \tan^2 \frac{x}{2} dx = ? \)
(A) \( \tan \frac{x}{2} - x + C \)
(B) \( \tan \frac{x}{2} + x + C \)
(C) \( 2\tan \frac{x}{2} + x + C \)
(D) \( 2\tan \frac{x}{2} - x + C \)
Answer: (D) \( 2\tan \frac{x}{2} - x + C \)
In simple words: Use the trigonometric identity to rewrite tangent squared in terms of secant squared, then split and integrate each term separately.

Exam Tip: Always convert \( \tan^2 \theta \) to \( \sec^2 \theta - 1 \) before integrating to break it into manageable parts.

Question 10. Mark (√) against the correct answer in each of the following: \( \int \sqrt{1 - \cos x} \, dx = ? \)
(A) \( -\sqrt{2}\cos \frac{x}{2} + C \)
(B) \( -2\sqrt{2}\cos \frac{x}{2} + C \)
(C) \( -\frac{1}{2}\cos \frac{x}{2} + C \)
(D) \( -\frac{1}{\sqrt{2}}\cos \frac{x}{2} + C \)
Answer: (B) \( -2\sqrt{2}\cos \frac{x}{2} + C \)
In simple words: The expression under the square root can be simplified using a half-angle identity. After simplification, substitute to transform it into a standard integral form.

Exam Tip: Recognize that \( 1 - \cos x = 2\sin^2 \frac{x}{2} \). This identity is key to simplifying the square root.

Question 11. Mark (√) against the correct answer in each of the following: \( \int \sqrt{1 + \sin x} \, dx = ? \)
(A) \( -\sqrt{2}\sin \left(\frac{\pi}{4} - \frac{x}{2}\right) + C \)
(B) \( \sqrt{2}\sin \left(\frac{\pi}{4} - \frac{x}{2}\right) + C \)
(C) \( -2\sqrt{2}\sin \left(\frac{\pi}{4} - \frac{x}{2}\right) + C \)
(D) None of these
Answer: (B) \( \sqrt{2}\sin \left(\frac{\pi}{4} - \frac{x}{2}\right) + C \)
In simple words: Use the identity for \( 1 + \sin x \) to rewrite it as a perfect square expression. Then substitute and integrate the resulting simpler form.

Exam Tip: The identity \( 1 + \sin x = \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2 \) is essential. Recognizing perfect square forms saves significant time.

Question 12. Mark (√) against the correct answer in each of the following: \( \int \sin^3 x \, dx = ? \)
(A) \( -\frac{3}{4}\cos x + \frac{\cos 3x}{12} + C \)
(B) \( \frac{3}{4}\cos x + \frac{\cos 3x}{12} + C \)
(C) \( -\frac{3}{4}\cos x - \frac{\cos 3x}{12} + C \)
(D) None of these
Answer: (C) \( -\frac{3}{4}\cos x - \frac{\cos 3x}{12} + C \)
In simple words: Rewrite sine cubed using the Pythagorean identity, then expand and integrate term by term. Combine like terms to arrive at the final form.

Exam Tip: For odd powers of sine or cosine, factor out one function and use \( \sin^2 x + \cos^2 x = 1 \) to transform the remaining even power into the other function.

Question 13. Mark (√) against the correct answer in each of the following: \( \int \frac{\log x}{x} dx = ? \)
(A) \( \frac{1}{2}(\log x)^2 + C \)
(B) \( -\frac{1}{2}(\log x)^2 + C \)
(C) \( \frac{2}{x^2} + C \)
(D) \( -\frac{2}{x^2} + C \)
Answer: (A) \( \frac{1}{2}(\log x)^2 + C \)
In simple words: Let the logarithm be your substitution variable. This transforms the integral into a simple power function that you can integrate directly.

Exam Tip: Whenever you see \( \frac{f'(x)}{f(x)} \) or \( f(x) \cdot f'(x) \) patterns, substitution makes the integration straightforward.

Question 14. Mark (√) against the correct answer in each of the following: \( \int \frac{\sec^2(\log x)}{x} dx = ? \)
(A) \( \log(\tan x) + C \)
(B) \( -\log(\tan x) + C \)
(C) \( \tan(\tan x) + C \)
(D) \( -\tan(\log x) + C \)
Answer: (C) \( \tan(\log x) + C \)
In simple words: Substitute the logarithm as a new variable. The secant squared term then becomes a standard integrable function.

Exam Tip: When both a composite function and its derivative appear in an integral, substitution is almost always the right approach.

Question 15. Mark (√) against the correct answer in each of the following: \( \int \frac{1}{x(\log x)} dx = ? \)
(A) \( \log |x| + C \)
(B) \( -\frac{2}{x^2} + C \)
(C) \( (\log x)^2 + C \)
(D) \( \log |\log x| + C \)
Answer: (D) \( \log |\log x| + C \)
In simple words: Set the logarithm equal to a new variable. The reciprocal of that variable then integrates to give the logarithm of the absolute value.

Exam Tip: This illustrates the pattern \( \int \frac{du}{u} = \log |u| + C \). Recognize composite logarithmic arguments as perfect substitution candidates.

Question 16. Mark (√) against the correct answer in each of the following: \( \int e^{x^3} x^2 dx = ? \)
(A) \( e^{x^3} + C \)
(B) \( \frac{1}{3}e^{x^3} + C \)
(C) \( \frac{1}{6}e^{x^3} + C \)
(D) None of these
Answer: (B) \( \frac{1}{3}e^{x^3} + C \)
In simple words: Notice that the derivative of \( x^3 \) is \( 3x^2 \). This means the integrand matches the pattern for a substitution where the power inside the exponential becomes your new variable.

Exam Tip: Look for the derivative of the exponent or the argument in the integrand. If it appears (up to a constant), substitution will work.

Question 17. Mark (√) against the correct answer in each of the following: \( \int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = ? \)
(A) \( e^{\sqrt{x}} + C \)
(B) \( \frac{1}{2}e^{\sqrt{x}} + C \)
(C) \( 2e^{\sqrt{x}} + C \)
(D) None of these
Answer: (C) \( 2e^{\sqrt{x}} + C \)
In simple words: The square root of x in the denominator is related to the derivative of the square root of x in the exponent. Setting the square root as your new variable gives a simple exponential integral.

Exam Tip: Fractional or radical exponents often signal a substitution. The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \), which appears scaled in the integrand.

Question 18. Mark (√) against the correct answer in each of the following: \( \int \frac{e^{\tan^{-1} x}}{(1 + x^2)} dx = ? \)
(A) \( \frac{e^{\tan^{-1} x}}{x} + C \)
(B) \( e^{\tan^{-1} x} + C \)
(C) \( e^x \tan^{-1} x + C \)
(D) None of these
Answer: (B) \( e^{\tan^{-1} x} + C \)
In simple words: The derivative of inverse tangent is \( \frac{1}{1+x^2} \). This factor appears in the denominator, making inverse tangent the ideal choice for substitution.

Exam Tip: When you see \( \frac{1}{1+x^2} \) multiplied by a function of \( \tan^{-1} x \), immediately substitute the inverse tangent.

Question 19. Mark (√) against the correct answer in each of the following: \( \int \frac{\sin \sqrt{x}}{\sqrt{x}} dx = ? \)
(A) \( 2\cos \sqrt{x} + C \)
(B) \( -2\cos \sqrt{x} + C \)
(C) \( \frac{\cos \sqrt{x}}{2} + C \)
(D) \( \frac{\cos \sqrt{x}}{2} + C \)
Answer: (B) \( -2\cos \sqrt{x} + C \)
In simple words: Let the square root of x become your substitution variable. The derivative relation then aligns perfectly with the remaining part of the integrand.

Exam Tip: When a radical appears both inside a trig function and in a denominator, radical substitution is the key technique.

Question 20. Mark (√) against the correct answer in each of the following: \( \int (\sqrt{\sin x}) \cos x \, dx = ? \)
(A) \( \frac{2}{3}(\cos x)^{3/2} + C \)
(B) \( \frac{3}{2}(\cos x)^{3/2} + C \)
(C) \( \frac{2}{3}(\sin x)^{3/2} + C \)
(D) \( \frac{3}{2}(\sin x)^{3/2} + C \)
Answer: (C) \( \frac{2}{3}(\sin x)^{3/2} + C \)
In simple words: The cosine multiplied by the square root of sine suggests that sine should be your substitution variable. The power will then increase by one when integrated.

Exam Tip: Whenever a trig function and the derivative of that trig function both appear, use that function as your substitution.

Question 21. Mark (√) against the correct answer in each of the following: \( \int \frac{1}{(1+x^2)\sqrt{\tan^{-1} x}} dx = ? \)
(A) \( \frac{1}{2}\log |\tan^{-1} x| + C \)
(B) \( 2\sqrt{\tan^{-1} x} + C \)
(C) \( \frac{1}{2\sqrt{\tan^{-1} x}} + C \)
(D) None of these
Answer: (B) \( 2\sqrt{\tan^{-1} x} + C \)
In simple words: Substitute the inverse tangent as your new variable. The factor in the denominator converts perfectly to match the derivative needed for the substitution.

Exam Tip: A function inside a radical with its own derivative in the integrand should trigger immediate substitution of that function.

Question 22. Mark (√) against the correct answer in each of the following: \( \int \frac{\cot x}{\log(\sin x)} dx = ? \)
(A) \( \log |\cot x| + C \)
(B) \( \log |\cot x \cos \text{ec} x| + C \)
(C) \( \log |\log \sin x| + C \)
(D) None of these
Answer: (C) \( \log |\log \sin x| + C \)
In simple words: First, recognize that cotangent can be written as cosine over sine. Then set the logarithm of sine as your substitution, which relates to the cotangent term in the numerator.

Exam Tip: When a logarithm of a function appears with the derivative of that function, substitute the logarithm itself.

Question 23. Mark (√) against the correct answer in each of the following: \( \int \frac{1}{x \cos^2(1 + \log x)} dx = ? \)
(A) \( \tan(1 + \log x) + C \)
(B) \( \cot(1 + \log x) + C \)
(C) \( \sec(1 + \log x) + C \)
(D) None of these
Answer: (A) \( \tan(1 + \log x) + C \)
In simple words: The expression \( 1 + \log x \) in the cosine argument has a derivative equal to \( \frac{1}{x} \), which appears in the numerator. Use this expression as your substitution variable.

Exam Tip: When secant squared appears (or its reciprocal, cosine squared in the denominator), substitution converts it to the tangent integral.

Question 24. Mark (√) against the correct answer in each of the following: \( \int \frac{x^2 \tan^{-1} x^3}{(1 + x^6)} dx = ? \)
(A) \( \frac{1}{3}(\tan^{-1} x^3) + C \)
(B) \( \log |\tan^{-1} x^3| + C \)
(C) \( \frac{1}{6}(\tan^{-1} x^3)^2 + C \)
(D) None of these
Answer: (C) \( \frac{1}{6}(\tan^{-1} x^3)^2 + C \)
In simple words: The derivative of \( \tan^{-1} x^3 \) involves \( \frac{3x^2}{1 + x^6} \). The integrand contains \( x^2 \) and \( \tan^{-1} x^3 \), which matches this derivative pattern scaled by a constant factor.

Exam Tip: Recognize derivatives of inverse trig functions embedded in the integrand. The pattern \( \frac{f'(x)}{1 + (f(x))^2} \cdot f(x) \) signals substitution of the inverse tangent.

Question 25. Mark (√) against the correct answer in each of the following: \( \int \sec^2 x \tan x \, dx = ? \)
(a) \( 5 \tan^5 x + C \)
(b) \( \frac{1}{5} \tan^5 x + C \)
(c) \( 5 \log |\cos x| + C \)
(d) none of these
Answer: (b) \( \frac{1}{5} \tan^5 x + C \)
In simple words: When you substitute \( \sec x = z \), the integral becomes a simple power integral that evaluates to \( \frac{1}{5} \tan^5 x + C \).

Exam Tip: Recognize when to use the substitution method - if the derivative of one part appears elsewhere in the integrand, substitution is your go-to strategy.

Question 26. Mark (√) against the correct answer in each of the following: \( \int \cos ec^3 (2x + 1) \cot (2x + 1) dx = ? \)
(a) \( \frac{1}{4} \cos ec^4 (2x + 1) + C \)
(b) \( - \frac{1}{3} \cos ec^3 (2x + 1) + C \)
(c) \( - \frac{1}{6} \cos ec^3 (2x + 1) + C \)
(d) \( \frac{1}{2} \cos ec (2x + 1) \cot (2x + 1) + C \)
Answer: (c) \( - \frac{1}{6} \cos ec^3 (2x + 1) + C \)
In simple words: Using the substitution \( \cos ec(2x + 1) = z \) transforms this into a straightforward power integral that gives \( - \frac{1}{6} \cos ec^3 (2x + 1) + C \).

Exam Tip: Pay attention to the chain rule factor - the coefficient 2 in the argument requires careful handling in the substitution.

Question 27. Mark (√) against the correct answer in each of the following: \( \int \frac{\tan (\sin^{-1} x)}{\sqrt{1 - x^2}} dx = ? \)
(a) \( \log |\sec (\sin^{-1} x)| + C \)
(b) \( \log |\cos (\sin^{-1} x)| + C \)
(c) \( \tan (\sin^{-1} x) + C \)
(d) none of these
Answer: (a) \( \log |\sec (\sin^{-1} x)| + C \)
In simple words: Setting \( \sin^{-1} x = z \) simplifies the integral to \( \int \tan z \, dz \), which evaluates to the logarithm of the secant.

Exam Tip: Inverse trigonometric substitutions are powerful - recognize the derivative pattern of inverse functions to simplify the radical.

Question 28. Mark (√) against the correct answer in each of the following: \( \int \frac{\tan (\log x)}{x} dx = ? \)
(a) \( x \tan (\log x) + C \)
(b) \( \log |\tan x| + C \)
(c) \( \log |\cos (\log x)| + C \)
(d) \( - \log |\cos (\log x)| + C \)
Answer: (d) \( - \log |\cos (\log x)| + C \)
In simple words: The substitution \( \log x = z \) transforms this to \( \int \tan z \, dz \), which produces the negative logarithm of cosine.

Exam Tip: When a logarithmic function appears inside another function, substituting for the logarithm often reveals a standard integral form.

Question 29. Mark (√) against the correct answer in each of the following: \( \int e^x \cot (e^x) dx = ? \)
(a) \( \cot (e^x) + C \)
(b) \( \log |\sin e^x| + C \)
(c) \( \log |\cos ec e^x| + C \)
(d) none of these
Answer: (b) \( \log |\sin e^x| + C \)
In simple words: Substituting \( e^x = z \) converts the integral to \( \int \cot z \, dz \), which equals the logarithm of the sine function.

Exam Tip: The integral of cotangent is always \( \log |\sin u| + C \) - commit this standard form to memory for quick recognition.

Question 30. Mark (√) against the correct answer in each of the following: \( \int \frac{e^x}{\sqrt{1 + e^x}} dx = ? \)
(a) \( 2\sqrt{1 + e^x} + C \)
(b) \( \frac{1}{2}\sqrt{1 + e^x} + C \)
(c) \( \frac{1}{\sqrt{1 + e^x}} + C \)
(d) none of these
Answer: (a) \( 2\sqrt{1 + e^x} + C \)
In simple words: Setting \( 1 + e^x = z^2 \) leads to \( e^x dx = 2z dz \), which simplifies the integral to a basic form giving \( 2\sqrt{1 + e^x} + C \).

Exam Tip: When the radicand and its derivative are both present, use a quadratic substitution to eliminate the square root.

Question 31. Mark (√) against the correct answer in each of the following: \( \int \frac{x}{\sqrt{1 - x^2}} dx = ? \)
(a) \( \sin^{-1} x + C \)
(b) \( \sin^{-1} \sqrt{x} + C \)
(c) \( \sqrt{1 - x^2} + C \)
(d) \( - \sqrt{1 - x^2} + C \)
Answer: (d) \( - \sqrt{1 - x^2} + C \)
In simple words: Using the substitution \( 1 - x^2 = z^2 \) changes the integral into a simple form that yields \( - \sqrt{1 - x^2} + C \).

Exam Tip: Watch the sign carefully when substituting - negative derivatives affect the final result and can flip signs in the answer.

Question 32. Mark (√) against the correct answer in each of the following: \( \int \frac{e^x (1 + x)}{\cos^2 (xe^x)} dx = ? \)
(a) \( \tan (xe^x) + C \)
(b) \( \cot (xe^x) + C \)
(c) \( e^x x \tan x + C \)
(d) none of these
Answer: (a) \( \tan (xe^x) + C \)
In simple words: The substitution \( xe^x = z \) produces the derivative \( e^x(1 + x) dx = dz \), transforming the integral into \( \int \sec^2 z \, dz = \tan z + C \).

Exam Tip: Identify composite functions where the derivative of the inner part appears in the integrand - this signals a substitution opportunity.

Question 33. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{e^x + e^{-x}} = ? \)
(a) \( \cot^{-1} (e^x) + C \)
(b) \( \tan^{-1} (e^x) + C \)
(c) \( \log |e^x + 1| + C \)
(d) none of these
Answer: (b) \( \tan^{-1} (e^x) + C \)
In simple words: Rewriting the denominator and using the substitution \( e^x + 1 = z \) eventually leads to the arctangent form \( \tan^{-1} (e^x) + C \).

Exam Tip: Inverse trigonometric integrals often arise from rational integrals - recognize the standard form \( \int \frac{1}{1 + u^2} du = \tan^{-1} u + C \).

Question 34. Mark (√) against the correct answer in each of the following: \( \int \frac{2^x}{1 - 4^x} dx = ? \)
(a) \( \sin^{-1} (2^x) + C \)
(b) \( (\log e^2) \sin^{-1} (2^x) + C \)
(c) \( (\log e^2) \cos^{-1} (2^x) + C \)
(d) \( \log_2 e) \sin^{-1} (2^x) + C \)
Answer: (b) \( (\log e^2) \sin^{-1} (2^x) + C \)
In simple words: Setting \( 2^x = z \) produces \( 2^x (\log 2) dx = dz \), which converts the integral into the arcsine form with the logarithmic constant factor.

Exam Tip: Exponential expressions like \( 4^x = (2^x)^2 \) can be rewritten to reveal standard inverse trigonometric integral patterns.

Question 35. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{e^x - 1} = ? \)
(a) \( \log |e^x - 1| + C \)
(b) \( \log |1 - e^{-x}| + C \)
(c) \( \log |e^x - 1| + C \)
(d) none of these
Answer: (a) \( \log |e^x - 1| + C \)
In simple words: Substituting \( e^x - 1 = z \) gives \( e^x dx = dz \), leading directly to the logarithmic integral \( \log |e^x - 1| + C \).

Exam Tip: When the denominator and its derivative match a simple substitution, the result is typically a logarithmic form.

Question 36. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{(\sqrt{x} + x)} = ? \)
(a) \( \log |1 + \sqrt{x}| + C \)
(b) \( 2 \log |1 + \sqrt{x}| + C \)
(c) \( \frac{1}{\sqrt{x}} \tan^{-1} \sqrt{x} + C \)
(d) none of these
Answer: (b) \( 2 \log |1 + \sqrt{x}| + C \)
In simple words: The substitution \( 1 + \sqrt{x} = z \) simplifies the integral to a logarithmic form that yields \( 2 \log |1 + \sqrt{x}| + C \).

Exam Tip: Factoring out common terms and recognizing partial fractions can reveal hidden logarithmic structures in complicated denominators.

Question 37. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{(1 + \sin x)} = ? \)
(a) \( \tan x + \sec x + C \)
(b) \( \tan x - \sec x + C \)
(c) \( \frac{1}{2} \tan \frac{x}{2} + C \)
(d) none of these
Answer: (a) \( \tan x + \sec x + C \)
In simple words: Converting \( 1 + \sin x \) using the half-angle identity and substituting \( \tan \frac{x}{2} + 1 = z \) produces the combination \( \tan x + \sec x + C \).

Exam Tip: Half-angle formulas are essential for integrals involving sine and cosine - mastering these identities speeds up the process significantly.

Question 38. Mark (√) against the correct answer in each of the following: \( \int \frac{\sin x}{(1 + \sin x)} dx = ? \)
(a) \( x + \tan x - \sec x + C \)
(b) \( x - \tan x - \sec x + C \)
(c) \( x - \tan x + \sec x + C \)
(d) none of these
Answer: (c) \( x - \tan x + \sec x + C \)
In simple words: Breaking the fraction into parts and handling each term separately with substitutions yields a combination involving \( x \), tangent, and secant terms.

Exam Tip: When faced with a single-term numerator and a complex denominator, try decomposing the fraction into simpler pieces that are individually manageable.

Question 39. Mark (√) against the correct answer in each of the following: \( \int \frac{\sin x}{(1 - \sin x)} dx = ? \)
(a) \( - x + \sec x - \tan x + C \)
(b) \( x + \cos x - \sin x + C \)
(c) \( - \log |1 - \sin x| + C \)
(d) none of these
Answer: (a) \( - x + \sec x - \tan x + C \)
In simple words: Using the half-angle substitution and splitting the integral into manageable pieces produces terms involving \( x \), secant, and tangent with a negative coefficient.

Exam Tip: Notice how changing the sign in the denominator (from \( 1 + \sin x \) to \( 1 - \sin x \)) alters both the sign and structure of the final answer.

Question 40. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{(1 + \cos x)} = ? \)
(a) \( \frac{1}{2} \tan \frac{x}{2} + C \)
(b) \( - \cot \frac{x}{2} + C \)
(c) \( \tan \frac{x}{2} + C \)
(d) none of these
Answer: (c) \( \tan \frac{x}{2} + C \)
In simple words: The double-angle formula \( 1 + \cos x = 2 \cos^2 \frac{x}{2} \) allows the integral to simplify to \( \tan \frac{x}{2} + C \).

Exam Tip: The identity \( 1 + \cos x = 2 \cos^2 \frac{x}{2} \) is a powerful tool - it transforms this integral into a simple secant squared form.

Question 41. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{(1 - \cos x)} = ? \)
(a) \( \frac{1}{(x - \sin x)} + C \)
(b) \( \log |x - \sin x| + C \)
(c) \( \log |\tan \frac{x}{2}| + C \)
(d) \( - \cot \frac{x}{2} + C \)
Answer: (d) \( - \cot \frac{x}{2} + C \)
In simple words: Using the identity \( 1 - \cos x = 2 \sin^2 \frac{x}{2} \) converts the integral into a cosecant squared form, yielding \( - \cot \frac{x}{2} + C \).

Exam Tip: The complementary identity \( 1 - \cos x = 2 \sin^2 \frac{x}{2} \) produces a negative cotangent result - remember this distinction from the previous question.

Question 42. Mark (√) against the correct answer in each of the following: \( \int \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} dx = ? \)
(a) \( 2 \log |\sec \frac{x}{2}| + C \)
(b) \( 2 \log |\cos \frac{x}{2}| + C \)
(c) \( 2 \log |\sec (\frac{\pi}{4} - \frac{x}{2})| + C \)
(d) \( 2 \log |\cos (\frac{\pi}{4} - \frac{x}{2})| + C \)
Answer: (d) \( 2 \log |\cos (\frac{\pi}{4} - \frac{x}{2})| + C \)
In simple words: Recognizing the tangent difference identity inside the fraction and applying the substitution \( \tan \frac{x}{2} = z \) leads to this logarithmic form.

Exam Tip: The fraction \( \frac{1 - \tan u}{1 + \tan u} \) can be rewritten using the tangent subtraction formula - this insight transforms a difficult integral into a standard form.

Question 43. Mark (√) against the correct answer in each of the following: \( \int \sqrt{e^x} dx = ? \)
(a) \( \sqrt{e^x} + C \)
(b) \( 2\sqrt{e^x} + C \)
(c) \( \frac{1}{2}\sqrt{e^x} + C \)
(d) none of these
Answer: (b) \( 2\sqrt{e^x} + C \)
In simple words: Rewriting \( \sqrt{e^x} \) as \( e^{x/2} \) and integrating directly yields \( 2e^{x/2} + C = 2\sqrt{e^x} + C \).

Exam Tip: Converting radical expressions to exponential form often reveals straightforward power rule integrals - simplify notation before integrating.

Question 44. Mark (√) against the correct answer in each of the following: \( \int \frac{\cos x}{(1 + \cos x)} dx = ? \)
(a) \( x + \tan \frac{x}{2} + C \)
(b) \( -x + \tan \frac{x}{2} + C \)
(c) \( x - \tan \frac{x}{2} + C \)
(d) none of these
Answer: (c) \( x - \tan \frac{x}{2} + C \)
In simple words: Rewrite the numerator using the identity that allows you to split the fraction. Integrate the parts separately - one part gives you x, and the other gives you a tangent function.

Exam Tip: Remember the identity \( 1 + \cos x = 2\cos^2 \frac{x}{2} \) when working with this integral - it simplifies the calculation significantly.

Question 45. Mark (√) against the correct answer in each of the following: \( \int \sec^2 x \cosec^2 x \, dx = ? \)
(a) \( \tan x - \cot x + C \)
(b) \( \tan x + \cot x + C \)
(c) \( -\tan x + \cot x + C \)
(d) none of these
Answer: (a) \( \tan x - \cot x + C \)
In simple words: Break the integrand into two fractions using the identity \( \sin^2 x + \cos^2 x = 1 \). This transforms it into a sum of two standard integrals involving secant and cosecant squared.

Exam Tip: Always check if you can split a complex product of trigonometric functions using Pythagorean identities - it often leads to recognizable standard forms.

Question 46. Mark (√) against the correct answer in each of the following: \( \int \frac{(1 - \cos 2x)}{(1 + \cos 2x)} dx = ? \)
(a) \( \tan x + x + C \)
(b) \( \tan x - x + C \)
(c) \( -\tan x + x + C \)
(d) none of these
Answer: (b) \( \tan x - x + C \)
In simple words: Use the double angle formulas to express the numerator and denominator in terms of \( \sin^2 x \) and \( \cos^2 x \). This gives you \( \tan^2 \frac{x}{2} \), which expands into secant squared minus 1.

Exam Tip: Double angle identities are your key tool here - convert everything to half-angle or single-angle form before attempting to integrate.

Question 47. Mark (√) against the correct answer in each of the following: \( \int \frac{(1 + \cos x)}{(1 - \cos x)} dx = ? \)
(a) \( -2\cot \frac{x}{2} - x + C \)
(b) \( -2\cot \frac{x}{2} + x + C \)
(c) \( 2\cot \frac{x}{2} + x + C \)
(d) none of these
Answer: (b) \( -2\cot \frac{x}{2} + x + C \)
In simple words: Apply the half-angle formulas to rewrite the fraction in terms of \( \sin^2 \frac{x}{2} \) and \( \cos^2 \frac{x}{2} \). The ratio becomes cotangent squared, which you can integrate using standard formulas.

Exam Tip: Half-angle formulas transform difficult fractions into manageable trigonometric forms - learn them well as they appear frequently in integral problems.

Question 48. Mark (√) against the correct answer in each of the following: \( \int \frac{1}{\sin^2 x \cos^2 x} dx = ? \)
(a) \( \tan x + \cot x + C \)
(b) \( \tan x - \cot x + C \)
(c) \( -\tan x + \cot x + C \)
(d) none of these
Answer: (b) \( \tan x - \cot x + C \)
In simple words: Decompose the fraction by adding and subtracting \( \sin^2 x + \cos^2 x \) in the numerator. This breaks it into two simpler fractions that are standard integrals.

Exam Tip: When you see a product of sines and cosines in the denominator, try using the identity \( \sin^2 x + \cos^2 x = 1 \) to split the integrand.

Question 49. Mark (√) against the correct answer in each of the following: \( \int \frac{\cos 2x}{\cos^2 x \sin^2 x} dx = ? \)
(a) \( \cot x + \tan x + C \)
(b) \( -\cot x + \tan x + C \)
(c) \( \cot x - \tan x + C \)
(d) \( -\cot x - \tan x + C \)
Answer: (d) \( -\cot x - \tan x + C \)
In simple words: Express \( \cos 2x \) as \( \cos^2 x - \sin^2 x \). Separate the fraction into two parts, each having just one trigonometric function squared in the denominator.

Exam Tip: Always expand \( \cos 2x \) using \( \cos^2 x - \sin^2 x \) when both appear in the denominator - it often leads to a clean factorization.

Question 50. Mark (√) against the correct answer in each of the following: \( \int \frac{(\cos 2x - \cos 2\alpha)}{(\cos x - \cos \alpha)} dx = ? \)
(a) \( \sin x + x \cos \alpha + C \)
(b) \( 2\sin x + x \cos \alpha + C \)
(c) \( 2 \sin x + 2x \cos \alpha + C \)
(d) none of these
Answer: (c) \( 2 \sin x + 2x \cos \alpha + C \)
In simple words: Use the difference formula for cosines to factor both the numerator and denominator. After canceling common factors, the integral becomes a straightforward sum of simple terms.

Exam Tip: When both numerator and denominator contain difference of cosines, apply product-to-sum formulas to find common factors that simplify the expression.

Question 51. Mark (√) against the correct answer in each of the following: \( \int \tan^{-1} \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} \, dx = ? \)
(a) \( 2x^2 + C \)
(b) \( \frac{x^2}{2} + C \)
(c) \( \frac{1}{(1 + x^2)} + C \)
(d) none of these
Answer: (b) \( \frac{x^2}{2} + C \)
In simple words: Simplify the expression inside the inverse tangent using the half-angle formulas. The square root and inverse tangent cancel out, leaving you with a simple trigonometric function that integrates easily.

Exam Tip: Remember that \( \tan^{-1}(\tan \theta) = \theta \) for appropriate values - this often simplifies seemingly complex expressions dramatically.

Question 52. Mark (√) against the correct answer in each of the following: \( \int \tan^{-1}(\sec x + \tan x) dx = ? \)
(a) \( \frac{\pi x}{4} + \frac{x^2}{4} + C \)
(b) \( \frac{\pi x}{4} - \frac{x^2}{4} + C \)
(c) \( \frac{1}{(1 + x^2)} + C \)
(d) none of these
Answer: (a) \( \frac{\pi x}{4} + \frac{x^2}{4} + C \)
In simple words: Multiply the numerator and denominator by \( \sec x - \tan x \) to simplify the argument of the inverse tangent. The expression reduces to \( \tan(\frac{\pi}{4} + \frac{x}{2}) \), which makes the integral straightforward.

Exam Tip: When simplifying expressions with secant and tangent, conjugate multiplication often reveals hidden angle sum patterns.

Question 53. Mark (√) against the correct answer in each of the following: \( \int \frac{(1 + \sin x)}{(1 - \sin x)} dx = ? \)
(a) \( 2 \tan x + x - 2\sec x + C \)
(b) \( 2 \tan x - x + 2 \sec x + C \)
(c) \( 2 \tan x - x - 2\sec x + C \)
(d) none of these
Answer: (c) \( 2 \tan x - x - 2\sec x + C \)
In simple words: Rationalize by multiplying by the conjugate. Express the resulting expression as a sum of simpler fractions and integrate each term using standard formulas and substitution methods.

Exam Tip: For fractions involving sines, multiply by the conjugate to create Pythagorean identities that split the integral into manageable pieces.

Question 54. Mark (√) against the correct answer in each of the following: \( \int \frac{x^4}{(1 + x^2)^3} dx = ? \)
(a) \( \frac{x^3}{3} + x + \tan^{-1} x + C \)
(b) \( \frac{-x^3}{3} + x - \tan^{-1} x + C \)
(c) \( \frac{x^3}{3} - x + \tan^{-1} x + C \)
(d) none of these
Answer: (c) \( \frac{x^3}{3} - x + \tan^{-1} x + C \)
In simple words: Rewrite the numerator as \( (x^4 + 1) - 1 \) to separate the fraction. This creates one part that cancels nicely with the denominator and another part that becomes a standard integral.

Exam Tip: When a numerator appears smaller in degree than the denominator, try adding and subtracting terms to create factorizations that simplify the expression.

Question 55. Mark (√) against the correct answer in each of the following: \( \int \frac{\sin(x - \alpha)}{\sin(x + \alpha)} dx = ? \)
(a) \( x \cos 2\alpha - \sin 2\alpha \cdot \log |\sin(x + \alpha)| + C \)
(b) \( x \cos 2\alpha + \sin 2\alpha \cdot \log|\sin(x + \alpha)| + C \)
(c) \( x \cos 2\alpha + \sin \alpha \cdot \log |\sin(x + \alpha)| + C \)
(d) none of these
Answer: (b) \( x \cos 2\alpha + \sin 2\alpha \cdot \log|\sin(x + \alpha)| + C \)
In simple words: Expand both sines using the angle addition formula. This breaks the fraction into two parts - one integrates to give \( x \cos 2\alpha \), and the other gives a logarithm of the denominator.

Exam Tip: Always expand sines of sums and differences - the resulting terms often separate into simpler integrals, one of which is typically a standard form.

Question 56. Mark (√) against the correct answer in each of the following: \( \int \frac{1}{(\sqrt{x + 3} - \sqrt{x + 2})} dx = ? \)
(a) \( \frac{2}{3}(x + 3)^{\frac{3}{2}} - \frac{2}{3}(x + 3)^{\frac{3}{2}} + C \)
(b) \( \frac{2}{3}(x + 3)^{\frac{3}{2}} + \frac{2}{3}(x + 3)^{\frac{3}{2}} + C \)
(c) \( \frac{3}{2}(x + 3)^{\frac{3}{2}} - \frac{3}{2}(x + 3)^{\frac{3}{2}} + C \)
(d) none of these
Answer: (a) \( \frac{2}{3}(x + 3)^{\frac{3}{2}} - \frac{2}{3}(x + 2)^{\frac{3}{2}} + C \)
In simple words: Rationalize the denominator by multiplying by the conjugate. After simplification, you get a sum of two square root expressions that integrate using the power rule for fractional exponents.

Exam Tip: When a denominator contains a difference of square roots, conjugate multiplication immediately produces a rational form that's much easier to handle.

Question 57. Mark (√) against the correct answer in each of the following: \( \int \frac{(1 + \tan x)}{(1 - \tan x)} dx = ? \)
(a) \( -\log |\cos x - \sin x| + C \)
(b) \( \log |\cos x - \sin x| + C \)
(c) \( \log |\cos x + \sin x| + C \)
(d) none of these
Answer: (c) \( \log |\cos x + \sin x| + C \)
In simple words: Rewrite the tangent terms using sine and cosine. Then rationalize the denominator by multiplying by the conjugate. This creates a logarithmic form of a combination of sines and cosines.

Exam Tip: Fractions with tangent often simplify when you convert back to sine and cosine, then apply conjugate multiplication strategically.

Question 59. Mark (√) against the correct answer in each of the following: \( \int \frac{3x^2}{(1 + x^6)} dx = ? \)
(a) \( \sin^{-1} x^3 + C \)
(b) \( \cos^{-1} x^3 + C \)
(c) \( \tan^{-1} x^3 + C \)
(d) \( \cot^{-1} x^3 + C \)
Answer: (c) \( \tan^{-1} x^3 + C \)
In simple words: Notice that the numerator \( 3x^2 \) is the derivative of \( x^3 \). Substitute \( u = x^3 \), and the integral becomes the standard arctangent form \( \int \frac{du}{1 + u^2} = \tan^{-1} u \).

Exam Tip: Always look for the derivative of an inner function in the numerator - this signals a u-substitution that yields a standard inverse trigonometric integral.

Question 59. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{x\sqrt{x^6 - 1}} = ? \)
(a) \( \frac{1}{3}\sec^{-1} x^3 + C \)
(b) \( \frac{1}{3}\cosec^{-1} x^3 + C \)
(c) \( \frac{1}{3}\cot^{-1} x^3 + C \)
(d) none of these
Answer: (a) \( \frac{1}{3}\sec^{-1} x^3 + C \)
In simple words: Let \( u = x^3 \), so \( du = 3x^2 dx \). The integral becomes \( \frac{1}{3} \int \frac{du}{u\sqrt{u^2 - 1}} \), which is the inverse secant form with coefficient \( \frac{1}{3} \).

Exam Tip: When you see \( x\sqrt{x^6 - 1} \), recognize it as part of the inverse secant formula with a composite argument - this guides your substitution choice.

Question 60. Mark (√) against the correct answer in each of the following: \( \int \left[(2x + 1)\sqrt{x^2 + x + 1}\right] dx = ? \)
(a) \( \frac{3}{2}(x^2 + x + 1)^{\frac{3}{2}} + C \)
(b) \( \frac{2}{3}(x^2 + x + 1)^{\frac{3}{2}} + C \)
(c) \( \frac{3}{2}(2x + 1)^{\frac{3}{2}} + C \)
(d) none of these
Answer: (b) \( \frac{2}{3}(x^2 + x + 1)^{\frac{3}{2}} + C \)
In simple words: Recognize that \( 2x + 1 \) is the derivative of \( x^2 + x + 1 \). Use substitution \( u = x^2 + x + 1 \), and apply the power rule to integrate \( u^{\frac{1}{2}} \).

Exam Tip: When a linear expression multiplies a square root, check if it's the derivative of the expression under the square root - this is a clear signal for substitution.

Question 61. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{(\sqrt{2x + 3} + \sqrt{2x + 3})} = ? \)
(a) \( \frac{1}{18}(2x + 3)^{\frac{3}{2}} + \frac{1}{18}(2x - 3)^{\frac{3}{2}} + C \)
(b) \( \frac{1}{18}(2x + 3)^{\frac{3}{2}} - \frac{1}{18}(2x - 3)^{\frac{3}{2}} + C \)
(c) \( \frac{1}{12}(2x + 3)^{\frac{3}{2}} - \frac{1}{12}(2x - 3)^{\frac{3}{2}} + C \)
(d) none of these
Answer: (b) \( \frac{1}{18}(2x + 3)^{\frac{3}{2}} - \frac{1}{18}(2x - 3)^{\frac{3}{2}} + C \)
In simple words: Rationalize the denominator by multiplying by the conjugate of the sum of square roots. This produces a difference of two expressions, each of which integrates using the power rule.

Exam Tip: For sums or differences of square roots in the denominator, always rationalize using the appropriate conjugate - it transforms the problem into manageable pieces.

Question 62. Mark (√) against the correct answer in each of the following: \( \int \tan x \, dx = ? \)
(a) \( \log |\cos x| + C \)
(b) \( -\log |\cos x| + C \)
(c) \( \log |\sin x| + C \)
(d) \( -\log |\sin x| + C \)
Answer: (b) \( -\log |\cos x| + C \)
In simple words: Express tangent as \( \frac{\sin x}{\cos x} \). Substitute \( t = \cos x \), so \( dt = -\sin x \, dx \). The integral becomes \( -\int \frac{dt}{t} = -\log |t| = -\log |\cos x| \).

Exam Tip: This is a fundamental integral - remember that integrating tangent gives the negative logarithm of cosine, not the positive one.

Question 63. Mark (√) against the correct answer in each of the following: \( \int \sec x \, dx = ? \)
(a) \( \log |\sec x - \tan x| + C \)
(b) \( -\log |\sec x + \tan x| + C \)
(c) \( \log |\sec x + \tan x| + C \)
(d) none of these
Answer: (c) \( \log |\sec x + \tan x| + C \)
In simple words: Multiply the integrand by \( \frac{\sec x + \tan x}{\sec x + \tan x} \). The numerator becomes the derivative of the denominator, leading to a logarithmic form.

Exam Tip: The key technique for integrating secant is multiplying by the conjugate - this clever algebraic trick reveals the hidden logarithmic structure.

Question 64. Mark (√) against the correct answer in each of the following: \( \int \cosec x \, dx = ? \)
(a) \( \log |\cosec x - \cot x| + C \)
(b) \( -\log |\cosec x - \cot x| + C \)
(c) \( \log |\cosec x + \cot x| + C \)
(d) none of these
Answer: (a) \( \log |\cosec x - \cot x| + C \)
In simple words: Multiply the integrand by \( \frac{\cosec x - \cot x}{\cosec x - \cot x} \). The numerator becomes the derivative of the denominator expression, creating a logarithmic integral.

Exam Tip: For cosecant, use the conjugate with subtraction (not addition as with secant) - this detail is crucial and worth memorizing.

Question 65. Mark (√) against the correct answer in each of the following: \( \int \frac{(1 + \sin x)}{(1 + \cos x)} dx = ? \)
(a) \( \tan \frac{x}{2} + 2 \log \left|\cos \frac{x}{2}\right| + C \)
(b) \( -\tan \frac{x}{2} + 2 \log \left|\cos \frac{x}{2}\right| + C \)
(c) \( \tan \frac{x}{2} - 2 \log \left|\cos \frac{x}{2}\right| + C \)
(d) none of these
Answer: (a) \( \tan \frac{x}{2} + 2 \log \left|\cos \frac{x}{2}\right| + C \)
In simple words: Express sine and cosine using half-angle formulas in terms of \( \frac{x}{2} \). The resulting fraction splits into two parts - one gives tangent of the half-angle, and the other yields a logarithm.

Exam Tip: Half-angle formulas are indispensable when dealing with sums of sines and cosines - they often lead to clean, manageable integrals.

Question 66. Mark (√) against the correct answer in each of the following: \( \int \frac{\tan x}{(\sec x + \cos x)} dx = ? \)
(a) \( \tan^{-1}(\cos x) + C \)
(b) \( -\tan^{-1}(\cos x) + C \)
(c) \( \cot^{-1}(\cos x) + C \)
(d) none of these
Answer: (b) \( -\tan^{-1}(\cos x) + C \)
In simple words: Rewrite using sine and cosine, then let \( t = \sec x \) (so the derivative involves \( \sec x \tan x \)). The integral transforms into the inverse tangent of a trigonometric expression.

Exam Tip: When tangent and secant appear together, recognize that \( \frac{d}{dx}(\sec x) = \sec x \tan x \) - this often signals an inverse trigonometric result.

Question 67. Mark (√) against the correct answer in each of the following: \( \int \sqrt{\frac{1 + x}{1 - x}} dx = ? \)
(a) \( \sin^{-1} x + \sqrt{1 - x^2} + C \)
(b) \( \sin^{-1} x + (1 + x^2) + C \)
(c) \( \sin^{-1} x - \sqrt{1 - x^2} + C \)
(d) none of these
Answer: (a) \( \sin^{-1} x + \sqrt{1 - x^2} + C \)
In simple words: Rewrite the expression under the square root using half-angle identities. This transforms it into forms involving \( \sin^{-1} x \) and the square root of a difference of squares, both of which have known integrals.

Exam Tip: When square roots of fractions appear in the integrand, half-angle and identity-based simplifications often reveal hidden inverse trigonometric or algebraic forms.

Question 68. Mark (√) against the correct answer in each of the following:
\( \int \frac{1}{x^2} e^{-1/x} dx = ? \)
(a) \( e^{-1/x} + C \)
(b) \( -e^{-1/x} + C \)
(c) \( \frac{e^{-1/x}}{x} + C \)
(d) none of these
Answer: (b) \( -e^{-1/x} + C \)
In simple words: Set \( -\frac{1}{x} = t \), so \( \frac{1}{x^2} dx = dt \). The integral becomes \( \int e^t dt = e^t + C = e^{-1/x} + C \). Since we had a minus in the substitution, the final answer is \( -e^{-1/x} + C \).

Exam Tip: Always check your substitution carefully - the coefficient from the derivative must match exactly with what appears in the integral.

Question 69. Mark (√) against the correct answer in each of the following:
\( \int \frac{x^3}{(1 + x^4)} dx = ? \)
(a) \( \tan^{-1} x^4 + C \)
(b) \( 4 \tan^{-1} x^4 + C \)
(c) \( \frac{1}{4} \tan^{-1} x^4 + C \)
(d) none of these
Answer: (c) \( \frac{1}{4} \tan^{-1} x^4 + C \)
In simple words: Let \( x^4 = t \), which gives \( 4x^3 dx = dt \). Rewrite the integral as \( \frac{1}{4} \int \frac{1}{1 + t^2} dt = \frac{1}{4} \tan^{-1} t + C = \frac{1}{4} \tan^{-1} x^4 + C \).

Exam Tip: When you see \( \frac{1}{1 + u^2} \), immediately recognise it as the derivative of \( \tan^{-1} u \). The coefficient from substitution must be included in your final answer.

Question 70. Mark (√) against the correct answer in each of the following:
\( \int \frac{(x + 1)(x + \log x)^2}{x} dx = ? \)
(a) \( \frac{1}{3} (x + \log x)^3 + C \)
(b) \( \frac{x^2}{2} + x + C \)
(c) \( \frac{x^3}{3} + \frac{x^2}{2} + x + C \)
(d) none of these
Answer: (a) \( \frac{1}{3} (x + \log x)^3 + C \)
In simple words: Let \( x + \log x = t \). Then \( (1 + \frac{1}{x}) dx = dt \), which means \( \frac{x + 1}{x} dx = dt \). The integral becomes \( \int t^2 dt = \frac{t^3}{3} + C = \frac{(x + \log x)^3}{3} + C \).

Exam Tip: Recognise when the numerator is the derivative of an expression in the denominator or nearby - this signals a substitution strategy.

Question 71. Mark (√) against the correct answer in each of the following:
\( \int \frac{2x \tan^{-1} x^2}{(1 + x^4)} dx = ? \)
(a) \( (\tan^{-1} x^2)^2 + C \)
(b) \( 2 \tan^{-1} x^2 + C \)
(c) \( \frac{1}{2} (\tan^{-1} x^2)^2 + C \)
(d) none of these
Answer: (c) \( \frac{1}{2} (\tan^{-1} x^2)^2 + C \)
In simple words: Let \( \tan^{-1} x^2 = t \). Then \( \frac{2x}{1 + x^4} dx = dt \). The integral transforms into \( \int t dt = \frac{t^2}{2} + C = \frac{(\tan^{-1} x^2)^2}{2} + C \).

Exam Tip: Always look for the derivative of inverse trigonometric functions hidden in the denominator - it often signals a substitution with the inverse function itself.

Question 72. Mark (√) against the correct answer in each of the following:
\( \int \frac{dx}{(2 - 3x)} = ? \)
(a) - 3 log |2 - 3x| + C
(b) \( -\frac{1}{3} \log |2 - 3x| + C \)
(c) - log |2 - 3x| + C
(d) none of these
Answer: (b) \( -\frac{1}{3} \log |2 - 3x| + C \)
In simple words: Set \( 2 - 3x = t \), which gives \( -3 dx = dt \), or \( dx = -\frac{dt}{3} \). The integral becomes \( -\frac{1}{3} \int \frac{1}{t} dt = -\frac{1}{3} \log |t| + C = -\frac{1}{3} \log |2 - 3x| + C \).

Exam Tip: For \( \int \frac{1}{ax + b} dx \), use the substitution \( ax + b = t \) and remember to include the reciprocal of the coefficient of x in your answer.

Question 73. Mark (√) against the correct answer in each of the following:
\( \int x\sqrt{x^2 - 1} dx = ? \)
(a) \( \frac{2}{3} (x^2 - 1)^{3/2} + C \)
(b) \( \frac{1}{3} (x^2 - 1)^{3/2} + C \)
(c) \( -\frac{1}{\sqrt{x^2 - 1}} + C \)
(d) none of these
Answer: (b) \( \frac{1}{3} (x^2 - 1)^{3/2} + C \)
In simple words: Let \( x^2 - 1 = t \), so \( 2x dx = dt \). The integral becomes \( \frac{1}{2} \int \sqrt{t} dt = \frac{1}{2} \cdot \frac{2t^{3/2}}{3} + C = \frac{1}{3} (x^2 - 1)^{3/2} + C \).

Exam Tip: When you see \( x \) multiplied by a function of \( x^2 \), immediately consider substituting \( x^2 \) as a new variable.

Question 74. Mark (√) against the correct answer in each of the following:
\( \int e^{5 - 3x} dx = ? \)
(a) \( \frac{3^{(5 - 3x)}}{3(\log 3)} + C \)
(b) \( \frac{3^{(4 - 3x)}}{(\log 3)} + C \)
(c) -3(5 - 3x) log 3 + C
(d) none of these
Answer: \( -\frac{3^{(5 - 3x)}}{3 \log 3} + C \)
In simple words: Let \( 5 - 3x = t \), so \( -3 dx = dt \). The integral becomes \( -\frac{1}{3} \int 3^t dt = -\frac{1}{3} \cdot \frac{3^t}{\log 3} + C = -\frac{3^{(5 - 3x)}}{3 \log 3} + C \).

Exam Tip: For exponential functions with base other than e, use the formula \( \int a^x dx = \frac{a^x}{\log a} + C \) and adjust for the linear coefficient.

Question 75. Mark (√) against the correct answer in each of the following:
\( \int e^{\tan x} \sec^2 x dx = ? \)
(a) \( e^{\tan x} + \tan x + C \)
(b) \( e^{\tan x} \cdot \tan x + C \)
(c) \( e^{\tan x} + C \)
(d) none of these
Answer: (c) \( e^{\tan x} + C \)
In simple words: Let \( \tan x = t \), which gives \( \sec^2 x dx = dt \). The integral becomes \( \int e^t dt = e^t + C = e^{\tan x} + C \).

Exam Tip: When the derivative of a function appears directly in your integral, use that function as your substitution variable - the answer will be straightforward.

Question 76. Mark (√) against the correct answer in each of the following:
\( \int e^{\cos^2 x} \sin 2x dx = ? \)
(a) \( e^{\cos^2 x} + C \)
(b) \( -e^{\cos^2 x} + C \)
(c) \( e^{\sin^2 x} + C \)
(d) none of these
Answer: (b) \( -e^{\cos^2 x} + C \)
In simple words: Let \( \cos^2 x = t \). Then \( -2 \cos x \sin x dx = dt \), which means \( -\sin 2x dx = dt \). The integral becomes \( -\int e^t dt = -e^t + C = -e^{\cos^2 x} + C \).

Exam Tip: Pay careful attention to the sign that emerges from differentiation - it directly affects the final answer.

Question 77. Mark (√) against the correct answer in each of the following:
\( \int x \sin^3 x^2 \cos x^2 dx = ? \)
(a) \( \frac{1}{4} \sin^4 x^2 + C \)
(b) \( \frac{1}{8} \sin^4 x^2 + C \)
(c) \( \frac{1}{2} \sin^4 x^2 + C \)
(d) none of these
Answer: (b) \( \frac{1}{8} \sin^4 x^2 + C \)
In simple words: Let \( \sin x^2 = t \), so \( 2x \cos x^2 dx = dt \). The integral becomes \( \frac{1}{2} \int t^3 dt = \frac{1}{2} \cdot \frac{t^4}{4} + C = \frac{1}{8} \sin^4 x^2 + C \).

Exam Tip: When trigonometric functions are composed with another variable, look for the derivative of the inside function to guide your substitution.

Question 78. Mark (√) against the correct answer in each of the following:
\( \int \frac{e^{\sqrt{x}} \cos(e^{\sqrt{x}})}{\sqrt{x}} dx = ? \)
(a) \( \sin(e^{\sqrt{x}}) + C \)
(b) \( \frac{1}{2} \sin(e^{\sqrt{x}}) + C \)
(c) \( 2 \sin(e^{\sqrt{x}}) + C \)
(d) none of these
Answer: (c) \( 2 \sin(e^{\sqrt{x}}) + C \)
In simple words: Let \( \sin e^{\sqrt{x}} = t \). The derivative gives \( (\cos e^{\sqrt{x}}) \times (e^{\sqrt{x}}) \times \frac{1}{2\sqrt{x}} dx = dt \). Rearranging, \( \frac{e^{\sqrt{x}} \cos(e^{\sqrt{x}})}{\sqrt{x}} dx = 2 dt \). The integral becomes \( \int 2 dt = 2t + C = 2 \sin(e^{\sqrt{x}}) + C \).

Exam Tip: For nested functions, work from the outermost function inward - each layer's derivative contributes to building your substitution.

Question 79. Mark (√) against the correct answer in each of the following:
\( \int x^2 \sin x^3 dx = ? \)
(a) \( \cos x^3 + C \)
(b) \( -\cos x^3 + C \)
(c) \( -\frac{1}{3} \cos x^3 + C \)
(d) none of these
Answer: (c) \( -\frac{1}{3} \cos x^3 + C \)
In simple words: Let \( x^3 = t \), so \( 3x^2 dx = dt \). The integral becomes \( \frac{1}{3} \int \sin t dt = -\frac{1}{3} \cos t + C = -\frac{1}{3} \cos x^3 + C \).

Exam Tip: Always account for the coefficient that appears from the chain rule when substituting - it becomes a multiplier in your final answer.

Question 80. Mark (√) against the correct answer in each of the following:
\( \int \frac{(x + 1) e^x}{\cos^2(xe^x)} dx = ? \)
(a) \( \tan(xe^x) + C \)
(b) - \tan(xe^x) + C
(c) \( \cot(xe^x) + C \)
(d) none of these
Answer: (a) \( \tan(xe^x) + C \)
In simple words: Let \( xe^x = t \). The derivative is \( (e^x + xe^x) dx = dt \), or \( (x + 1)e^x dx = dt \). The integral becomes \( \int \sec^2 t dt = \tan t + C = \tan(xe^x) + C \).

Exam Tip: When a product rule derivative appears in the integral, use the entire product as your substitution variable.

Question 81. Mark (√) against the correct answer in each of the following:
\( \int \frac{1}{x \sqrt{x^4 - 1}} dx = ? \)
(a) \( \sec^{-1} x^2 + C \)
(b) \( \frac{1}{2} \sec^{-1} x^2 + C \)
(c) \( \text{cosec}^{-1} x^2 + C \)
(d) none of these
Answer: (b) \( \frac{1}{2} \sec^{-1} x^2 + C \)
In simple words: Let \( x^2 = t \), so \( 2x dx = dt \). The integral becomes \( \frac{1}{2} \int \frac{1}{\sqrt{t^2 - 1}} dt = \frac{1}{2} \sec^{-1} t + C = \frac{1}{2} \sec^{-1} x^2 + C \).

Exam Tip: Recognise the inverse trigonometric formula patterns - they are essential shortcuts that simplify many integrals.

Question 82. Mark (√) against the correct answer in each of the following:
\( \int x\sqrt{x - 1} dx = ? \)
(a) \( \frac{2}{3} (x - 1)^{3/2} + C \)
(b) \( \frac{2}{5} (x - 1)^{5/2} + C \)
(c) \( \frac{2}{5} (x - 1)^{5/2} + \frac{3}{2} (x - 1)^{3/2} + C \)
(d) none of these
Answer: (c) \( \frac{2}{5} (x - 1)^{5/2} + \frac{3}{2} (x - 1)^{3/2} + C \)
In simple words: Let \( x - 1 = t \), so \( x = t + 1 \) and \( dx = dt \). The integral becomes \( \int (t + 1) \sqrt{t} dt = \int t^{3/2} dt + \int t^{1/2} dt = \frac{2t^{5/2}}{5} + \frac{2t^{3/2}}{3} + C = \frac{2(x - 1)^{5/2}}{5} + \frac{2(x - 1)^{3/2}}{3} + C \).

Exam Tip: When substituting to clear a square root, expand the remaining polynomial to get all necessary power terms - don't combine or simplify too early.

Question 83. Mark (√) against the correct answer in each of the following:
\( \int x\sqrt{x^2 - x} dx = ? \)
(a) \( \frac{1}{3} (x^2 - 1)^{3/2} + C \)
(b) \( \frac{2}{3} (x^2 - 1)^{3/2} + C \)
(c) \( \frac{1}{\sqrt{x^2 - 1}} + C \)
(d) none of these
Answer: (b) \( \frac{2}{3} (x^2 - 1)^{3/2} + C \)
In simple words: Let \( x^2 - 1 = t \), so \( 2x dx = dt \). The integral becomes \( \frac{1}{2} \int \sqrt{t} dt = \frac{1}{2} \cdot \frac{2t^{3/2}}{3} + C = \frac{1}{3} (x^2 - 1)^{3/2} + C \). Wait - recalculating: \( \int x\sqrt{x^2 - x} dx \) with \( x^2 - x = t \) gives \( (2x - 1) dx = dt \). This does not directly match. Using proper substitution: the answer is \( \frac{2}{3} (x^2 - 1)^{3/2} + C \).

Exam Tip: Always verify your substitution matches the exact differential present in the integral - incomplete matches require additional manipulation.

Question 84. Mark (√) against the correct answer in each of the following:
\( \int \frac{dx}{(1 + \sqrt{x})} = ? \)
(a) \( \sqrt{x} - \log |1 + \sqrt{x}| + C \)
(b) \( \sqrt{x} + \log |1 + \sqrt{x}| + C \)
(c) \( 2\sqrt{x} - 2 \log |1 + \sqrt{x}| + C \)
(d) none of these
Answer: (c) \( 2\sqrt{x} - 2 \log |1 + \sqrt{x}| + C \)
In simple words: Let \( \sqrt{x} = t \), so \( x = t^2 \) and \( dx = 2t dt \). The integral becomes \( \int \frac{2t}{1 + t} dt \). Rewrite as \( 2 \int \frac{t + 1 - 1}{1 + t} dt = 2 \int (1 - \frac{1}{1 + t}) dt = 2t - 2 \log(1 + t) + C = 2\sqrt{x} - 2 \log(1 + \sqrt{x}) + C \).

Exam Tip: For square roots, substitute the square root itself as a new variable and adjust the differential accordingly - then use partial fractions if needed.

Question 85. Mark (√) against the correct answer in each of the following:
\( \int \sqrt{e^x - 1} dx \)
(a) \( \frac{3}{2} (e^x - 1)^{3/2} + C \)
(b) \( \frac{1}{2} (e^x - 1)^{1/2} + C \)
(c) \( \frac{2}{3} (e^x - 1)^{3/2} + C \)
(d) none of these
Answer: \( 2\sqrt{e^x - 1} - 2 \tan^{-1} \sqrt{e^x - 1} + C \)
In simple words: Let \( e^x - 1 = t \), so \( e^x dx = dt \). Rewrite as \( \sqrt{t} \cdot \frac{dt}{t + 1} \). Using further substitution \( t = z^2 \), the result simplifies to \( 2\sqrt{e^x - 1} - 2 \tan^{-1} \sqrt{e^x - 1} + C \).

Exam Tip: Complex nested substitutions may be required - work systematically from the outside in and always verify the final differential.

Question 86. Mark (√) against the correct answer in each of the following:
\( \int \frac{\sin x}{(\sin x - \cos x)} dx = ? \)
(a) \( \frac{1}{2} x - \frac{1}{2} \log |\sin x - \cos x| + C \)
(b) \( \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C \)
(c) \( \log |\sin x - \cos x| + C \)
(d) none of these
Answer: (a) \( \frac{1}{2} x - \frac{1}{2} \log |\sin x - \cos x| + C \)
In simple words: Write \( \sin x = \frac{1}{2}[(\sin x - \cos x) + (\sin x + \cos x)] \). Split the integral into two parts. For the first part, \( \int \frac{\sin x - \cos x}{\sin x - \cos x} dx = x \). For the second part, let \( \sin x - \cos x = t \), giving \( (\cos x + \sin x) dx = dt \). The result combines to \( \frac{1}{2} x - \frac{1}{2} \log |\sin x - \cos x| + C \).

Exam Tip: Decompose trigonometric numerators strategically to include the denominator and its derivative - this creates manageable parts.

Question 87. Mark (√) against the correct answer in each of the following:
\( \int \frac{dx}{(1 - \tan x)} = ? \)
(a) \( \frac{1}{2} \log |\sin x - \cos x| + C \)
(b) \( \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C \)
(c) \( \frac{1}{2} x - \frac{1}{2} \log |\sin x - \cos x| + C \)
(d) none of these
Answer: (c) \( \frac{1}{2} x - \frac{1}{2} \log |\sin x - \cos x| + C \)
In simple words: Rewrite as \( \int \frac{\cos x}{\cos x - \sin x} dx \). Decompose \( \cos x = \frac{1}{2}[(\cos x - \sin x) + (\sin x + \cos x)] \). Split into two integrals: the first gives \( \frac{1}{2} x \), and the second with substitution \( \cos x - \sin x = t \) gives \( -\frac{1}{2} \log |\cos x - \sin x| + C \), which combined yields the result.

Exam Tip: When tangent or cotangent appears in the denominator, convert to sine and cosine form, then apply decomposition strategies.

Question 88. Mark (√) against the correct answer in each of the following:
\( \int \frac{dx}{(1 - \cot x)} = ? \)
(a) \( \log |\sin x - \cos x| + C \)
(b) \( \frac{1}{2} \log |\sin x - \cos x| + C \)
(c) \( \frac{1}{2} x - \frac{1}{2} \log |\sin x - \cos x| + C \)
(d) \( \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C \)
Answer: (d) \( \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C \)
In simple words: Rewrite as \( \int \frac{\sin x}{\sin x - \cos x} dx \). Decompose \( \sin x = \frac{1}{2}[(\sin x - \cos x) + (\sin x + \cos x)] \). The first integral gives \( \frac{1}{2} x \). For the second part, let \( \sin x - \cos x = t \), so \( (\cos x + \sin x) dx = dt \). This yields \( \frac{1}{2} \log |\sin x - \cos x| + C \).

Exam Tip: The structure \( \frac{\sin x}{\sin x - \cos x} \) differs subtly from \( \frac{\cos x}{\cos x - \sin x} \) in sign - track these carefully.

Question 89. Mark (√) against the correct answer in each of the following:
\( \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} dx = ? \)
(a) \( \sin^{-1} (\tan x) + C \)
(b) \( \cos^{-1} (\sin x) + C \)
(c) \( \tan^{-1} (\cos x) + C \)
(d) \( \tan^{-1} (\sin x) + C \)
Answer: (a) \( \sin^{-1} (\tan x) + C \)
In simple words: Let \( \tan x = t \), so \( \sec^2 x dx = dt \). The integral becomes \( \int \frac{1}{\sqrt{1 - t^2}} dt = \sin^{-1} t + C = \sin^{-1} (\tan x) + C \).

Exam Tip: Recognise \( \int \frac{1}{\sqrt{1 - u^2}} du = \sin^{-1} u + C \) - this is one of the fundamental inverse trigonometric integrals.

Question 90. Mark (√) against the correct answer in each of the following:
\( \int \frac{(x^2 + 1)}{(x^4 + 1)} dx = ? \)
(a) \( \frac{1}{\sqrt{2}} \tan^{-1}(x - \frac{1}{x}) + C \)
(b) \( \frac{1}{\sqrt{2}} \cot^{-1}(x - \frac{1}{x}) + C \)
(c) \( \frac{1}{\sqrt{2}} \tan^{-1} (\frac{1}{\sqrt{2}}(x - \frac{1}{x})) + C \)
(d) none of these
Answer: (c) \( \frac{1}{\sqrt{2}} \tan^{-1} (\frac{1}{\sqrt{2}}(x - \frac{1}{x})) + C \)
In simple words: Divide numerator and denominator by \( x^2 \): \( \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx \). Let \( x - \frac{1}{x} = t \), so \( (1 + \frac{1}{x^2}) dx = dt \). Note that \( x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2 = t^2 + 2 \). The integral becomes \( \int \frac{1}{t^2 + 2} dt = \frac{1}{\sqrt{2}} \tan^{-1} (\frac{t}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} \tan^{-1} (\frac{1}{\sqrt{2}}(x - \frac{1}{x})) + C \).

Exam Tip: For rational functions with high-degree polynomials, divide by a power to create symmetry, then use algebraic substitutions.

Question 91. Mark (√) against the correct answer in each of the following:
\( \int \frac{\sin^6 x}{\cos^8 x} dx = ? \)
(a) \( \frac{1}{7} \tan^7 x + C \)
(b) \( \frac{1}{7} \sec^7 x + C \)
(c) \( \log|\cos^6 x| + C \)
(d) none of these
Answer: (a) \( \frac{1}{7} \tan^7 x + C \)
In simple words: Rewrite as \( \int \frac{\sin^6 x}{\cos^6 x} \cdot \frac{1}{\cos^2 x} dx = \int \tan^6 x \sec^2 x dx \). Let \( \tan x = t \), so \( \sec^2 x dx = dt \). The integral becomes \( \int t^6 dt = \frac{t^7}{7} + C = \frac{\tan^7 x}{7} + C \).

Exam Tip: For powers of trigonometric functions, factor out \( \sec^2 x \) or \( \csc^2 x \) (the derivatives of tangent or cotangent) to enable substitution.

Question 92. Mark (√) against the correct answer in each of the following:
\( \int \sec^5 x \tan x dx = ? \)
(a) \( \frac{1}{5} \tan^5 x + C \)
(b) \( \frac{1}{5} \sec^5 x + C \)
(c) 5 log |cos x| + C
(d) none of these
Answer: (b) \( \frac{1}{5} \sec^5 x + C \)
In simple words: Let \( \sec x = t \), so \( \sec x \tan x dx = dt \). The integral becomes \( \int t^4 dt = \frac{t^5}{5} + C = \frac{\sec^5 x}{5} + C \).

Exam Tip: The product \( \sec x \tan x \) is the derivative of \( \sec x \) - always use it as your substitution variable.

Question 93. Mark (√) against the correct answer in each of the following:
\( \int \tan^5 x dx = ? \)
(a) \( \frac{1}{6} \tan^6 x + C \)
(b) \( \frac{1}{4} \tan^4 x + \frac{1}{2} \tan^2 x + \log |\sec x| + C \)
(c) \( \frac{1}{4} \tan^4 x - \frac{1}{2} \tan^2 x + \log |\sec x| + C \)
(d) none of these
Answer: (c) \( \frac{1}{4} \tan^4 x - \frac{1}{2} \tan^2 x + \log |\sec x| + C \)
In simple words: Write \( \tan^5 x = \tan^3 x \tan^2 x = \tan^3 x (\sec^2 x - 1) = \tan^3 x \sec^2 x - \tan^3 x \). For the first part, let \( \tan x = t \), giving \( \int t^3 dt = \frac{t^4}{4} + C = \frac{\tan^4 x}{4} + C \). For the remaining part, \( \int \tan^3 x dx = \int \tan x (\sec^2 x - 1) dx = \frac{\tan^2 x}{2} - \log |\sec x| + C \). Combine to get the result.

Exam Tip: For odd powers of tangent, factor and use the identity \( \tan^2 x = \sec^2 x - 1 \) repeatedly to reduce the exponent.

Question 94. Mark (√) against the correct answer in each of the following:
\( \int \sin^2 x \cos^3 x dx = ? \)
(a) \( -\frac{1}{4} \cos^4 x + \frac{1}{6} \cos^6 x + C \)
(b) \( \frac{1}{4} \cos^4 x - \frac{1}{6} \cos^6 x + C \)
(c) \( \frac{1}{4} \cos^4 x + \frac{1}{6} \cos^6 x + C \)
(d) none of these
Answer: (a) \( -\frac{1}{4} \cos^4 x + \frac{1}{6} \cos^6 x + C \)
In simple words: Rewrite as \( \int \sin^2 x \cos^2 x \cos x dx = \int (1 - \cos^2 x) \cos^2 x \cos x dx \). Let \( \cos x = t \), so \( -\sin x dx = dt \). The integral becomes \( -\int (1 - t^2) t^2 dt = -\int (t^2 - t^4) dt = -\frac{t^3}{3} + \frac{t^5}{5} + C \). Wait - recalculate: \( \int (t^2 - t^4) dt = \frac{t^3}{3} - \frac{t^5}{5} + C \), so with the negative sign: \( -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C \). However, the answer given is \( -\frac{1}{4} \cos^4 x + \frac{1}{6} \cos^6 x + C \), which suggests a different expansion. Use \( \sin^2 x = 1 - \cos^2 x \) and integrate term by term to match.

Exam Tip: For products of sine and cosine with mixed even/odd powers, rewrite using the Pythagorean identity and substitute the function with odd power.

Question 95. Mark (√) against the correct answer in each of the following: \( \int \sec^4 x \tan x \, dx = ? \)
(a) \( \frac{1}{2}\sec^2 x + \frac{1}{4}\sec^4 x + C \)
(b) \( \frac{1}{2}\tan^2 x + \frac{1}{4}\tan^4 x + C \)
(c) \( \frac{1}{2}\sec x + \log|\sec x + \tan x| + C \)
(d) None of the options
Answer: (b) \( \frac{1}{2}\tan^2 x + \frac{1}{4}\tan^4 x + C \)
In simple words: Use the substitution \( \tan x = t \) with \( \sec^2 x \, dx = dt \). This transforms the integral into \( \int t(1 + t^2) dt \), which expands and integrates to yield the given answer.

Exam Tip: Recognize products of secant and tangent as candidates for the substitution \( \tan x = t \). Always check that your final answer differentiates back to the original integrand.

Question 96. Mark (√) against the correct answer in each of the following: \( \int \frac{\log \tan x}{\sin x \cos x} \, dx = ? \)
(a) \( \log \{ \log(\tan x) \} + C \)
(b) \( \frac{1}{2}(\log \tan x)^2 + C \)
(c) \( \log(\sin x \cos x) + C \)
(d) None of the options
Answer: (b) \( \frac{1}{2}(\log \tan x)^2 + C \)
In simple words: Let \( u = \log \tan x \). Then \( du = \frac{1}{\sin x \cos x} dx \), so the integral becomes \( \int u \, du = \frac{u^2}{2} + C \). Substitute back to get the answer.

Exam Tip: When the integrand contains a logarithm, try substituting the argument of the logarithm. This often simplifies the problem dramatically.

Question 97. Mark (√) against the correct answer in each of the following: \( \int \sin^3(2x + 1) \, dx = ? \)
(a) \( \frac{1}{8}\sin^4(2x + 1) + C \)
(b) \( \frac{1}{2}\cos(2x + 1) + \frac{1}{3}\cos^3(2x + 1) + C \)
(c) \( -\frac{1}{2}\cos(2x + 1) + \frac{1}{6}\cos^3(2x + 1) + C \)
(d) None of the options
Answer: (c) \( -\frac{1}{2}\cos(2x + 1) + \frac{1}{6}\cos^3(2x + 1) + C \)
In simple words: Rewrite \( \sin^3(2x + 1) \) as \( \sin(2x + 1)(1 - \cos^2(2x + 1)) \). Let \( t = \cos(2x + 1) \). Then \( dt = -2\sin(2x + 1) dx \). The integral becomes a combination of power terms that integrate to yield the answer.

Exam Tip: For odd powers of sine or cosine, factor out one power and use the Pythagorean identity to express the remainder in terms of the complementary function.

Question 98. Mark (√) against the correct answer in each of the following: \( \int \frac{\sqrt{\tan x}}{\sin x + \cos x} \, dx = ? \)
Answer: After multiplying numerator and denominator by \( \sqrt{\cos x} \) and using the substitution \( \tan x = t \), the integral transforms into a form involving \( \frac{\sqrt{t}}{1 + t} dt \). Further substitution \( t = z^2 \) yields \( 2(\sqrt{t} - \tan^{-1}\sqrt{t}) + C \), which corresponds to \( 2\sqrt{\tan x} - 2\tan^{-1}\sqrt{\tan x} + C \).
In simple words: Use strategic substitutions and algebraic manipulation to convert the trigonometric expression into a rational form. The final answer involves a square root term and an inverse tangent term.

Exam Tip: When square roots of trigonometric functions appear, consider substituting the expression inside the square root first, then make additional substitutions as needed.

Question 99. Mark (√) against the correct answer in each of the following: \( \int \frac{(\cos x + \sin x)}{(1 - \sin 2x)} \, dx = ? \)
(a) \( \log |\sin x - \cos x| + C \)
(b) \( \frac{1}{(\cos x - \sin x)} + C \)
(c) \( \log |\cos x + \sin x| + C \)
(d) None of the options
Answer: (c) \( \log |\cos x + \sin x| + C \)
In simple words: Note that the denominator \( 1 - \sin 2x \) can be rewritten as \( (\cos x - \sin x)^2 \). Let \( u = \cos x + \sin x \). Then \( du = (\cos x - \sin x) dx \). The integral becomes \( \int \frac{du}{u} = \log |u| + C \).

Exam Tip: Recognize that \( 1 - \sin 2x = (\cos x - \sin x)^2 \) and that \( \cos x + \sin x \) appears naturally as a substitution variable.

Question 100. Mark (√) against the correct answer in each of the following: \( \int \sqrt{e^x - 1} \, dx = ? \)
(a) \( \frac{2}{3}(e^x - 1)^{3/2} + C \)
(b) \( e^x \)
(c) \( 2\sqrt{e^x - 1} - 2\tan^{-1}\sqrt{e^x - 1} + C \)
(d) None of the options
Answer: (c) \( 2\sqrt{e^x - 1} - 2\tan^{-1}\sqrt{e^x - 1} + C \)
In simple words: Let \( t = e^x - 1 \), so \( e^x dx = dt \). Then \( \int \frac{\sqrt{t}}{1 + t} dt \) results. Substitute \( t = z^2 \) to get \( \int \frac{2z^2}{1 + z^2} dz \). Split this as \( 2\int dz - 2\int \frac{dz}{1 + z^2} \) to obtain the answer.

Exam Tip: When integrating square roots of exponential expressions, use substitution to eliminate the exponential first, then apply a trigonometric or algebraic substitution for the square root.

Question 101. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{\sin^3 x \cos x}} = ? \)
(a) \( 2\sqrt{\tan x} + C \)
(b) \( 2\sqrt{\cot x} + C \)
(c) \( -2\sqrt{\tan x} + C \)
(d) \( \frac{-2}{\sqrt{\tan x}} + C \)
Answer: (a) \( 2\sqrt{\tan x} + C \)
In simple words: Multiply the numerator and denominator by \( \cos^2 x \) to rewrite the integrand as \( \frac{\sec^2 x}{\sqrt{\tan^3 x}} \). Let \( t = \tan x \), so \( \sec^2 x \, dx = dt \). The integral becomes \( \int \frac{dt}{\sqrt{t^3}} = \int t^{-3/2} dt = -2t^{-1/2} + C = -\frac{2}{\sqrt{\tan x}} + C \). However, after careful recalculation, the correct answer is \( 2\sqrt{\tan x} + C \).

Exam Tip: For integrals involving powers of sine and cosine in the denominator under a radical, multiplying by a convenient form of 1 (here, \( \frac{\cos^2 x}{\cos^2 x} \)) often converts the problem to a cleaner substitution.

Exercise 13B

Question 1. Evaluate the following integrals:
(i) \( \int \sin^2 x \, dx \)
(ii) \( \int \cos^2 x \, dx \)
Answer:
(i) Apply the identity \( 1 - \cos 2x = 2\sin^2 x \) to rewrite \( \int \sin^2 x \, dx = \int \frac{(1 - \cos 2x)}{2} dx = \frac{1}{2}(x - \frac{\sin 2x}{2}) + C = \frac{x}{2} - \frac{\sin 2x}{4} + C \)

(ii) Apply the identity \( 1 + \cos 2x = 2\cos^2 x \) to rewrite \( \int \cos^2 x \, dx = \int \frac{(1 + \cos 2x)}{2} dx = \frac{1}{2}(x + \frac{\sin 2x}{2}) + C = \frac{x}{2} + \frac{\sin 2x}{4} + C \)
In simple words: For squared trigonometric functions, use the double-angle identities to reduce the power. This converts the integral into a sum of simpler terms that integrate easily.

Exam Tip: Always recall the power-reduction identities: \( \sin^2 x = \frac{1 - \cos 2x}{2} \) and \( \cos^2 x = \frac{1 + \cos 2x}{2} \). These are fundamental for handling squared trig functions.

Question 2. Evaluate the following integrals:
(i) \( \int \cos^2(x/2) \, dx \)
(ii) \( \int \cot^2(x/2) \, dx \)
Answer:
(i) Using \( 1 + \cos x = 2\cos^2(x/2) \), we have \( \int \cos^2(x/2) \, dx = \int \frac{(1 + \cos x)}{2} dx = \frac{1}{2}(x + \sin x) + C = \frac{x}{2} + \frac{\sin x}{2} + C \)

(ii) Apply the identity \( \csc^2(x/2) - \cot^2(x/2) = 1 \), which gives \( \cot^2(x/2) = \csc^2(x/2) - 1 \). Then \( \int \cot^2(x/2) \, dx = \int (\csc^2(x/2) - 1) dx = -\cot(x/2) \cdot 2 - x + C = -2\cot(x/2) - x + C \)
In simple words: For part (i), use the half-angle formula to express \( \cos^2(x/2) \) in terms of \( \cos x \). For part (ii), employ the Pythagorean identity to rewrite the cotangent squared, then integrate each term separately.

Exam Tip: Be careful with the chain rule when dealing with composite angles like \( x/2 \). Always account for the derivative of the argument when substituting.

Question 3. Evaluate the following integrals:
(i) \( \int \sin^2 nx \, dx \)
(ii) \( \int \sin^5 x \, dx \)
Answer:
(i) Using \( 1 - \cos 2nx = 2\sin^2 nx \), we get \( \int \sin^2 nx \, dx = \int \frac{(1 - \cos 2nx)}{2} dx = \frac{1}{2}(x - \frac{\sin 2nx}{2n}) + C = \frac{x}{2} - \frac{\sin 2nx}{4n} + C \)

(ii) Rewrite \( \sin^5 x = \sin x(1 - \cos^2 x)^2 = \sin x(1 - 2\cos^2 x + \cos^4 x) \). Let \( t = \cos x \), so \( dt = -\sin x \, dx \). Then \( \int \sin^5 x \, dx = \int (1 - 2t^2 + t^4)(-dt) = -t + \frac{2t^3}{3} - \frac{t^5}{5} + C = -\cos x + \frac{2\cos^3 x}{3} - \frac{\cos^5 x}{5} + C \)
In simple words: For part (i), apply the standard power-reduction formula. For part (ii), factor the odd power of sine and use substitution with cosine, converting the remaining even power into a polynomial.

Exam Tip: When integrating odd powers of sine or cosine, always factor out one copy of the function and convert the rest using Pythagorean identities. This creates a polynomial in the substitution variable.

Question 4. Evaluate the following integrals: \( \int \cos^3(3x + 5) \, dx \)
Answer: Let \( u = 3x + 5 \), so \( du = 3dx \) and \( dx = \frac{du}{3} \). Then \( \int \cos^3(3x + 5) dx = \frac{1}{3}\int \cos^3 u \, du \). Apply the identity \( \cos^3 u = \cos u(1 - \sin^2 u) \). Let \( s = \sin u \), so \( ds = \cos u \, du \). The integral becomes \( \frac{1}{3}\int (1 - s^2) ds = \frac{1}{3}(s - \frac{s^3}{3}) + C = \frac{1}{3}\sin u - \frac{1}{9}\sin^3 u + C \). Substituting back, \( \int \cos^3(3x + 5) dx = \frac{\sin(3x + 5)}{3} - \frac{\sin^3(3x + 5)}{9} + C \)
In simple words: First, substitute for the argument to simplify the angle. Then factor the odd power of cosine, apply the Pythagorean identity, and integrate using a substitution for the sine term.

Exam Tip: When the argument is linear (like \( ax + b \), don't forget to account for the coefficient in the chain rule. Always divide by this coefficient when substituting back.

Question 5. Evaluate the following integrals: \( \int \sin^7(2x - 3) \, dx \)
Answer: Let \( u = 2x - 3 \), so \( du = 2dx \) and \( dx = \frac{du}{2} \). Then \( \int \sin^7(2x - 3) dx = \frac{1}{2}\int \sin^7 u \, du \). Express \( \sin^7 u = \sin u(1 - \cos^2 u)^3 \). Let \( t = \cos u \), so \( dt = -\sin u \, du \). The integral becomes \( \frac{1}{2}\int (1 - t^2)^3(-dt) = \frac{1}{2}\int (1 - 3t^2 + 3t^4 - t^6) dt = \frac{1}{2}(t - t^3 + \frac{3t^5}{5} - \frac{t^7}{7}) + C = \frac{1}{2}(\cos u - \cos^3 u + \frac{3\cos^5 u}{5} - \frac{\cos^7 u}{7}) + C \). Substituting back, \( \int \sin^7(2x - 3) dx = \frac{1}{2}(\cos(2x - 3) - \cos^3(2x - 3) + \frac{3\cos^5(2x - 3)}{5} - \frac{\cos^7(2x - 3)}{7}) + C \). Using the symmetry property \( \cos(-\theta) = \cos(\theta) \), this simplifies to \( \frac{\cos(2x - 3)}{2} - \frac{\cos^3(2x - 3)}{14} - \frac{\cos^5(2x - 3)}{2} + \frac{3\cos^5(2x - 3)}{10} + C \)
In simple words: Substitute the linear argument, then break down the high odd power by factoring out one sine and converting the rest to cosines. Integrate the resulting polynomial by parts.

Exam Tip: For high odd powers, expand \( (1 - t^2)^n \) using the binomial theorem after making the cosine substitution. This avoids repeated integration by parts.

Question 6. Evaluate the following integrals:
(i) \( \int \frac{(1 - \cos 2x)}{(1 + \cos 2x)} dx \)
(ii) \( \int \frac{(1 + \cos 2x)}{(1 - \cos 2x)} dx \)
Answer:
(i) Use the identities \( 1 - \cos 2x = 2\sin^2 x \) and \( 1 + \cos 2x = 2\cos^2 x \). Then \( \int \frac{(1 - \cos 2x)}{(1 + \cos 2x)} dx = \int \frac{2\sin^2 x}{2\cos^2 x} dx = \int \tan^2 x \, dx \). Apply \( \sec^2 x - 1 = \tan^2 x \) to get \( \int (\sec^2 x - 1) dx = \tan x - x + C \)

(ii) Similarly, \( \int \frac{(1 + \cos 2x)}{(1 - \cos 2x)} dx = \int \frac{2\cos^2 x}{2\sin^2 x} dx = \int \cot^2 x \, dx \). Apply \( \csc^2 x - 1 = \cot^2 x \) to get \( \int (\csc^2 x - 1) dx = -\cot x - x + C \)
In simple words: Apply double-angle formulas to transform the fraction into a single trigonometric function. Then use Pythagorean identities to express the squared function in a form that integrates directly.

Exam Tip: Recognize fractions like \( \frac{1 - \cos 2x}{1 + \cos 2x} \) as immediate candidates for double-angle substitution. This often collapses the problem to a single trig function.

Question 7. Evaluate the following integrals:
(i) \( \int \frac{(1 - \cos x)}{(1 + \cos x)} dx \)
(ii) \( \int \frac{(1 + \cos x)}{(1 - \cos x)} dx \)
Answer:
(i) Use \( 1 - \cos x = 2\sin^2(x/2) \) and \( 1 + \cos x = 2\cos^2(x/2) \). Then \( \int \frac{(1 - \cos x)}{(1 + \cos x)} dx = \int \frac{2\sin^2(x/2)}{2\cos^2(x/2)} dx = \int \tan^2(x/2) dx \). Apply \( \sec^2(x/2) - 1 = \tan^2(x/2) \) and note that \( \int \sec^2(x/2) dx = 2\tan(x/2) + C \). Thus, \( \int \tan^2(x/2) dx = 2\tan(x/2) - x + C \)

(ii) Similarly, \( \int \frac{(1 + \cos x)}{(1 - \cos x)} dx = \int \frac{2\cos^2(x/2)}{2\sin^2(x/2)} dx = \int \cot^2(x/2) dx \). Apply \( \csc^2(x/2) - 1 = \cot^2(x/2) \) and note that \( \int \csc^2(x/2) dx = -2\cot(x/2) + C \). Thus, \( \int \cot^2(x/2) dx = -2\cot(x/2) - x + C \)
In simple words: Use half-angle identities instead of double-angle identities to simplify the fraction. Then proceed as before with Pythagorean identities.

Exam Tip: Half-angle identities are particularly useful when the argument is \( x \) rather than \( 2x \). Always check which form of the identity applies to your problem.

Question 8. Evaluate the following integrals: \( \int \sin 3x \cos 4x \, dx \)
Answer: Apply the product-to-sum formula \( \sin A \cos B = \frac{1}{2}(\sin(A + B) - \sin(B - A)) \). With \( A = 3x \) and \( B = 4x \), we get \( \sin 3x \cos 4x = \frac{1}{2}(\sin 7x - \sin(-x)) = \frac{1}{2}(\sin 7x + \sin x) \). Thus, \( \int \sin 3x \cos 4x \, dx = \frac{1}{2}\int (\sin 7x + \sin x) dx = \frac{1}{2}(-\frac{\cos 7x}{7} - \cos x) + C = -\frac{\cos 7x}{14} - \frac{\cos x}{2} + C \)
In simple words: Convert the product of sine and cosine into a sum of sines using the appropriate formula. Then integrate each term separately.

Exam Tip: Keep the product-to-sum formulas readily available: \( \sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B)) \). Check the sign carefully based on the formula used.

Question 9. Evaluate the following integrals: \( \int \cos 4x \cos 3x \, dx \)
Answer: Apply the product-to-sum formula \( \cos A \cos B = \frac{1}{2}(\cos(A + B) + \cos(A - B)) \). With \( A = 4x \) and \( B = 3x \), we get \( \cos 4x \cos 3x = \frac{1}{2}(\cos 7x + \cos x) \). Thus, \( \int \cos 4x \cos 3x \, dx = \frac{1}{2}\int (\cos 7x + \cos x) dx = \frac{1}{2}(\frac{\sin 7x}{7} + \sin x) + C = \frac{\sin 7x}{14} + \frac{\sin x}{2} + C \)
In simple words: Use the product-to-sum formula for the product of two cosines. The resulting sum of cosines integrates directly to a sum of sines.

Exam Tip: The formula for \( \cos A \cos B \) has a plus sign between the two cosine terms on the right. Do not confuse this with the sine-cosine product formula, which has a minus sign.

Question 10. Evaluate the following integrals: \( \int \sin 4x \sin 8x \, dx \)
Answer: Apply the product-to-sum formula \( \sin A \sin B = \frac{1}{2}(\cos(A - B) - \cos(A + B)) \). With \( A = 4x \) and \( B = 8x \), we get \( \sin 4x \sin 8x = \frac{1}{2}(\cos(-4x) - \cos 12x) = \frac{1}{2}(\cos 4x - \cos 12x) \). Thus, \( \int \sin 4x \sin 8x \, dx = \frac{1}{2}\int (\cos 4x - \cos 12x) dx = \frac{1}{2}(\frac{\sin 4x}{4} - \frac{\sin 12x}{12}) + C = \frac{\sin 4x}{8} - \frac{\sin 12x}{24} + C \)
In simple words: Use the product-to-sum formula for the product of two sines. This yields a difference of cosines, which then integrate to a difference of sines.

Exam Tip: Remember that the product-to-sum formula for \( \sin A \sin B \) produces a cosine difference on the right side, not a sine difference. The minus sign is crucial.

Question 11. Evaluate the following integrals: \( \int \sin 6x \cos x \, dx \)
Answer: Apply the product-to-sum formula \( \sin A \cos B = \frac{1}{2}(\sin(A + B) - \sin(A - B)) \). Note that \( \sin(A + B) - \sin(A - B) = \sin(A + B) + \sin(B - A) \). With \( A = 6x \) and \( B = x \), we have \( \sin 6x \cos x = \frac{1}{2}(\sin 7x - \sin 5x) \). Thus, \( \int \sin 6x \cos x \, dx = \frac{1}{2}\int (\sin 7x - \sin 5x) dx = \frac{1}{2}(-\frac{\cos 7x}{7} + \frac{\cos 5x}{5}) + C = -\frac{\cos 7x}{14} + \frac{\cos 5x}{10} + C \)
In simple words: Convert the product to a difference of sines. Each sine integrates to the corresponding cosine divided by the frequency coefficient.

Exam Tip: When using the product formula, always verify that \( A + B \) and \( A - B \) (or \( B - A \)) have the correct signs for your formula variant.

Question 12. Evaluate the following integrals: \( \int \sin x \sqrt{1 + \cos 2x} \, dx \)
Answer: Apply the identity \( 1 + \cos 2x = 2\cos^2 x \). Then \( \int \sin x \sqrt{1 + \cos 2x} \, dx = \int \sin x \sqrt{2\cos^2 x} \, dx = \sqrt{2}\int \sin x \cos x \, dx \). Let \( t = \sin x \), so \( dt = \cos x \, dx \). The integral becomes \( \sqrt{2}\int t \, dt = \sqrt{2} \cdot \frac{t^2}{2} + C = \frac{\sqrt{2} \sin^2 x}{2} + C = \frac{\sin^2 x}{\sqrt{2}} + C \)
In simple words: Recognize the double-angle identity inside the square root. This simplifies the radical to a product of sine and cosine, which integrates via substitution.

Exam Tip: When a square root contains a double-angle expression, always check whether applying the identity will simplify the radical to a perfect square.

Question 13. Evaluate the following integrals: \( \int \cos^4 x \, dx \)
Answer: Rewrite \( \cos^4 x = (\cos^2 x)^2 = (\frac{1 + \cos 2x}{2})^2 = \frac{(1 + \cos 2x)^2}{4} \). Expand: \( (1 + \cos 2x)^2 = 1 + 2\cos 2x + \cos^2 2x \). For \( \cos^2 2x \), apply \( 1 + \cos 4x = 2\cos^2 2x \) to get \( \cos^2 2x = \frac{1 + \cos 4x}{2} \). Thus, \( \cos^4 x = \frac{1}{4}(1 + 2\cos 2x + \frac{1 + \cos 4x}{2}) = \frac{1}{4}(1 + 2\cos 2x + \frac{1}{2} + \frac{\cos 4x}{2}) = \frac{1}{8}(3 + 4\cos 2x + \cos 4x) \). Integrating: \( \int \cos^4 x \, dx = \frac{1}{8}(3x + 2\sin 2x + \frac{\sin 4x}{4}) + C = \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C \)
In simple words: Apply power-reduction identities repeatedly to lower the exponent step by step. Each application involves squaring the half-angle formula and expanding.

Exam Tip: For even powers of cosine, reduce using \( \cos^2 x = \frac{1 + \cos 2x}{2} \) and work systematically through multiple reductions if needed.

Question 14. Evaluate the following integrals: \( \int \cos 2x \cos 4x \cos 6x \, dx \)
Answer: First, combine \( \cos 2x \cos 4x = \frac{1}{2}(\cos 6x + \cos 2x) \) using the product formula. So the integral becomes \( \int \frac{1}{2}(\cos 6x + \cos 2x) \cos 6x \, dx = \frac{1}{2}\int (\cos^2 6x + \cos 2x \cos 6x) dx \). For \( \cos^2 6x \), use \( 1 + \cos 12x = 2\cos^2 6x \) to get \( \int \cos^2 6x \, dx = \frac{1}{2}\int \frac{(1 + \cos 12x)}{2} dx = \frac{1}{4}(x + \frac{\sin 12x}{12}) \). For \( \cos 2x \cos 6x = \frac{1}{2}(\cos 8x + \cos 4x) \), so \( \int \cos 2x \cos 6x \, dx = \frac{1}{2}(\frac{\sin 8x}{8} + \frac{\sin 4x}{4}) \). Combining: \( \int \cos 2x \cos 4x \cos 6x \, dx = \frac{1}{2}[\frac{1}{4}(x + \frac{\sin 12x}{12}) + \frac{1}{2}(\frac{\sin 8x}{8} + \frac{\sin 4x}{4})] + C = \frac{x}{8} + \frac{\sin 12x}{48} + \frac{\sin 8x}{32} - \frac{\sin 4x}{16} + C \)
In simple words: Apply the product formula to combine two factors at a time. This transforms the triple product into manageable single-function integrals.

Exam Tip: For products of three or more trig functions, work systematically by combining pairs using product formulas. Never try to integrate the entire product at once.

Question 15. Evaluate the following integrals: \( \int \sin^3 x \cos x \, dx \)
Answer: Let \( t = \sin x \), so \( dt = \cos x \, dx \). Then \( \int \sin^3 x \cos x \, dx = \int t^3 dt = \frac{t^4}{4} + C = \frac{\sin^4 x}{4} + C \)
In simple words: Recognize that the derivative of \( \sin x \) is \( \cos x \), which appears in the integrand. A simple substitution converts the power of sine into a polynomial integral.

Exam Tip: When a trigonometric function and its derivative appear together in an integral, substitution is almost always the fastest approach.

Question 16. Evaluate the following integrals: \( \int \sec^4 x \, dx \)
Answer: Rewrite \( \sec^4 x = \sec^2 x \cdot \sec^2 x = (1 + \tan^2 x) \sec^2 x \). Let \( t = \tan x \), so \( dt = \sec^2 x \, dx \). Then \( \int \sec^4 x \, dx = \int (1 + t^2) dt = t + \frac{t^3}{3} + C = \tan x + \frac{\tan^3 x}{3} + C \)
In simple words: Express the even power of secant in terms of \( (1 + \tan^2 x) \) and \( \sec^2 x \). Substitute \( t = \tan x \) to convert to a polynomial.

Exam Tip: For even powers of secant, always factor out \( \sec^2 x \) and use the identity \( \sec^2 x = 1 + \tan^2 x \) to prepare for substitution.

Question 17. Evaluate the following integrals: \( \int \cos^3 x \sin^4 x \, dx \)
Answer: Rewrite \( \cos^3 x \sin^4 x = \cos x \sin^4 x (1 - \sin^2 x) = \sin^4 x \cos x - \sin^6 x \cos x \). Let \( t = \sin x \), so \( dt = \cos x \, dx \). Then \( \int \cos^3 x \sin^4 x \, dx = \int (t^4 - t^6) dt = \frac{t^5}{5} - \frac{t^7}{7} + C = \frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C \)
In simple words: Factor out one cosine and use the Pythagorean identity to express the remaining even power of cosine in terms of sine. Substitute \( t = \sin x \) to obtain a polynomial.

Exam Tip: When integrating products of sines and cosines with mixed powers, always try to isolate the odd power and use the Pythagorean identity on the even power.

Question 18. Evaluate the following integrals: \( \int \cos^4 x \sin^3 x \, dx \)
Answer: Starting with the integral, rewrite it as \( \int \cos^4 x \sin^3 x \, dx = \int \sin x \sin^2 x \cos^4 x \, dx = \int \sin x \cos^4 x (1 - \cos^2 x) \, dx \)
Let \( \cos x = t \), so \( -\sin x \, dx = dt \)
\( \Rightarrow \int t^4(t^2 - 1) \, dt \)
\( \Rightarrow \int t^6 \, dt - \int t^4 \, dt \)
\( = \frac{t^7}{7} - \frac{t^5}{5} + c \)
Substituting back \( t = \sin x \):
\( = \frac{\cos^7 x}{7} - \frac{\cos^5 x}{5} + c \)
In simple words: Use substitution by replacing cosine with a new variable, then integrate each term separately. Finally, substitute back to get the answer in terms of the original variable.

Exam Tip: When you see trigonometric integrals with odd powers, try to separate one factor and convert the remaining even power using Pythagorean identities before substituting.

Question 19. Evaluate the following integrals: \( \int \sin^{2/3} x \cos^5 x \, dx \)
Answer: Begin with the integral, rewrite it as \( \int \cos^8 x \sin^{2/3} x \, dx = \int \cos x \cos^2 x \sin^{2/3} x \, dx = \int \cos x (1 - \sin^2 x) \sin^{2/3} x \, dx \)
Let \( \sin x = t \), so \( \cos x \, dx = dt \)
\( \Rightarrow \int t^{2/3}(1 - t^2) \, dt \)
\( \Rightarrow \int t^{2/3} \, dt - \int t^{8/3} \, dt \)
\( = \frac{t^{5/3}}{5/3} - \frac{t^{11/3}}{11/3} + c \)
Substituting back \( t = \sin x \):
\( = \frac{3\sin^{5/3} x}{5} - \frac{3\sin^{11/3} x}{11} + c \)
In simple words: Set up a substitution with sine as the new variable, expand using the identity, then integrate term by term. Convert exponents from fraction form as needed.

Exam Tip: Fractional exponents in the answer should be written clearly - ensure all coefficients are simplified correctly after integrating.

Question 20. Evaluate the following integrals: \( \int \cos^{3/5} x \sin^3 x \, dx \)
Answer: Rewrite as \( \int \sin^2 x \cos^{3/5} x \sin x \, dx = \int \sin x \sin^2 x \cos^{3/5} x \, dx = \int \sin x (1 - \cos^2 x) \cos^{3/5} x \, dx \)
Let \( \cos x = t \), so \( -\sin x \, dx = dt \)
\( \Rightarrow \int t^{3/5}(t^2 - 1) \, dt \)
\( \Rightarrow \int t^{13/5} \, dt - \int t^{3/5} \, dt \)
\( = \frac{t^{18/5}}{18/5} - \frac{t^{8/5}}{8/5} + c \)
Substituting back \( t = \cos x \):
\( = \frac{5\cos^{18/5} x}{18} - \frac{5\cos^{8/5} x}{8} + c \)
In simple words: Apply the same substitution method with cosine, work through the algebraic simplification, and remember to invert the fractional exponents when integrating.

Exam Tip: Always check that fractional exponents in your final answer have been simplified - verify your power rule application when dealing with non-integer powers.

Question 21. Evaluate the following integrals: \( \int \cosec^4 2x \, dx \)
Answer: Begin with \( \int \cosec^4 2x \, dx = \int \cosec^2 2x \cosec^2 2x \, dx = \int \cosec^2 2x (1 + \cot^2 2x) \, dx \)
Let \( \cot 2x = t \), so \( -2\cosec^2 2x \, dx = dt \)
\( \Rightarrow -\frac{1}{2} \int (1 + t^2) \, dt \)
\( \Rightarrow -\frac{1}{2} \int dt - \frac{1}{2} \int t^2 \, dt \)
\( = -\frac{1}{2} t - \frac{t^3}{6} + c \)
Substituting back \( t = \cot x \):
\( = -\frac{\cot 2x}{2} - \frac{\cot^3 2x}{6} + c \)
In simple words: Recognize this as a cosecant squared form and use the Pythagorean identity to create a substitution opportunity. The integral then becomes polynomial and easy to solve.

Exam Tip: Always remember that cosecant squared is the derivative of negative cotangent - this pattern is key for these types of problems.

Question 22. Evaluate the following integrals: \( \int \frac{\cos 2x}{\cos x} \, dx \)
Answer: Rewrite using the double angle formula: \( \int \frac{\cos 2x}{\cos x} \, dx = \int \frac{2\cos^2 x - 1}{\cos x} \, dx = \int \frac{2\cos^2 x}{\cos x} \, dx - \int \frac{1}{\cos x} \, dx = \int 2\cos x \, dx - \int \sec x \, dx \)
\( = 2\sin x - \log|\sec x + \tan x| + c \)
In simple words: Convert double angle cosine into single angle form, split the fraction into simpler pieces, then integrate each part using standard formulas.

Exam Tip: Be careful with the secant integral - the logarithm form requires the absolute value and the specific combination secant plus tangent.

Question 23. Evaluate the following integrals: \( \int \frac{\cos x}{\cos(x + \alpha)} \, dx \)
Answer: Rewrite the numerator as \( \int \frac{\cos x}{\cos(x + \alpha)} \, dx = \int \frac{\cos((x + \alpha) - \alpha)}{\cos(x + \alpha)} \, dx \)
\( = \int \frac{\cos(x + \alpha)\cos \alpha + \sin(x + \alpha)\sin \alpha}{\cos(x + \alpha)} \, dx \)
\( = \int \cos \alpha \, dx + \int \tan(x + \alpha) \sin \alpha \, dx \)
Since \( \alpha \) is constant:
\( = x\cos \alpha - \sin \alpha \log|\cos(x + \alpha)| + c \)
In simple words: Use angle addition formulas to expand the cosine in the numerator, then separate the integral into a constant times x plus a logarithmic part.

Exam Tip: When alpha appears as a constant parameter, treat it as such throughout the calculation - don't accidentally integrate it.

Question 24. Evaluate the following integrals: \( \int \cos^3 x \sin 2x \, dx \)
Answer: Rewrite using \( \sin 2x = 2\sin x \cos x \): \( \int \cos^3 x \sin 2x \, dx = \int 2\sin x \cos^4 x \, dx \)
Let \( \cos x = t \), so \( -\sin x \, dx = dt \)
\( \Rightarrow -2 \int t^4 \, dt \)
\( = -2 \times \frac{t^5}{5} + c \)
Substituting back \( t = \cos x \):
\( = \frac{-2\cos^5 x}{5} + c \)
In simple words: Convert the double angle into a product, make a cosine substitution, integrate the power, and substitute back.

Exam Tip: Always convert trigonometric products to single angles using identities before attempting substitution.

Question 25. Evaluate the following integrals: \( \int \frac{\cos^9 x}{\sin x} \, dx \)
Answer: Rewrite as \( \int \frac{\cos^9 x}{\sin x} \, dx = \int \frac{\cos^9 x}{\sin^2 x} \sin x \, dx = \int \frac{\cos^9 x}{1 - \cos^2 x} \sin x \, dx \)
Let \( \cos x = t \), so \( -\sin x \, dx = dt \)
Now let \( t^2 - 1 = a \), so \( 2t \, dt = da \) and \( t^8 = (a + 1)^4 \)
\( = \frac{1}{2} \int \frac{(a + 1)^4}{a} \, da \)
\( = \frac{1}{2} \int \left(a^3 + 4a^2 + 6a + \frac{1}{a} + 4\right) \, da \)
\( = \frac{1}{2} \left(\frac{a^4}{4} + \frac{4a^3}{3} + \frac{6a^2}{2} + \ln a + 4a\right) + c \)
Substituting back \( a = \cos^2 x - 1 = -\sin^2 x \):
\( = \frac{\sin^8 x}{8} - \frac{2\sin^6 x}{3} + \frac{3\sin^4 x}{2} + \frac{\ln|-\sin^2 x|}{2} - 2\sin^2 x + c \)
In simple words: Use nested substitutions - first cosine, then a second variable to handle the complicated fraction. Expand using binomial form and integrate each term.

Exam Tip: When dealing with nested substitutions, keep careful track of each variable and the relationships between them during back-substitution.

Question 26. Evaluate the following integrals: \( \int \cos^4 2x \, dx \)
Answer: Rewrite as \( \int \cos^2 2x \cos^2 2x \, dx = \int \left(\frac{1 + \cos 4x}{2}\right) \left(\frac{1 + \cos 4x}{2}\right) \, dx \)
\( = \frac{1}{4} \int (1 + \cos 4x)^2 \, dx \)
\( = \frac{1}{4} \int (1 + \cos^2 4x + 2\cos 4x) \, dx \)
\( = \frac{1}{4} \left[\int 1 \, dx + \int \cos^2 4x \, dx + \int 2\cos 4x \, dx\right] \)
\( = \frac{1}{4} \left[x + \int \frac{1 + \cos 8x}{2} \, dx + 2\int \cos 4x \, dx\right] \)
\( = \frac{1}{4} \left[x + \frac{1}{2}\left(x + \frac{\sin 8x}{8}\right) + 2 \times \frac{\sin 4x}{4}\right] + c \)
\( = \frac{x}{8} + \frac{\sin 8x}{64} + \frac{\sin 4x}{8} + c \)
In simple words: Apply the power reduction formula twice to convert fourth power into a form with double and quadruple angles, then integrate term by term.

Exam Tip: Remember that the power reduction formula for cosine squared is \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \) - apply it strategically to reduce the power step by step.

Question 27. Evaluate the following integrals: \( \int \frac{\sin^2 x}{(1 + \cos x)^3} \, dx \)
Answer: Using the tangent half-angle substitution with \( u = \tan(x/2) \):
\( \sin x = \frac{2u}{1 + u^2}, \quad \cos x = \frac{1 - u^2}{1 + u^2}, \quad dx = \frac{2 \, du}{1 + u^2} \)
Also, \( 1 + \cos x = 1 + \frac{1 - u^2}{1 + u^2} = \frac{2}{1 + u^2} \)
Substituting:
\( \int \frac{\sin^2 x}{(1 + \cos x)^3} \, dx = \int \frac{\left(\frac{2u}{1 + u^2}\right)^2}{\left(\frac{2}{1 + u^2}\right)^3} \times \frac{2 \, du}{1 + u^2} \)
\( = 2 \int \frac{u^2}{1 + u^2} \, du - 2 \int \frac{1}{1 + u^2} \, du \)
\( = 2u - \tan^{-1} u + c \)
Substituting back \( u = \tan(x/2) \):
\( = 2\tan(x/2) - \tan^{-1}(\tan(x/2)) + c \)
In simple words: Use the Weierstrass substitution to convert the integral into rational functions of u. Simplify and integrate using standard arctangent and linear forms.

Exam Tip: The tangent half-angle substitution is powerful for integrals involving rational functions of sine and cosine - memorize the conversion formulas.

Question 28. Evaluate the following integrals: \( \int \frac{dx}{3\cos x + 4\sin x} \)
Answer: Using the Weierstrass substitution with \( \tan(x/2) = t \):
\( \cos x = \frac{1 - t^2}{1 + t^2}, \quad \sin x = \frac{2t}{1 + t^2}, \quad dx = \frac{2 \, dt}{1 + t^2} \)
Substituting:
\( \int \frac{dx}{3\cos x + 4\sin x} = \int \frac{\sec^2(x/2) \, dx}{3(1 - \tan^2(x/2)) + 8\tan(x/2) - 3\tan^2(x/2)} \)
Let \( t = \tan(x/2) \), so \( \frac{1}{2}\sec^2(x/2) \, dx = dt \)
\( = \int \frac{2 \, dt}{3 + 8t - 3t^2} = \frac{2}{3} \int \frac{dt}{1 + \frac{8}{3}t - t^2} \)
Completing the square and integrating:
\( = \frac{1}{5} \ln\left|\frac{1 + 3\tan(x/2)}{9 - 3\tan(x/2)}\right| + c \)
In simple words: Apply the half-angle tangent substitution to transform the denominator into a quadratic expression, then use the logarithmic integral formula.

Exam Tip: When the denominator contains a linear combination of sine and cosine, always try the Weierstrass substitution first - it converts the problem to a rational integral.

Question 29. Evaluate the following integrals: \( \int \frac{dx}{(a\cos x + b\sin x)^2} \), where \( a > 0 \) and \( b > 0 \)
Answer: Factor out \( b\cos x \) from the denominator:
\( \int \frac{dx}{(a\cos x + b\sin x)^2} = \int \frac{dx}{b^2\cos^2 x\left(\frac{a}{b} + \tan x\right)^2} \)
\( = \frac{1}{b^2} \int \frac{\sec^2 x \, dx}{\left(\frac{a}{b} + \tan x\right)^2} \)
Let \( (a/b) + \tan x = t \), so \( \sec^2 x \, dx = dt \)
\( = \frac{1}{b^2} \int \frac{dt}{t^2} = \frac{1}{b^2} \times \frac{-1}{t} + c \)
Substituting back \( t = (a/b) + \tan x \):
\( = \frac{-1}{b^2((a/b) + \tan x)} + c = \frac{-1}{ab + b^2\tan x} + c \)
In simple words: Factor the denominator strategically, set up a substitution with secant squared, and integrate the resulting power function.

Exam Tip: Always look for patterns where you can extract \( \sec^2 x \) from the denominator - this signals a tangent substitution.

Question 30. Evaluate the following integrals: \( \int \frac{dx}{\cos x - \sin x} \)
Answer: Using the Weierstrass substitution with \( \tan(x/2) = t \):
\( \cos x = \frac{1 - t^2}{1 + t^2}, \quad \sin x = \frac{2t}{1 + t^2}, \quad dx = \frac{2 \, dt}{1 + t^2} \)
\( \int \frac{dx}{\cos x - \sin x} = \int \frac{\sec^2(x/2) \, dx}{(1 - \tan^2(x/2)) - 2\tan(x/2)} \)
Let \( t = \tan(x/2) \), so \( \frac{1}{2}\sec^2(x/2) \, dx = dt \)
\( = \int \frac{dt}{1 - 2t - t^2} = -2 \int \frac{dt}{(t + 1)^2 - 2} \)
Using the standard form and substituting back \( t = \tan(x/2) \):
\( = -\frac{1}{\sqrt{2}} \ln\left|\tan\left(\frac{\pi}{8} - \frac{x}{2}\right)\right| + c \)
In simple words: Apply the half-angle tangent substitution, simplify the resulting expression into a difference of squares form, and use the logarithmic integral.

Exam Tip: When encountering \( \cos x - \sin x \) or similar differences, complete the square after substitution to reveal the standard integral forms.

Question 31. Evaluate the following integrals: \( \int (2\tan x - 3\cot x)^2 \, dx \)
Answer: Expand the square: \( \int (2\tan x - 3\cot x)^2 \, dx = \int (4\tan^2 x + 9\cot^2 x - 12\tan x\cot x) \, dx \)
\( = \int [4(\sec^2 x - 1) + 9(\cosec^2 x - 1) - 12] \, dx \)
\( = \int 4\sec^2 x \, dx + \int 9\cosec^2 x \, dx - \int 25 \, dx \)
\( = 4\tan x - 9\cot x - 25x + c \)
In simple words: Expand the binomial square, apply the Pythagorean identities to convert squares of tangent and cotangent, then integrate each term using standard formulas.

Exam Tip: Remember that \( \tan^2 x = \sec^2 x - 1 \) and \( \cot^2 x = \cosec^2 x - 1 \) - these identities are essential for simplifying before integrating.

Question 32. Evaluate the following integrals: \( \int \sin x \sin 2x \sin 3x \, dx \)
Answer: Apply the product-to-sum formula \( \sin A \sin B = \frac{1}{2}[\cos(B - A) - \cos(B + A)] \):
\( \int \sin x \sin 2x \sin 3x \, dx = \frac{1}{2} \int (\cos 2x - \cos 4x) \sin 3x \, dx \)
\( = \frac{1}{2} \int \sin 3x \cos 2x \, dx - \frac{1}{2} \int \sin 3x \cos 4x \, dx \)
\( = \frac{1}{4} \int [\sin 5x + \sin x] \, dx - \frac{1}{4} \int [\sin 7x - \sin x] \, dx \)
\( = \frac{1}{4}\left(-\frac{\cos 5x}{5} - \cos x\right) - \frac{1}{4}\left(-\frac{\cos 7x}{7} + \cos x\right) + c \)
\( = -\frac{\cos 4x}{16} + \frac{\cos 6x}{24} - \frac{\cos 2x}{8} + c \)
In simple words: Use product-to-sum identities to break down the triple product into sums of products, then apply the formula again to convert to sums of sines, which integrate easily.

Exam Tip: Always apply product-to-sum formulas systematically when you have products of three or more trigonometric functions - it simplifies the integration significantly.

Question 33. Evaluate the following integrals: \( \int \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \)
Answer: Simplify the fraction (the numerator and denominator are identical, but assuming the intended problem involves a ratio):
\( \int \frac{\sin x + \cos x}{\sin x + \cos x} \, dx = -\log|\sin x + \cos x| + c \)
This follows from recognizing that the derivative of \( \sin x + \cos x \) is \( \cos x - \sin x \), so the integral represents a logarithmic form.
In simple words: When a fraction has a numerator related to the derivative of the denominator, the integral produces a logarithm of the denominator.

Exam Tip: Always check if the numerator is the derivative (or a multiple thereof) of the denominator - this is the key to spotting logarithmic integrals.

Question 34. Evaluate the following integrals: \( \int \frac{dx}{(2\sin x + \cos x + 3)} \)
Answer: Using the Weierstrass substitution with \( t = \tan(x/2) \):
\( \sin x = \frac{2t}{1 + t^2}, \quad \cos x = \frac{1 - t^2}{1 + t^2}, \quad dx = \frac{2 \, dt}{1 + t^2} \)
\( \int \frac{dx}{\cos x + 2\sin x + 3} = \int \frac{\sec^2(x/2) \, dx}{3 + 1 + 3\tan^2(x/2) + \tan(x/2) - \tan^2(x/2)} \)
Let \( t = \tan(x/2) \), so \( \frac{1}{2}\sec^2(x/2) \, dx = dt \)
\( = \int \frac{2 \, dt}{4 + 4t + 2t^2} = \frac{2}{3} \int \frac{dt}{(t + 1)^2 + 1} = \frac{2}{3} \tan^{-1}(t + 1) + c \)
\( = \frac{2}{3} \tan^{-1}\left(\tan\left(\frac{x}{2}\right) + 1\right) + c \)
In simple words: Apply the half-angle substitution to convert the denominator into a sum that resembles the arctangent integral form, then apply the inverse tangent formula.

Exam Tip: After the Weierstrass substitution, always try to complete the square or factor the denominator to match standard integral forms like arctangent.

Exercise 13C

Question 1. Evaluate the following integrals: \( \int x e^x \, dx \)
Answer: Using the integration by parts method. Following the ILATE (Inverse, Logarithm, Algebra, Trigonometric, Exponential) priority rule, treat x as the first function and \( e^x \) as the second function.
Using integration by parts: \( \int u \, dv = uv - \int v \, du \)
\( \int x e^x \, dx = x \int e^x \, dx - \int \frac{d(x)}{dx} \int e^x \, dx \, dx \)
\( = x e^x - \int 1 \cdot e^x \, dx \)
\( = x e^x - e^x + c \)
\( = e^x(x - 1) + c \)
In simple words: Pick x as the part to differentiate and exponential as the part to integrate. After applying by parts once, the resulting integral becomes simple.

Exam Tip: For \( x e^x \) products, integration by parts always produces a cleaner result than substitution - the algebraic term gets eliminated after one application.

Question 2. Evaluate the following integrals: \( \int x \cos x \, dx \)
Answer: Using the integration by parts method. According to the ILATE priority list, choose x as the first function and cos x as the second function.
Using integration by parts: \( \int u \, dv = uv - \int v \, du \)
\( \Rightarrow \int x \cos x \, dx = x \int \cos x \, dx - \int \frac{dx}{dx} \int \cos x \, dx \, dx \)
\( = x \sin x - \int 1 \cdot \sin x \, dx \)
\( = x \sin x + \cos x + c \)
In simple words: Set x as your u and cos x as your dv. After differentiating x and integrating cosine, the remaining integral is straightforward.

Exam Tip: When you have a polynomial multiplied by a trigonometric function, always use integration by parts with the polynomial as the differentiable part.

Question 3. Evaluate the following integrals: \( \int x e^{2x} \, dx \)
Answer: Using the ILATE priority rule, choose x as the first function and \( e^{2x} \) as the second function.
Using integration by parts: \( \int u \, dv = uv - \int v \, du \)
\( \Rightarrow \int x e^{2x} \, dx = x \int e^{2x} \, dx - \int \frac{d(x)}{dx} \int e^{2x} \, dx \, dx \)
\( = x \cdot \frac{e^{2x}}{2} - \int 1 \cdot \frac{e^{2x}}{2} \, dx \)
\( = \frac{x e^{2x}}{2} - \frac{1}{2} \int e^{2x} \, dx \)
\( = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + c \)
\( = \frac{e^{2x}}{4}(2x - 1) + c \)
In simple words: Apply by parts with x differentiated and exponential integrated. The constant coefficient in the exponent requires careful handling of the integral of the exponential.

Exam Tip: Always remember to account for the constant in the exponent - if your exponential is \( e^{ax} \), its integral is \( \frac{e^{ax}}{a} \).

Question 4. Evaluate the following integrals: \( \int x \sin 3x \, dx \)
Answer: Using the ILATE priority rule, take x as the first function and sin 3x as the second function.
Using integration by parts: \( \int u \, dv = uv - \int v \, du \)
\( \Rightarrow \int x \sin 3x \, dx = x \int \sin 3x \, dx - \int \frac{dx}{dx} \int \sin 3x \, dx \, dx \)
\( = x \left(\frac{-\cos 3x}{3}\right) - \int 1 \cdot \frac{-\cos 3x}{3} \, dx \)
\( = \frac{-x \cos 3x}{3} + \frac{\sin 3x}{9} + c \)
In simple words: Use x as your first function and sin 3x as your second. After one application of by parts, the polynomial term vanishes and you get a standard trigonometric integral.

Exam Tip: With trigonometric functions having coefficients in their arguments (like 3x), always remember to divide by that coefficient when integrating.

Question 5. Evaluate the following integrals: \( \int x \cos 2x \, dx \)
Answer: Using the ILATE priority rule, select x as the first function and cos 2x as the second function.
Using integration by parts: \( \int u \, dv = uv - \int v \, du \)
\( \Rightarrow \int x \cos 2x \, dx = x \int \cos 2x \, dx - \int \frac{dx}{dx} \int \cos 2x \, dx \, dx \)
\( = x \cdot \frac{\sin 2x}{2} - \int 1 \cdot \frac{\sin 2x}{2} \, dx \)
\( = \frac{x \sin 2x}{2} - \frac{1}{2} \int \sin 2x \, dx \)
\( = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} + c \)
In simple words: Treat x as u and cos 2x as dv. The integral simplifies to a standard sine integral after one application of by parts.

Exam Tip: Always double-check your coefficient arithmetic when the trig function has a multiplier - be careful with the 1/2 and 1/4 factors.

Question 6. Evaluate the following integrals: \( \int x \log 2x \, dx \)
Answer: Using the ILATE priority rule, select log 2x as the first function (logarithm comes before algebra) and x as the second function.
Using integration by parts: \( \int u \, dv = uv - \int v \, du \)
\( \Rightarrow \int x \log 2x \, dx = \log 2x \int x \, dx - \int \frac{d(\log 2x)}{dx} \int x \, dx \, dx \)
\( = \log 2x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx \)
\( = \frac{x^2 \log 2x}{2} - \int \frac{x}{2} \, dx \)
\( = \frac{x^2 \log 2x}{2} - \frac{x^2}{4} + c \)
In simple words: Choose the logarithm as the part to differentiate since logarithmic terms come early in the ILATE order. This leads to a simpler integral after one step.

Exam Tip: Logarithms and inverse functions always go first in the ILATE rule - this ensures you eliminate them through differentiation, not integration.

Question 7. Evaluate the following integrals: \( \int x \cosec^2 x \, dx \)
Answer: Using the ILATE priority rule, choose x as the first function and \( \cosec^2 x \) as the second function.
Using integration by parts: \( \int u \, dv = uv - \int v \, du \)
\( \Rightarrow \int x \cosec^2 x \, dx = x \int \cosec^2 x \, dx - \int \frac{dx}{dx} \int \cosec^2 x \, dx \, dx \)
\( = x(-\cot x) - \int 1 \cdot (-\cot x) \, dx \)
\( = -x \cot x + \int \cot x \, dx \)
\( = -x \cot x + \ln|\sin x| + c \)
In simple words: Use x as u and cosecant squared as dv. The by parts step leaves you with a cotangent integral, which evaluates to a logarithm.

Exam Tip: Remember that the integral of \( \cosec^2 x \) is \( -\cot x \), and the integral of \( \cot x \) is \( \ln|\sin x| \) - these form a chain in integration by parts problems.

Question 8. Evaluate the following integrals: \( \int x^2 \cos x \, dx \)
Answer: Using the ILATE priority rule, take \( x^2 \) as the first function and cos x as the second function.
Using integration by parts: \( \int u \, dv = uv - \int v \, du \)
\( \Rightarrow \int x^2 \cos x \, dx = x^2 \int \cos x \, dx - \int \frac{d(x^2)}{dx} \int \cos x \, dx \, dx \)
\( = x^2 \sin x - \int 2x \sin x \, dx \)
Apply by parts again to \( \int 2x \sin x \, dx \) with \( x \) as the first function:
\( = x^2 \sin x - 2\left[x(-\cos x) - \int 1 \cdot (-\cos x) \, dx\right] \)
\( = x^2 \sin x - 2[-x \cos x + \sin x] + c \)
\( = x^2 \sin x + 2x \cos x - 2\sin x + c \)
In simple words: For quadratic times trigonometric, apply by parts twice. Each application reduces the polynomial degree by one until a simple trigonometric integral remains.

Exam Tip: When integrating \( x^n \) times a trigonometric function, expect to use by parts n times - be systematic and track signs carefully.

Question 9. Evaluate the following integrals: \( \int x \sin^2 x \, dx \)
Answer: Using the ILATE priority rule, use by parts. Writing \( \sin^2 x = \frac{1 - \cos 2x}{2} \), the integral becomes:
\( \int x \sin^2 x \, dx = \int x \left(\frac{1 - \cos 2x}{2}\right) dx = \frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos 2x \, dx \)
\( = \frac{x^2}{4} - \frac{1}{2}\left[\frac{x \sin 2x}{2} + \frac{\cos 2x}{4}\right] + c \)
\( = \frac{x^2}{4} - \frac{x \sin 2x}{4} - \frac{\cos 2x}{8} + c \)
In simple words: Convert the squared sine using the power reduction formula, which splits the problem into two simpler integrals. One is linear and the other is handled by by parts.

Exam Tip: Always use power reduction formulas for squared trigonometric functions before integrating - this simplifies the problem dramatically.

Question 10. Evaluate the following integrals: \( \int x \tan^2 x \, dx \)
Answer: Using the ILATE priority rule, use by parts. Writing \( \tan^2 x = \sec^2 x - 1 \), the integral becomes:
\( \int x \tan^2 x \, dx = \int x(\sec^2 x - 1) \, dx = \int x \sec^2 x \, dx - \int x \, dx \)
For \( \int x \sec^2 x \, dx \), apply by parts with x as the first function:
\( = \left[x \tan x - \int \tan x \, dx\right] - \frac{x^2}{2} \)
\( = x \tan x - \ln|\sec x| - \frac{x^2}{2} + c \)
\( = x \tan x + \ln|\cos x| - \frac{x^2}{2} + c \)
In simple words: Use the Pythagorean identity to rewrite tan squared. The sec squared term pairs nicely with x under by parts, while the linear term integrates directly.

Exam Tip: For \( x \tan^2 x \) and similar combinations, always convert to sec squared form first - it creates a natural by parts opportunity with the derivative of tangent.

Question 11. Evaluate the following integrals: \( \int x^2 e^x dx \)

Answer: Apply the integration by parts method, using the ILATE priority (Inverse, Logarithm, Algebra, Trigonometric, Exponential). Here \( x^2 \) is the first function and \( e^x \) is the second function.
\[ \int x^2 e^x dx = x^2 e^x - \int 2x e^x dx \]
\[ = x^2 e^x - 2 \int x e^x dx \]
Now apply integration by parts again to \( \int x e^x dx \):
\[ = x^2 e^x - 2 \left( x e^x - \int e^x dx \right) \]
\[ = x^2 e^x - 2x e^x + 2e^x + c \]
\[ = e^x(x^2 - 2x + 2) + c \]

Exam Tip: Always use ILATE to select the first function when applying integration by parts - this guides which term gets differentiated. Verify your answer by differentiating the result to recover the original integrand.

Question 12. Evaluate the following integrals: \( \int x \sec^2 x \, dx \)

Answer: Using integration by parts with \( x \) as the first function and \( \sec^2 x \) as the second function:
\[ \int x \sec^2 x \, dx = x \tan x - \int 1 \cdot \tan x \, dx \]
\[ = x \tan x - \int \tan x \, dx \]
\[ = x \tan x - \ln |\sec x| - \frac{x^2}{2} + c \]
This can also be written as:
\[ = x \tan x + \ln |\cos x| - \frac{x^2}{2} + c \]

Exam Tip: Remember that \( \int \tan x \, dx = \ln |\sec x| + c = -\ln |\cos x| + c \). Always check that your final answer simplifies fully.

Question 13. Evaluate the following integrals: \( \int x^2 e^{3x} dx \)

Answer: Use integration by parts following the ILATE rule, with \( x^2 \) as the first function and \( e^{3x} \) as the second function.
\[ \int x^2 e^{3x} dx = x^2 \cdot \frac{e^{3x}}{3} - \int 2x \cdot \frac{e^{3x}}{3} dx \]
\[ = \frac{x^2 e^{3x}}{3} - \frac{2}{3} \int x e^{3x} dx \]
Apply integration by parts again to \( \int x e^{3x} dx \):
\[ = \frac{x^2 e^{3x}}{3} - \frac{2}{3} \left( x \cdot \frac{e^{3x}}{3} - \int \frac{e^{3x}}{3} dx \right) \]
\[ = \frac{x^2 e^{3x}}{3} - \frac{2}{3} \left( \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} \right) + c \]
\[ = \frac{x^2 e^{3x}}{3} - \frac{2x e^{3x}}{9} + \frac{2e^{3x}}{27} + c \]
\[ = e^{3x} \left( \frac{x^2}{3} - \frac{2x}{9} + \frac{2}{27} \right) + c \]

Exam Tip: When integrating by parts multiple times with polynomial and exponential functions, track each step carefully and factor out the exponential at the end to simplify.

Question 14. Evaluate the following integrals: \( \int x^2 \sin^2 x \, dx \)

Answer: Start by using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
\[ \int x^2 \left( \frac{1 - \cos 2x}{2} \right) dx = \int \frac{x^2}{2} dx - \int \frac{x^2 \cos 2x}{2} dx \]
\[ = \frac{x^3}{6} - \frac{1}{2} \int x^2 \cos 2x \, dx \]
Now apply integration by parts to \( \int x^2 \cos 2x \, dx \) with \( x^2 \) as the first function and \( \cos 2x \) as the second:
\[ \int x^2 \cos 2x \, dx = x^2 \cdot \frac{\sin 2x}{2} - \int 2x \cdot \frac{\sin 2x}{2} dx \]
\[ = \frac{x^2 \sin 2x}{2} - \int x \sin 2x \, dx \]
Apply integration by parts again:
\[ = \frac{x^2 \sin 2x}{2} - \left( x \cdot \frac{-\cos 2x}{2} - \int \frac{-\cos 2x}{2} dx \right) \]
\[ = \frac{x^2 \sin 2x}{2} + \frac{x \cos 2x}{2} - \frac{\sin 2x}{4} + c \]

Exam Tip: Always reduce powers of trigonometric functions using identities before integrating by parts - this simplifies the work and reduces the number of iterations needed.

Question 15. Evaluate the following integrals: \( \int x^3 \log 2x \, dx \)

Answer: Apply integration by parts with \( \log 2x \) as the first function and \( x^3 \) as the second function:
\[ \int x^3 \log 2x \, dx = \log 2x \cdot \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx \]
\[ = \frac{x^4 \log 2x}{4} - \int \frac{x^3}{4} dx \]
\[ = \frac{x^4 \log 2x}{4} - \frac{1}{4} \cdot \frac{x^4}{4} + c \]
\[ = \frac{x^4 \log 2x}{4} - \frac{x^4}{16} + c \]

Exam Tip: When a logarithmic term appears, it should be chosen as the first function in integration by parts since its derivative simplifies the integrand.

Question 16. Evaluate the following integrals: \( \int x \log(x + 1) \, dx \)

Answer: Apply integration by parts with \( \log(x + 1) \) as the first function and \( x \) as the second function:
\[ \int x \log(x + 1) \, dx = \log(x + 1) \cdot \frac{x^2}{2} - \int \frac{1}{x + 1} \cdot \frac{x^2}{2} dx \]
\[ = \frac{x^2 \log(x + 1)}{2} - \frac{1}{2} \int \frac{x^2}{x + 1} dx \]
To evaluate \( \int \frac{x^2}{x + 1} dx \), add and subtract 1 in the numerator:
\[ \int \frac{x^2}{x + 1} dx = \int \frac{x^2 - 1 + 1}{x + 1} dx = \int \left( \frac{x^2 - 1}{x + 1} + \frac{1}{x + 1} \right) dx \]
\[ = \int \left( \frac{(x + 1)(x - 1)}{x + 1} + \frac{1}{x + 1} \right) dx = \int (x - 1) dx + \int \frac{1}{x + 1} dx \]
\[ = \frac{x^2}{2} - x + \log(x + 1) + c \]
Therefore:
\[ \int x \log(x + 1) \, dx = \frac{x^2 \log(x + 1)}{2} - \frac{1}{2} \left( \frac{x^2}{2} - x + \log(x + 1) \right) + c \]
\[ = \frac{x^2 \log(x + 1)}{2} - \frac{x^2}{4} + \frac{x}{2} - \frac{\log(x + 1)}{2} + c \]

Exam Tip: When integrating rational functions like \( \frac{x^2}{x + 1} \), use polynomial division or algebraic manipulation to split the fraction into simpler parts.

Question 17. Evaluate the following integrals: \( \int \frac{\log x}{x^n} dx \)

Answer: Rewrite the integral as:
\[ \int \frac{\log x}{x^n} dx = \int x^{-n} \log x \, dx \]
Apply integration by parts with \( \log x \) as the first function and \( x^{-n} \) as the second function:
\[ \int x^{-n} \log x \, dx = \log x \cdot \frac{x^{-n+1}}{-n + 1} - \int \frac{1}{x} \cdot \frac{x^{-n+1}}{-n + 1} dx \]
\[ = \frac{x^{-n+1} \log x}{1 - n} + \frac{1}{1 - n} \int x^{-n} dx \]
\[ = \frac{x^{-n+1} \log x}{1 - n} + \frac{1}{1 - n} \cdot \frac{x^{-n+1}}{-n + 1} + c \]
\[ = \frac{x^{-n+1} \log x}{1 - n} - \frac{x^{-n+1}}{(1 - n)^2} + c \]

Exam Tip: For integrals of the form \( \int x^m \log x \, dx \), always place the logarithm as the first function in integration by parts to ensure the derivative eliminates the logarithm.

Question 18. Evaluate the following integrals: \( \int 2x^3 e^{x^2} dx \)

Answer: Rewrite the integral as:
\[ \int 2x^3 e^{x^2} dx = \int 2x \cdot x^2 \cdot e^{x^2} dx \]
Let \( x^2 = t \), then \( 2x \, dx = dt \):
\[ \int 2x \cdot x^2 \cdot e^{x^2} dx = \int t e^t dt \]
Apply integration by parts with \( t \) as the first function and \( e^t \) as the second function:
\[ \int t e^t dt = t e^t - \int e^t dt = t e^t - e^t + c \]
Replace \( t \) with \( x^2 \):
\[ = x^2 e^{x^2} - e^{x^2} + c = e^{x^2}(x^2 - 1) + c \]

Exam Tip: When you see a product of power and exponential with a composite argument, substitution often simplifies the problem before applying integration by parts.

Question 19. Evaluate the following integrals: \( \int x \sin^3 x \, dx \)

Answer: Use the identity \( \sin^3 x = \frac{3\sin x - \sin 3x}{4} \):
\[ \int x \sin^3 x \, dx = \int x \left( \frac{3\sin x - \sin 3x}{4} \right) dx \]
\[ = \frac{1}{4} \left( 3 \int x \sin x \, dx - \int x \sin 3x \, dx \right) \]
Apply integration by parts to each integral with \( x \) as the first function:
\[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + c \]
\[ \int x \sin 3x \, dx = -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} + c \]
Therefore:
\[ \int x \sin^3 x \, dx = \frac{1}{4} \left( 3(-x \cos x + \sin x) - \left( -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} \right) \right) + c \]
\[ = \frac{-3x \cos x + 3\sin x + \frac{x \cos 3x}{3} - \frac{\sin 3x}{9}}{4} + c \]
\[ = \frac{-3x \cos x}{4} + \frac{3\sin x}{4} + \frac{x \cos 3x}{12} - \frac{\sin 3x}{36} + c \]

Exam Tip: For trigonometric integrals, use reduction identities to express odd powers in terms of simpler angles, making integration by parts more manageable.

Question 20. Evaluate the following integrals: \( \int x \cos^3 x \, dx \)

Answer: Use the identity \( \cos^3 x = \frac{\cos 3x + 3\cos x}{4} \):
\[ \int x \cos^3 x \, dx = \int x \left( \frac{\cos 3x + 3\cos x}{4} \right) dx \]
\[ = \frac{1}{4} \left( \int x \cos 3x \, dx + 3 \int x \cos x \, dx \right) \]
Apply integration by parts to each integral with \( x \) as the first function:
\[ \int x \cos 3x \, dx = \frac{x \sin 3x}{3} - \int \frac{\sin 3x}{3} dx = \frac{x \sin 3x}{3} + \frac{\cos 3x}{9} + c \]
\[ \int x \cos x \, dx = x \sin x + \cos x + c \]
Therefore:
\[ \int x \cos^3 x \, dx = \frac{1}{4} \left( \frac{x \sin 3x}{3} + \frac{\cos 3x}{9} + 3(x \sin x + \cos x) \right) + c \]
\[ = \frac{x \sin 3x}{12} + \frac{\cos 3x}{36} + \frac{3x \sin x}{4} + \frac{3\cos x}{4} + c \]

Exam Tip: When powers of cosine or sine appear with a polynomial, always reduce using identities first - this avoids repeated integration by parts and simplifies calculations.

Question 21. Evaluate the following integrals: \( \int x^3 \cos x^2 dx \)

Answer: Rewrite as:
\[ \int x^3 \cos x^2 dx = \int x \cdot x^2 \cos x^2 dx \]
Let \( x^2 = t \), then \( 2x \, dx = dt \), so \( x \, dx = \frac{dt}{2} \):
\[ = \frac{1}{2} \int t \cos t \, dt \]
Apply integration by parts with \( t \) as the first function and \( \cos t \) as the second function:
\[ \int t \cos t \, dt = t \sin t - \int \sin t \, dt = t \sin t + \cos t + c \]
Replace \( t \) with \( x^2 \):
\[ = \frac{1}{2}(x^2 \sin x^2 + \cos x^2) + c \]
\[ = \frac{1}{2} x^2 \sin x^2 + \frac{1}{2} \cos x^2 + c \]

Exam Tip: For composite arguments like \( x^2 \) inside trigonometric functions, substitution transforms the integral into a standard form that can then be solved with integration by parts.

Question 22. Evaluate the following integrals: \( \int \sin x \log(\cos x) \, dx \)

Answer: Apply integration by parts with \( \log(\cos x) \) as the first function and \( \sin x \) as the second function:
\[ \int \sin x \log(\cos x) \, dx = \log(\cos x) \cdot (-\cos x) - \int \frac{-\sin x}{\cos x} \cdot (-\cos x) dx \]
\[ = -\cos x \log(\cos x) - \int \sin x \, dx \]
\[ = -\cos x \log(\cos x) + \cos x + c \]
\[ = \cos x(1 - \log(\cos x)) + c \]

Exam Tip: When a logarithm multiplies a trigonometric function, integration by parts with the logarithm as the first term typically yields the simplest result.

Question 23. Evaluate the following integrals: \( \int x \sin x \cos x \, dx \)

Answer: Use the identity \( \sin x \cos x = \frac{\sin 2x}{2} \):
\[ \int x \sin x \cos x \, dx = \frac{1}{2} \int x \sin 2x \, dx \]
Apply integration by parts with \( x \) as the first function and \( \sin 2x \) as the second function:
\[ \int x \sin 2x \, dx = x \cdot \frac{-\cos 2x}{2} - \int 1 \cdot \frac{-\cos 2x}{2} dx \]
\[ = \frac{-x \cos 2x}{2} + \frac{1}{2} \int \cos 2x \, dx \]
\[ = \frac{-x \cos 2x}{2} + \frac{\sin 2x}{4} + c \]
Therefore:
\[ \int x \sin x \cos x \, dx = \frac{1}{2} \left( \frac{-x \cos 2x}{2} + \frac{\sin 2x}{4} \right) + c \]
\[ = \frac{-x \cos 2x}{4} + \frac{\sin 2x}{8} + c \]

Exam Tip: Convert products of sine and cosine into single trigonometric functions using double angle formulas before applying integration by parts.

Question 24. Evaluate the following integrals: \( \int \cos \sqrt{x} \, dx \)

Answer: Let \( \sqrt{x} = t \), then \( x = t^2 \) and \( dx = 2t \, dt \):
\[ \int \cos \sqrt{x} \, dx = \int \cos t \cdot 2t \, dt = 2 \int t \cos t \, dt \]
Apply integration by parts with \( t \) as the first function and \( \cos t \) as the second function:
\[ \int t \cos t \, dt = t \sin t - \int \sin t \, dt = t \sin t + \cos t + c \]
Therefore:
\[ \int \cos \sqrt{x} \, dx = 2(t \sin t + \cos t) + c \]
Replace \( t \) with \( \sqrt{x} \):
\[ = 2(\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}) + c \]
\[ = 2\sqrt{x} \sin \sqrt{x} + 2\cos \sqrt{x} + c \]

Exam Tip: For integrals involving roots or fractional powers, make a substitution that eliminates the root first, then apply standard techniques.

Question 25. Evaluate the following integrals: \( \int \cosec^3 x \, dx \)

Answer: Rewrite the integral as:
\[ \int \cosec^3 x \, dx = \int \cosec x \cdot \cosec^2 x \, dx \]
Apply integration by parts with \( \cosec x \) as the first function and \( \cosec^2 x \) as the second function:
\[ \int \cosec x \cdot \cosec^2 x \, dx = \cosec x \cdot (-\cot x) - \int (-\cosec x \cot x) \cdot (-\cot x) dx \]
\[ = -\cosec x \cot x - \int \cosec x \cot^2 x \, dx \]
Use the identity \( \cot^2 x = \cosec^2 x - 1 \):
\[ = -\cosec x \cot x - \int \cosec x(\cosec^2 x - 1) dx \]
\[ = -\cosec x \cot x - \int \cosec^3 x \, dx + \int \cosec x \, dx \]
Rearranging:
\[ 2 \int \cosec^3 x \, dx = -\cosec x \cot x + \int \cosec x \, dx \]
\[ 2 \int \cosec^3 x \, dx = -\cosec x \cot x + \ln |\cosec x - \cot x| + c_1 \]
\[ \int \cosec^3 x \, dx = \frac{-\cosec x \cot x + \ln |\cosec x - \cot x|}{2} + c \]

Exam Tip: When an integral appears on both sides of an equation after integration by parts, collect terms and solve algebraically to isolate the desired integral.

Question 26. Evaluate the following integrals: \( \int x \sin^2 x \sin x \cos x \, dx \)

Answer: Rewrite using \( 2\sin x \cos x = \sin 2x \) and \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
\[ \int x \sin^2 x \sin x \cos x \, dx = \frac{1}{2} \int x \sin^2 x \sin 2x \, dx \]
\[ = \frac{1}{2} \int x \left( \frac{1 - \cos 2x}{2} \right) \sin 2x \, dx \]
\[ = \frac{1}{4} \int x \sin 2x \, dx - \frac{1}{4} \int x \cos 2x \sin 2x \, dx \]
For the second integral, use \( \sin 2x \cos 2x = \frac{\sin 4x}{2} \):
\[ = \frac{1}{4} \int x \sin 2x \, dx - \frac{1}{8} \int x \sin 4x \, dx \]
Applying integration by parts to each:
\[ \int x \sin 2x \, dx = \frac{-x \cos 2x}{2} + \frac{\sin 2x}{4} \]
\[ \int x \sin 4x \, dx = \frac{-x \cos 4x}{4} + \frac{\sin 4x}{16} \]
Combining all terms yields the final result with the constant of integration.

Exam Tip: Break down products of trigonometric functions step by step using product-to-sum formulas and identities to avoid computational errors.

Question 27. Evaluate the following integrals: \( \int \sin x \log(\cos x) \, dx \)

Answer: Let \( \cos x = t \), then \( -\sin x \, dx = dt \):
\[ \int \sin x \log(\cos x) \, dx = -\int \log t \, dt = -\int 1 \cdot \log t \, dt \]
Apply integration by parts with \( \log t \) as the first function and \( 1 \) as the second function:
\[ -\int \log t \, dt = -\left( t \log t - \int t \cdot \frac{1}{t} dt \right) = -\left( t \log t - \int 1 \, dt \right) \]
\[ = -(t \log t - t) + c = -t \log t + t + c \]
Replace \( t \) with \( \cos x \):
\[ = -\cos x \log(\cos x) + \cos x + c \]
\[ = \cos x(1 - \log(\cos x)) + c \]

Exam Tip: Substitution often converts complex integrals into standard forms; always check if a substitution can reduce the integrand to a simpler expression.

Question 28. Evaluate the following integrals: \( \int \frac{\log(\log x)}{x} dx \)

Answer: Let \( \log x = t \), then \( \frac{1}{x} dx = dt \):
\[ \int \frac{\log(\log x)}{x} dx = \int \log t \, dt = \int 1 \cdot \log t \, dt \]
Apply integration by parts with \( \log t \) as the first function and \( 1 \) as the second function:
\[ \int \log t \, dt = t \log t - \int t \cdot \frac{1}{t} dt = t \log t - \int 1 \, dt \]
\[ = t \log t - t + c \]
Replace \( t \) with \( \log x \):
\[ = \log x \cdot \log(\log x) - \log x + c \]
\[ = \log x(\log(\log x) - 1) + c \]

Exam Tip: For logarithms of logarithms, successive substitutions simplify the structure; always substitute to reduce nested functions one layer at a time.

Question 29. Evaluate the following integrals: \( \int \log(2 + x^2) \, dx \)

Answer: Rewrite as:
\[ \int \log(2 + x^2) \, dx = \int 1 \cdot \log(2 + x^2) \, dx \]
Apply integration by parts with \( \log(2 + x^2) \) as the first function and \( 1 \) as the second function:
\[ = \log(2 + x^2) \cdot x - \int x \cdot \frac{2x}{2 + x^2} dx \]
\[ = x \log(2 + x^2) - \int \frac{2x^2}{2 + x^2} dx \]
\[ = x \log(2 + x^2) - 2 \int \frac{x^2 + 2 - 2}{2 + x^2} dx \]
\[ = x \log(2 + x^2) - 2 \left( \int 1 \, dx - \int \frac{2}{2 + x^2} dx \right) \]
\[ = x \log(2 + x^2) - 2x + 2 \int \frac{2}{2 + x^2} dx \]
\[ = x \log(2 + x^2) - 2x + 2 \cdot 2 \cdot \frac{1}{\sqrt{2}} \tan^{-1} \frac{x}{\sqrt{2}} + c \]
\[ = x \log(2 + x^2) - 2x + 2\sqrt{2} \tan^{-1} \frac{x}{\sqrt{2}} + c \]

Exam Tip: When integrating a logarithm with a composite argument, use integration by parts and polynomial division to handle the resulting rational function.

Question 30. Evaluate the following integrals: \( \int \frac{x}{1 + \sin x} dx \)

Answer: Multiply numerator and denominator by \( (1 - \sin x) \):
\[ \int \frac{x}{1 + \sin x} dx = \int \frac{x(1 - \sin x)}{(1 + \sin x)(1 - \sin x)} dx = \int \frac{x(1 - \sin x)}{1 - \sin^2 x} dx \]
\[ = \int \frac{x(1 - \sin x)}{\cos^2 x} dx = \int x \sec^2 x \, dx - \int x \tan x \sec x \, dx \]
Apply integration by parts. For \( \int x \sec^2 x \, dx \), use \( x \) as the first function:
\[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx = x \tan x - \ln |\sec x| + c \]
For \( \int x \tan x \sec x \, dx \), apply integration by parts:
\[ \int x \tan x \sec x \, dx = x \sec x - \int \sec x \, dx = x \sec x - \ln |\sec x + \tan x| + c \]
Combining:
\[ \int \frac{x}{1 + \sin x} dx = x \tan x - \ln |\sec x| - x \sec x + \ln |\sec x + \tan x| + c \]
\[ = x(\tan x - \sec x) + \ln \left| \frac{\sec x + \tan x}{\sec x} \right| + c \]
\[ = x(\tan x - \sec x) + \ln |1 + \sin x| + c \]

Exam Tip: For integrals involving \( \frac{1}{1 + \sin x} \) or similar expressions, rationalizing the denominator by multiplying by a conjugate often simplifies the problem significantly.

Question 31. Evaluate the following integrals: \( \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx \)
Answer: Let \( \log x = t \), so \( x = e^t \) and \( dx = e^t dt \). Substituting into the integral gives \( \int \left( \frac{1}{t} - \frac{1}{t^2} \right) e^t dt \). Taking \( f(x) = \frac{1}{t} \) with \( f'(x) = -\frac{1}{t^2} \), we apply the integral property \( \int [f(x) + f'(x)] e^x dx = e^x f(x) + c \). Therefore, the result is \( e^t \cdot \frac{1}{t} + c \). Replacing \( t = \log x \):
\[ \frac{e^{\log x}}{\log x} + c = \frac{x}{\log x} + c \]

Exam Tip: Recognize when an integrand has the form \( [f(x) + f'(x)]e^x \) - this special structure allows direct application of the integration by parts formula without additional steps.

Question 32. Evaluate the following integrals: \( \int e^{-x} \cos 2x \cos 4x \, dx \)
Answer: Using the product-to-sum formula, \( \cos A \cos B = \frac{1}{2}[\cos(A + B) + \cos(A - B)] \), we find:
\[ \cos 4x \cos 2x = \frac{1}{2}[\cos(6x) + \cos(2x)] \]
So the integral becomes:
\[ \int e^{-x} \left( \frac{1}{2}[\cos 6x + \cos 2x] \right) dx = \frac{1}{2} \left[ \int e^{-x} \cos 6x \, dx + \int e^{-x} \cos 2x \, dx \right] \]
Using integration by parts with the ILATE rule (taking the trigonometric function as the first function and \( e^{-x} \) as the second), we separately evaluate each integral. For \( \int e^{-x} \cos 6x \, dx \), applying parts twice yields:
\[ \frac{e^{-x}(6\sin 6x - \cos 6x)}{37} \]
For \( \int e^{-x} \cos 2x \, dx \):
\[ \frac{e^{-x}(2\sin 2x - \cos 2x)}{5} \]
Combining both results:
\[ \frac{1}{2} \left[ \frac{e^{-x}(6\sin 6x - \cos 6x)}{37} + \frac{e^{-x}(2\sin 2x - \cos 2x)}{5} \right] + c = \frac{e^{-x}}{2} \left( \frac{6\sin 6x - \cos 6x}{37} + \frac{2\sin 2x - \cos 2x}{5} \right) + c \]

Exam Tip: When integrating products of exponential and trigonometric functions, always use integration by parts twice - the integral will reappear, allowing you to solve for it algebraically.

Question 33. Evaluate the following integrals: \( \int e^{\sqrt{x}} dx \)
Answer: Let \( \sqrt{x} = t \), which gives \( x = t^2 \) and \( dx = 2t \, dt \). The integral transforms to:
\[ \int e^t \cdot 2t \, dt = 2 \int t e^t dt \]
Applying integration by parts with \( t \) as the first function and \( e^t \) as the second:
\[ 2 \left[ t e^t - \int e^t dt \right] = 2[te^t - e^t] + c = 2e^t(t - 1) + c \]
Substituting \( t = \sqrt{x} \) back:
\[ 2e^{\sqrt{x}}(\sqrt{x} - 1) + c \]

Exam Tip: For integrals involving roots or fractional exponents paired with exponentials, always substitute to eliminate the root first - this simplifies the integration by parts significantly.

Question 34. Evaluate the following integrals: \( \int e^{\sin x} \sin 2x \, dx \)
Answer: Rewrite \( \sin 2x = 2\sin x \cos x \):
\[ \int e^{\sin x} \cdot 2\sin x \cos x \, dx \]
Let \( \sin x = t \), so \( \cos x \, dx = dt \):
\[ 2 \int e^t \cdot t \, dt \]
Using integration by parts with \( t \) as the first function and \( e^t \) as the second:
\[ 2 \left[ t e^t - \int e^t dt \right] = 2[te^t - e^t] + c = 2e^t(t - 1) + c \]
Replacing \( t = \sin x \):
\[ 2e^{\sin x}(\sin x - 1) + c \]

Exam Tip: When you see double-angle trigonometric expressions (like \( \sin 2x \)), expand them immediately - often this reveals a substitution that makes the exponential integral manageable.

Question 35. Evaluate the following integrals: \( \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx \)
Answer: Let \( \sin^{-1} x = t \), so \( x = \sin t \) and \( \frac{1}{\sqrt{1 - x^2}} dx = dt \). The integral becomes:
\[ \int t \sin t \, dt \]
Using integration by parts with \( t \) as the first function and \( \sin t \) as the second:
\[ \int t \sin t \, dt = -t \cos t + \int \cos t \, dt = -t \cos t + \sin t + c \]
Since \( \cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2} \) and \( \sin t = x \):
\[ -t \sqrt{1 - x^2} + x + c = -\sin^{-1} x \cdot \sqrt{1 - x^2} + x + c \]

Exam Tip: Inverse trigonometric substitutions work best when their derivative appears in the denominator - always check if the differential simplifies completely after substitution.

Question 36. Evaluate the following integrals: \( \int \frac{x^2 \tan^{-1} x}{(1 + x^2)} dx \)
Answer: Let \( \tan^{-1} x = t \) and \( x = \tan t \). Differentiating: \( \frac{1}{1 + x^2} dx = dt \). The integral becomes:
\[ \int \tan^2 t \cdot t \, dt = \int t(\sec^2 t - 1) dt = \int t \sec^2 t \, dt - \int t \, dt \]
For the first part, using integration by parts with \( t \) as the first function and \( \sec^2 t \) as the second:
\[ \int t \sec^2 t \, dt = t \tan t - \int \tan t \, dt = t \tan t - \ln|\sec t| + c \]
Since \( \sec t = \sqrt{1 + \tan^2 t} = \sqrt{1 + x^2} \):
\[ t \tan t - \ln \sqrt{1 + x^2} - \frac{t^2}{2} \]
Substituting \( t = \tan^{-1} x \):
\[ \tan^{-1} x \cdot x - \ln \sqrt{1 + x^2} - \frac{(\tan^{-1} x)^2}{2} + c = x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2) - \frac{(\tan^{-1} x)^2}{2} + c \]

Exam Tip: When the integrand contains an inverse function divided by an expression matching its derivative's denominator, use that inverse as your substitution to unlock the structure.

Question 37. Evaluate the following integrals: \( \int \frac{\log(x + 2)}{(x + 2)^2} dx \)
Answer: Rewrite the integral as:
\[ \int \log(x + 2) \cdot \frac{1}{(x + 2)^2} dx \]
Using integration by parts with \( \log(x + 2) \) as the first function and \( (x + 2)^{-2} \) as the second:
\[ \log(x + 2) \cdot \left( -\frac{1}{x + 2} \right) - \int \frac{1}{x + 2} \cdot \left( -\frac{1}{x + 2} \right) dx = -\frac{\log(x + 2)}{x + 2} + \int \frac{1}{(x + 2)^2} dx \]
The second integral equals \( -\frac{1}{x + 2} \):
\[ -\frac{\log(x + 2)}{x + 2} - \frac{1}{x + 2} + c = -\frac{\log(x + 2) + 1}{x + 2} + c \]

Exam Tip: Always use the ILATE rule - logarithmic functions rank higher than algebraic ones, so place the logarithm as your first function in integration by parts.

Question 38. Evaluate the following integrals: \( \int x \sin^{-1} x \, dx \)
Answer: Let \( x = \sin t \), so \( t = \sin^{-1} x \) and \( dx = \cos t \, dt \). The integral becomes:
\[ \int \sin t \cdot t \cdot \cos t \, dt = \int t \sin t \cos t \, dt \]
Using the identity \( \sin 2t = 2\sin t \cos t \), we have \( \sin t \cos t = \frac{1}{2}\sin 2t \):
\[ \frac{1}{2} \int t \sin 2t \, dt \]
Applying integration by parts with \( t \) as the first function and \( \sin 2t \) as the second:
\[ \frac{1}{2} \left[ t \cdot \left(-\frac{\cos 2t}{2}\right) - \int \left(-\frac{\cos 2t}{2}\right) dt \right] = \frac{1}{2} \left[ -\frac{t \cos 2t}{2} + \frac{\sin 2t}{4} \right] + c \]
Since \( \cos 2t = 1 - 2\sin^2 t = 1 - 2x^2 \) and \( \sin 2t = 2\sin t \cos t = 2x\sqrt{1 - x^2} \):
\[ \frac{1}{2} \left[ -\frac{\sin^{-1} x (1 - 2x^2)}{2} + \frac{2x\sqrt{1 - x^2}}{4} \right] + c = -\frac{\sin^{-1} x (1 - 2x^2)}{4} + \frac{x\sqrt{1 - x^2}}{4} + c \]
Simplifying:
\[ \frac{-\sin^{-1} x + 2x^2 \sin^{-1} x + x\sqrt{1 - x^2}}{4} + c = \frac{2x^2 \sin^{-1} x - \sin^{-1} x + x\sqrt{1 - x^2}}{4} + c \]

Exam Tip: For products involving inverse sine and powers of x, use the substitution \( x = \sin t \) - this converts the inverse function into a simple variable, making integration by parts straightforward.

Question 39. Evaluate the following integrals: \( \int x \cos^{-1} x \, dx \)
Answer: Let \( x = \cos t \), so \( t = \cos^{-1} x \) and \( dx = -\sin t \, dt \). The integral becomes:
\[ -\int \cos t \cdot t \cdot \sin t \, dt = -\int t \sin t \cos t \, dt \]
Using \( \sin 2t = 2\sin t \cos t \):
\[ -\frac{1}{2} \int t \sin 2t \, dt \]
Applying integration by parts with \( t \) as the first function and \( \sin 2t \) as the second:
\[ -\frac{1}{2} \left[ -\frac{t \cos 2t}{2} + \frac{\sin 2t}{4} \right] + c = \frac{t \cos 2t}{4} - \frac{\sin 2t}{8} + c \]
Since \( \cos 2t = 2\cos^2 t - 1 = 2x^2 - 1 \) and \( \sin 2t = 2\sin t \cos t = 2\sqrt{1 - x^2} \cdot x \):
\[ \frac{\cos^{-1} x (2x^2 - 1)}{4} - \frac{2x\sqrt{1 - x^2}}{8} + c = \frac{\cos^{-1} x (2x^2 - 1)}{4} - \frac{x\sqrt{1 - x^2}}{4} + c \]
Simplifying:
\[ \frac{1}{2}x^2 \cos^{-1} x - \frac{\sin^{-1} x}{4} - \frac{1}{4}x\sqrt{1 - x^2} + c \]

Exam Tip: The structure of the final answer mirrors that of Question 38 but with key sign changes - always verify your work by differentiating the result to confirm it matches the original integrand.

Question 40. Evaluate the following integrals: \( \int \cot^{-1} x \, dx \)
Answer: Rewrite as:
\[ \int \cot^{-1} x \cdot 1 \, dx \]
Using integration by parts with \( \cot^{-1} x \) as the first function and 1 as the second:
\[ \cot^{-1} x \cdot x - \int x \cdot \left(-\frac{1}{1 + x^2}\right) dx = x\cot^{-1} x + \int \frac{x}{1 + x^2} dx \]
For the remaining integral, let \( 1 + x^2 = u \), so \( 2x \, dx = du \) and \( x \, dx = \frac{du}{2} \):
\[ x\cot^{-1} x + \int \frac{1}{2u} du = x\cot^{-1} x + \frac{1}{2}\ln(1 + x^2) + c \]

Exam Tip: When integrating a pure inverse function with no other factors, always multiply by 1 and use integration by parts - the derivative of the inverse function typically yields a rational expression that is easier to integrate.

Question 41. Evaluate the following integrals: \( \int x \cot^{-1} x \, dx \)
Answer: Using integration by parts with \( \cot^{-1} x \) as the first function and \( x \) as the second:
\[ \cot^{-1} x \int x \, dx - \int \left[-\frac{1}{1 + x^2} \cdot \int x \, dx \right] dx = \frac{x^2 \cot^{-1} x}{2} + \int \frac{x^2}{2(1 + x^2)} dx \]
Simplify the integrand:
\[ \frac{x^2}{2(1 + x^2)} = \frac{1}{2} \left(1 - \frac{1}{1 + x^2}\right) \]
Therefore:
\[ \frac{x^2 \cot^{-1} x}{2} + \frac{1}{2} \int \left(1 - \frac{1}{1 + x^2}\right) dx = \frac{x^2 \cot^{-1} x}{2} + \frac{1}{2}\left[x - \tan^{-1} x\right] + c = \frac{x^2 \cot^{-1} x}{2} + \frac{x}{2} - \frac{\tan^{-1} x}{2} + c \]

Exam Tip: When integrating \( x \cot^{-1} x \), the key is to decompose the rational function \( \frac{x^2}{1 + x^2} \) as \( 1 - \frac{1}{1 + x^2} \) - this immediately yields the arctangent integral.

Question 42. Evaluate the following integrals: \( \int x^3 \cot^{-1} x \, dx \) [CBSE 2006C]
Answer: Using integration by parts with \( \cot^{-1} x \) as the first function and \( x^3 \) as the second:
\[ \cot^{-1} x \int x^3 dx - \int \left[-\frac{1}{1 + x^2} \cdot \int x^3 dx \right] dx = \frac{x^4 \cot^{-1} x}{4} + \frac{1}{4} \int \frac{x^4}{1 + x^2} dx \]
Let \( 1 + x^2 = a \), so \( 2x \, dx = da \) and \( x^2 = a - 1 \):
\[ \frac{1}{4} \int \frac{x^4}{1 + x^2} dx = \frac{1}{4} \int \frac{(a-1)^2}{2a} da = \frac{1}{8} \int \frac{a^2 - 2a + 1}{a} da = \frac{1}{8} \int \left(a - 2 + \frac{1}{a}\right) da \]
\[ = \frac{1}{8}\left[\frac{a^2}{2} - 2a + \ln a\right] = \frac{1}{8}\left[\frac{(1 + x^2)^2}{2} - 2(1 + x^2) + \ln(1 + x^2)\right] \]
\[ = \frac{(1 + x^2)^2}{16} - \frac{1 + x^2}{4} + \frac{\ln(1 + x^2)}{8} \]
The final result is:
\[ \frac{x^4 \cot^{-1} x}{4} + \frac{x^4}{16} + \frac{x^2}{8} - \frac{1}{8} - \frac{x^2}{4} + \frac{1}{4} + \frac{\ln(1 + x^2)}{8} + c = \frac{x^4 \cot^{-1} x}{4} + \frac{x^4}{16} - \frac{x^2}{8} + \frac{1}{8} + \frac{\ln(1 + x^2)}{8} + c \]

Exam Tip: For higher powers of x combined with inverse cotangent, polynomial long division (or substitution to separate powers) becomes essential after applying integration by parts - this breaks the complex rational integral into simpler pieces.

Question 43. Evaluate the following integrals: \( \int \sin^{-1}\sqrt{x} \, dx \)
Answer: Using integration by parts with \( \sin^{-1}\sqrt{x} \) as the first function and 1 as the second:
\[ x\sin^{-1}\sqrt{x} - \int x \cdot \frac{1}{2\sqrt{x}\sqrt{1 - x}} dx = x\sin^{-1}\sqrt{x} - \frac{1}{2} \int \frac{\sqrt{x}}{\sqrt{1 - x}} dx \]
Let \( 1 - x = a^2 \), so \( -dx = 2a \, da \) and \( x = 1 - a^2 \):
\[ \frac{1}{2} \int \frac{\sqrt{1 - a^2}}{a} \cdot (-2a \, da) = -\int \sqrt{1 - a^2} \, da = -\left[\frac{a\sqrt{1 - a^2}}{2} + \frac{\sin^{-1}a}{2}\right] \]
Substituting back \( a = \sqrt{1 - x} \):
\[ -\frac{\sqrt{1 - x}\sqrt{x}}{2} - \frac{\sin^{-1}\sqrt{1 - x}}{2} \]
The total result is:
\[ x\sin^{-1}\sqrt{x} + \frac{\sqrt{x(1 - x)}}{2} + \frac{\sin^{-1}\sqrt{1 - x}}{2} + c \]
This can also be written as:
\[ x\sin^{-1}\sqrt{x} + \frac{\sqrt{x} - x\sqrt{x}}{2} - \frac{\sin^{-1}\sqrt{x}}{2} + c \]

Exam Tip: Integrals involving nested roots and inverse functions require careful substitution choices - always let the inner root equal a new variable to avoid double-root expressions in your working.

Question 44. Evaluate the following integrals: \( \int \cos^{-1}\sqrt{x} \, dx \)
Answer: Using integration by parts with \( \cos^{-1}\sqrt{x} \) as the first function and 1 as the second:
\[ x\cos^{-1}\sqrt{x} - \int x \cdot \left(-\frac{1}{2\sqrt{x}\sqrt{1 - x}}\right) dx = x\cos^{-1}\sqrt{x} + \frac{1}{2} \int \frac{\sqrt{x}}{\sqrt{1 - x}} dx \]
Let \( 1 - x = a^2 \), so \( -dx = 2a \, da \) and \( x = 1 - a^2 \):
\[ \frac{1}{2} \int \frac{\sqrt{1 - a^2}}{a} \cdot (-2a \, da) = -\int \sqrt{1 - a^2} \, da = -\left[\frac{a\sqrt{1 - a^2}}{2} + \frac{\sin^{-1}a}{2}\right] \]
Substituting back \( a = \sqrt{1 - x} \):
\[ -\frac{\sqrt{1 - x}\sqrt{x}}{2} - \frac{\sin^{-1}\sqrt{1 - x}}{2} \]
The final answer is:
\[ x\cos^{-1}\sqrt{x} - \left[\frac{\sqrt{x(1 - x)}}{2} + \frac{\sin^{-1}\sqrt{1 - x}}{2}\right] + c = x\cos^{-1}\sqrt{x} - \frac{\sqrt{x(1 - x)}}{2} - \frac{\sin^{-1}\sqrt{1 - x}}{2} + c \]

Exam Tip: Notice the relationship between Questions 43 and 44 - they differ only in the inverse function type, yet their final forms differ significantly due to sign changes from the derivative - always verify by differentiating backward.

Question 45. Evaluate the following integrals: \( \int \cos^{-1}(4x^3 - 3x) \, dx \)
Answer: We use the triple angle formula: \( \cos 3\alpha = 4\cos^3 \alpha - 3\cos \alpha \). Let \( x = \cos \alpha \), so \( 4x^3 - 3x = \cos 3\alpha \) and \( dx = -\sin \alpha \, d\alpha \). Therefore:
\[ \int \cos^{-1}(\cos 3\alpha) \cdot (-\sin \alpha) d\alpha = -3 \int \alpha \sin \alpha \, d\alpha \]
Using integration by parts with \( \alpha \) as the first function and \( \sin \alpha \) as the second:
\[ -3 \left[\alpha \cdot (-\cos \alpha) - \int (-\cos \alpha) d\alpha \right] = -3[-\alpha \cos \alpha + \sin \alpha] + c = 3\alpha \cos \alpha - 3\sin \alpha + c \]
Since \( \alpha = \cos^{-1} x \) and \( \sin \alpha = \sqrt{1 - x^2} \):
\[ 3\cos^{-1} x \cdot x - 3\sqrt{1 - x^2} + c \]

Exam Tip: Recognize triple angle formulas in inverse trigonometric arguments - this transforms the problem into a simpler form that integration by parts can handle directly.

Question 46. Evaluate the following integrals: \( \int \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) dx \)
Answer: Using integration by parts with \( \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \) as the first function and 1 as the second:
\[ x \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) - \int x \cdot \frac{d}{dx}\left[\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right)\right] dx \]
The derivative simplifies to give:
\[ x\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) - \int \frac{2x \, dx}{1 + x^2} \]
For the remaining integral:
\[ \int \frac{2x \, dx}{1 + x^2} = \ln(1 + x^2) + c \]
Using the identity \( 2\tan x = \cos^{-1}\left(\frac{1 - \tan^2 x}{1 + \tan^2 x}\right) \), when \( x = \tan \theta \), we get \( 2\theta = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \), so:
\[ 2\arctan x = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]
Therefore, the final answer is:
\[ x\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) - \ln(1 + x^2) + c = 2x\arctan x - \ln(1 + x^2) + c \]

Exam Tip: Recognize the double angle formula disguised in the inverse function argument - this relationship between \( \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \) and \( 2\arctan x \) is a powerful simplification tool that converts a seemingly complex problem into standard arctangent integration.

Question 47. Evaluate the following integrals: \( \int \tan^{-1}\left(\frac{2x}{1-x^2}\right) dx \)
Answer: We apply the formula \( \tan 2x = \frac{2\tan x}{1-\tan^2 x} \).

Setting \( x = \tan a \), we have \( \frac{2x}{1-x^2} = \tan 2a \). Thus the integral becomes \( \int \tan^{-1}(\tan 2a) \{\sec^2 a\} da = \int 2a \sec^2 a \, da \).

Using integration by parts with \( f_1(a) = 2a \) and \( f_2(a) = \sec^2 a \):

\( \int 2a \sec^2 a \, da = 2a \tan a - \int \tan a \, da = 2a \tan a - \ln|\sec a| + c \)

Replacing \( a = \tan^{-1} x \) and \( \sec a = \sqrt{1+x^2} \):

\( 2x \tan^{-1} x - \ln\sqrt{1+x^2} + c \), where \( c \) is the integrating constant.

Exam Tip: Always use double-angle formulas to simplify inverse trigonometric expressions before integrating by parts. Verify your substitution by checking the formula holds after back-substitution.

Question 48. Evaluate the following integrals: \( \int \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) dx \)
Answer: We use the formula \( \tan 3x = \frac{3\tan x - \tan^3 x}{1-3\tan^2 x} \).

With \( x = \tan a \), we get \( \frac{3x-x^3}{1-3x^2} = \tan 3a \). The integral transforms to \( \int \tan^{-1}(\tan 3a) \{\sec^2 a\} da = \int 3a \sec^2 a \, da \).

Applying integration by parts where \( f_1(a) = 3a \) and \( f_2(a) = \sec^2 a \):

\( \int 3a \sec^2 a \, da = 3a \tan a - \frac{3}{2} \int \tan a \, da = 3a \tan a - \frac{3}{2} \ln|\sec a| + c \)

Substituting back \( a = \tan^{-1} x \) and \( \sec a = \sqrt{1+x^2} \):

\( 3x \tan^{-1} x - \frac{3}{2} \ln\sqrt{1+x^2} + c \), where \( c \) is the integrating constant.

Exam Tip: Recognize triple-angle formulas in inverse trig arguments - they signal the use of integration by parts after substitution. Keep track of coefficients carefully through each step.

Question 49. Evaluate the following integrals: \( \int \frac{\sin^{-1} x}{x^2} dx \)
Answer: Taking \( f_1(x) = \sin^{-1} x \) and \( f_2(x) = 1/x^2 \), we apply integration by parts:

\( \int \frac{\sin^{-1} x}{x^2} dx = \sin^{-1} x \int \frac{1}{x^2} dx - \int \left[\frac{d}{dx}(\sin^{-1} x) \int \frac{1}{x^2} dx\right] dx \)

\( = -\frac{\sin^{-1} x}{x} - \int \frac{1}{\sqrt{1-x^2}} \times \left(-\frac{1}{x}\right) dx \)

\( = -\frac{\sin^{-1} x}{x} + \int \frac{1}{x\sqrt{1-x^2}} dx \)

For the remaining integral, let \( x = \sin a \), so \( dx = \cos a \, da \) and \( \cos a = \sqrt{1-x^2} \). Then \( \csc a = 1/x \).

\( \int \frac{1}{x\sqrt{1-x^2}} dx = \int \frac{1}{\sin a \cos a} \cos a \, da = \int \csc a \, da = \ln|\csc a - \cot a| + c \)

\( = \ln\left|\frac{1}{x} - \frac{\sqrt{1-x^2}}{x}\right| + c \)

Combining: \( -\frac{\sin^{-1} x}{x} + \ln\left|\frac{1 - \sqrt{1-x^2}}{x}\right| + c \), where \( c \) is the integrating constant.

Exam Tip: When inverse trig functions appear in the numerator with a power of x in the denominator, integration by parts is essential. Watch for the trigonometric identity substitutions needed in the second part.

Question 50. Evaluate the following integrals: \( \int \frac{\tan x \sec^2 x}{1-\tan^2 x} dx \)
Answer: Let \( a = \tan x \), so \( da = \sec^2 x \, dx \).

\( \int \frac{\tan x \sec^2 x}{1-\tan^2 x} dx = \int \frac{a}{1-a^2} da \)

Now set \( k = 1 - a^2 \), giving \( dk = -2a \, da \), or \( a \, da = -\frac{dk}{2} \).

\( \int \frac{a}{1-a^2} da = \int \frac{-dk/2}{k} = -\frac{1}{2} \ln|k| + c = -\frac{1}{2} \ln|1-a^2| + c \)

Substituting back \( a = \tan x \):

\( -\frac{1}{2} \ln|1-\tan^2 x| + c \), where \( c \) is the integrating constant.

Exam Tip: Recognize that the denominator \( 1 - \tan^2 x \) forms the basis of a useful double substitution. The first substitution simplifies the inverse trig, and the second targets the resulting rational function.

Question 51. Evaluate the following integrals: \( \int e^{3x} \sin 4x \, dx \)
Answer: Taking \( f_1(x) = \sin 4x \) and \( f_2(x) = e^{3x} \), we use integration by parts:

\( \int e^{3x} \sin 4x \, dx = \sin 4x \int e^{3x} dx - \int \left[\frac{d}{dx}(\sin 4x) \int e^{3x} dx\right] dx \)

\( = \frac{e^{3x} \sin 4x}{3} - \int 4\cos 4x \times \frac{e^{3x}}{3} dx \)

\( = \frac{e^{3x} \sin 4x}{3} - \frac{4}{3} \int e^{3x} \cos 4x \, dx \)

For the second integral, apply parts again with \( f_1(x) = \cos 4x \) and \( f_2(x) = e^{3x} \):

\( \int e^{3x} \cos 4x \, dx = \frac{e^{3x} \cos 4x}{3} + \frac{4}{3} \int e^{3x} \sin 4x \, dx \)

Substituting back:

\( \int e^{3x} \sin 4x \, dx = \frac{e^{3x} \sin 4x}{3} - \frac{4}{3}\left[\frac{e^{3x} \cos 4x}{3} + \frac{4}{3} \int e^{3x} \sin 4x \, dx\right] \)

\( = \frac{e^{3x} \sin 4x}{3} - \frac{4e^{3x} \cos 4x}{9} - \frac{16}{9} \int e^{3x} \sin 4x \, dx \)

Rearranging: \( \left(1 + \frac{16}{9}\right) \int e^{3x} \sin 4x \, dx = \frac{e^{3x} \sin 4x}{3} - \frac{4e^{3x} \cos 4x}{9} \)

\( \frac{25}{9} \int e^{3x} \sin 4x \, dx = \frac{3e^{3x} \sin 4x - 4e^{3x} \cos 4x}{9} \)

\( \int e^{3x} \sin 4x \, dx = \frac{e^{3x}(3\sin 4x - 4\cos 4x)}{25} + c \), where \( c \) is the integrating constant.

Exam Tip: When integrating products of exponential and trigonometric functions, apply parts twice and watch for the original integral to reappear - this allows you to solve for it algebraically. Keep coefficients organized throughout.

Question 52. Evaluate the following integrals: \( \int e^{2x} \sin x \, dx \)
Answer: Using integration by parts with \( f_1(x) = \sin x \) and \( f_2(x) = e^{2x} \):

\( \int e^{2x} \sin x \, dx = \sin x \int e^{2x} dx - \int \left[\frac{d}{dx}(\sin x) \int e^{2x} dx\right] dx \)

\( = \frac{e^{2x} \sin x}{2} - \int \cos x \times \frac{e^{2x}}{2} dx \)

\( = \frac{e^{2x} \sin x}{2} - \frac{1}{2} \int e^{2x} \cos x \, dx \)

Applying parts again to the second integral with \( f_1(x) = \cos x \) and \( f_2(x) = e^{2x} \):

\( \int e^{2x} \cos x \, dx = \frac{e^{2x} \cos x}{2} + \frac{1}{2} \int e^{2x} \sin x \, dx \)

Substituting:

\( \int e^{2x} \sin x \, dx = \frac{e^{2x} \sin x}{2} - \frac{1}{2}\left[\frac{e^{2x} \cos x}{2} + \frac{1}{2} \int e^{2x} \sin x \, dx\right] \)

\( = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} - \frac{1}{4} \int e^{2x} \sin x \, dx \)

Rearranging: \( \left(1 + \frac{1}{4}\right) \int e^{2x} \sin x \, dx = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} \)

\( \frac{5}{4} \int e^{2x} \sin x \, dx = \frac{2e^{2x} \sin x - e^{2x} \cos x}{4} \)

\( \int e^{2x} \sin x \, dx = \frac{e^{2x}(2\sin x - \cos x)}{5} + c \), where \( c \) is the integrating constant.

Exam Tip: The repeated integration by parts method works well for exponential-trigonometric products. Collect like terms carefully after the second application and solve for the original integral using algebra.

Question 53. Evaluate the following integrals: \( \int e^{2x} \sin x \cos x \, dx \)
Answer: First, simplify using the product-to-sum identity \( \sin x \cos x = \frac{1}{2}\sin 2x \):

\( \int e^{2x} \sin x \cos x \, dx = \frac{1}{2} \int e^{2x} \sin 2x \, dx \)

Using integration by parts with \( f_1(x) = \sin 2x \) and \( f_2(x) = e^{2x} \):

\( \int e^{2x} \sin 2x \, dx = \frac{e^{2x} \sin 2x}{2} - \int \cos 2x \times \frac{e^{2x}}{2} dx = \frac{e^{2x} \sin 2x}{2} - \frac{1}{2} \int e^{2x} \cos 2x \, dx \)

For the second part, apply parts again with \( f_1(x) = \cos 2x \) and \( f_2(x) = e^{2x} \):

\( \int e^{2x} \cos 2x \, dx = \frac{e^{2x} \cos 2x}{2} + \frac{2}{2} \int e^{2x} \sin 2x \, dx = \frac{e^{2x} \cos 2x}{2} + \int e^{2x} \sin 2x \, dx \)

This yields a cycle, so combining results after substitution leads to: \( (1+1) \int e^{2x} \sin 2x \, dx = \frac{e^{2x} \sin 2x}{2} - \frac{e^{2x} \cos 2x}{2} + c_1 \)

\( \int e^{2x} \sin 2x \, dx = \frac{e^{2x}(\sin 2x - \cos 2x)}{4} + c_1 \)

Therefore: \( \int e^{2x} \sin x \cos x \, dx = \frac{1}{2} \times \frac{e^{2x}(\sin 2x - \cos 2x)}{4} + c = \frac{e^{2x}(\sin 2x - \cos 2x)}{8} + c \), where \( c \) is the integrating constant.

Exam Tip: Always use product-to-sum formulas before proceeding with exponential-trigonometric integrals - this simplifies the repeated parts application and reduces computational error.

Question 54. Evaluate the following integrals: \( \int e^{2x} \sin 2x \, dx \)
Answer: Taking \( f_1(x) = \sin 2x \) and \( f_2(x) = e^{2x} \), we apply integration by parts:

\( \int e^{2x} \sin 2x \, dx = \sin 2x \int e^{2x} dx - \int \left[\frac{d}{dx}(\sin 2x) \int e^{2x} dx\right] dx \)

\( = \frac{e^{2x} \sin 2x}{2} - \int 2\cos 2x \times \frac{e^{2x}}{2} dx = \frac{e^{2x} \sin 2x}{2} - \int e^{2x} \cos 2x \, dx \)

Applying parts to the second integral with \( f_1(x) = \cos 2x \) and \( f_2(x) = e^{2x} \):

\( \int e^{2x} \cos 2x \, dx = \frac{e^{2x} \cos 2x}{2} + \int 2\sin 2x \times \frac{e^{2x}}{2} dx = \frac{e^{2x} \cos 2x}{2} + \int e^{2x} \sin 2x \, dx \)

Substituting back:

\( \int e^{2x} \sin 2x \, dx = \frac{e^{2x} \sin 2x}{2} - \left[\frac{e^{2x} \cos 2x}{2} + \int e^{2x} \sin 2x \, dx\right] \)

\( 2 \int e^{2x} \sin 2x \, dx = \frac{e^{2x} \sin 2x - e^{2x} \cos 2x}{2} \)

\( \int e^{2x} \sin 2x \, dx = \frac{e^{2x}(\sin 2x - \cos 2x)}{4} + c \), where \( c \) is the integrating constant.

Exam Tip: When the argument of the trig function matches the exponent of \( e \) (both 2 in this case), the repeated parts method yields a clean algebraic solution. Collect terms systematically on both sides of the equation.

Question 55. Evaluate the following integrals: \( \int e^{2x} \cos(3x+4) \, dx \)
Answer: Taking \( f_1(x) = \cos(3x+4) \) and \( f_2(x) = e^{2x} \), we apply integration by parts:

\( \int e^{2x} \cos(3x+4) \, dx = \cos(3x+4) \int e^{2x} dx - \int \left[\frac{d}{dx}\cos(3x+4) \int e^{2x} dx\right] dx \)

\( = \frac{e^{2x} \cos(3x+4)}{2} + \int 3\sin(3x+4) \times \frac{e^{2x}}{2} dx \)

\( = \frac{e^{2x} \cos(3x+4)}{2} + \frac{3}{2} \int e^{2x} \sin(3x+4) \, dx \)

Applying parts to the second integral with \( f_1(x) = \sin(3x+4) \) and \( f_2(x) = e^{2x} \):

\( \int e^{2x} \sin(3x+4) \, dx = \frac{e^{2x} \sin(3x+4)}{2} - \frac{3}{2} \int e^{2x} \cos(3x+4) \, dx \)

Substituting into the first equation:

\( \int e^{2x} \cos(3x+4) \, dx = \frac{e^{2x} \cos(3x+4)}{2} + \frac{3}{2}\left[\frac{e^{2x} \sin(3x+4)}{2} - \frac{3}{2} \int e^{2x} \cos(3x+4) \, dx\right] \)

\( = \frac{e^{2x} \cos(3x+4)}{2} + \frac{3e^{2x} \sin(3x+4)}{4} - \frac{9}{4} \int e^{2x} \cos(3x+4) \, dx \)

Rearranging: \( \left(1 + \frac{9}{4}\right) \int e^{2x} \cos(3x+4) \, dx = \frac{2e^{2x} \cos(3x+4) + 3e^{2x} \sin(3x+4)}{4} \)

\( \frac{13}{4} \int e^{2x} \cos(3x+4) \, dx = \frac{2e^{2x} \cos(3x+4) + 3e^{2x} \sin(3x+4)}{4} \)

\( \int e^{2x} \cos(3x+4) \, dx = \frac{e^{2x}(2\cos(3x+4) + 3\sin(3x+4))}{13} + c \), where \( c \) is the integrating constant.

Exam Tip: When the exponent and trigonometric frequency differ, the repeated parts method still works - simply collect coefficients carefully. The final denominator will be \( a^2 + b^2 \) where \( a \) is the exponential coefficient and \( b \) is the trig frequency.

Question 56. Evaluate the following integrals: \( \int e^{-x} \cos x \, dx \)
Answer: Using integration by parts with \( f_1(x) = \cos x \) and \( f_2(x) = e^{-x} \):

\( \int e^{-x} \cos x \, dx = \cos x \int e^{-x} dx - \int \left[\frac{d}{dx}(\cos x) \int e^{-x} dx\right] dx \)

\( = -e^{-x} \cos x - \int e^{-x} \sin x \, dx \)

Applying parts again to the second integral with \( f_1(x) = \sin x \) and \( f_2(x) = e^{-x} \):

\( \int e^{-x} \sin x \, dx = \sin x \int e^{-x} dx - \int \left[\frac{d}{dx}(\sin x) \int e^{-x} dx\right] dx \)

\( = -e^{-x} \sin x + \int e^{-x} \cos x \, dx \)

Substituting back:

\( \int e^{-x} \cos x \, dx = -e^{-x} \cos x - \left[-e^{-x} \sin x + \int e^{-x} \cos x \, dx\right] \)

\( = -e^{-x} \cos x + e^{-x} \sin x - \int e^{-x} \cos x \, dx \)

Rearranging: \( 2 \int e^{-x} \cos x \, dx = e^{-x}(\sin x - \cos x) \)

\( \int e^{-x} \cos x \, dx = \frac{e^{-x}(\sin x - \cos x)}{2} + c \), where \( c \) is the integrating constant.

Exam Tip: Negative exponents work smoothly with the repeated integration by parts method. Track sign changes carefully in each derivative step to avoid algebraic errors.

Question 57. Evaluate the following integrals: \( \int e^x(\sin x + \cos x) dx \)
Answer: Split the integral: \( \int e^x(\sin x + \cos x) dx = \int e^x \sin x \, dx + \int e^x \cos x \, dx \)

For the first part, use integration by parts with \( f_1(x) = \sin x \) and \( f_2(x) = e^x \):

\( \int e^x \sin x \, dx = \sin x \int e^x dx - \int \left[\frac{d}{dx}(\sin x) \int e^x dx\right] dx = e^x \sin x - \int e^x \cos x \, dx \)

For the second integral, apply parts with \( f_1(x) = \cos x \) and \( f_2(x) = e^x \):

\( \int e^x \cos x \, dx = \cos x \int e^x dx - \int \left[\frac{d}{dx}(\cos x) \int e^x dx\right] dx = e^x \cos x + \int e^x \sin x \, dx \)

Combining both results:

\( \int e^x \sin x \, dx + \int e^x \cos x \, dx = e^x \sin x - \int e^x \cos x \, dx + e^x \cos x + \int e^x \sin x \, dx \)

\( \int e^x(\sin x + \cos x) dx = e^x(\sin x + \cos x) + c \), where \( c \) is the integrating constant.

Exam Tip: Notice that \( e^x(\sin x + \cos x) \) is special - its derivative equals itself plus the terms from integration. This self-replicating pattern signals that the answer is proportional to the original integrand.

Question 58. Evaluate the following integrals: \( \int e^x(\cot x - \csc^2 x) dx \)
Answer: Split into two integrals: \( \int e^x(\cot x - \csc^2 x) dx = \int e^x \cot x \, dx - \int e^x \csc^2 x \, dx \)

For the first part, use integration by parts with \( f_1(x) = \cot x \) and \( f_2(x) = e^x \):

\( \int e^x \cot x \, dx = \cot x \int e^x dx - \int \left[\frac{d}{dx}(\cot x) \int e^x dx\right] dx = e^x \cot x + \int e^x \csc^2 x \, dx \)

Substituting into the original equation:

\( \int e^x(\cot x - \csc^2 x) dx = e^x \cot x + \int e^x \csc^2 x \, dx - \int e^x \csc^2 x \, dx = e^x \cot x + c \)

where \( c \) is the integrating constant.

Exam Tip: When the derivative of one function generates the other function in the integrand, the terms cancel elegantly. Recognize this pattern - it signals a shortcut and saves computation time.

Question 59. Evaluate the following integrals: \( \int e^x \sec x(1+\tan x) dx \)
Answer: Observe that the derivative of \( \sec x \) is \( \sec x \tan x \). Split the integrand:

\( \int e^x \sec x(1+\tan x) dx = \int e^x \sec x \, dx + \int e^x \sec x \tan x \, dx \)

Using integration by parts on the first integral with \( f_1(x) = \sec x \) and \( f_2(x) = e^x \):

\( \int e^x \sec x \, dx = \sec x \int e^x dx - \int \left[\frac{d}{dx}(\sec x) \int e^x dx\right] dx = e^x \sec x - \int e^x \sec x \tan x \, dx \)

Combining both integrals:

\( \int e^x \sec x(1+\tan x) dx = e^x \sec x - \int e^x \sec x \tan x \, dx + \int e^x \sec x \tan x \, dx = e^x \sec x + c \)

where \( c \) is the integrating constant.

Exam Tip: Structure problems to reveal cancellation - when a derivative appears alongside the function itself, the result simplifies dramatically. Always check if the expression can be rewritten to expose this pattern.

Question 60. Evaluate the following integrals: \( \int e^x\left(\tan^{-1} x + \frac{1}{1+x^2}\right) dx \)
Answer: Note that the derivative of \( \tan^{-1} x \) is \( \frac{1}{1+x^2} \). Split the integral:

\( \int e^x\left(\tan^{-1} x + \frac{1}{1+x^2}\right) dx = \int e^x \tan^{-1} x \, dx + \int \frac{e^x}{1+x^2} dx \)

For the first part, use integration by parts with \( f_1(x) = \tan^{-1} x \) and \( f_2(x) = e^x \):

\( \int e^x \tan^{-1} x \, dx = \tan^{-1} x \int e^x dx - \int \left[\frac{d}{dx}(\tan^{-1} x) \int e^x dx\right] dx = e^x \tan^{-1} x - \int \frac{e^x}{1+x^2} dx \)

Adding the second integral:

\( \int e^x\left(\tan^{-1} x + \frac{1}{1+x^2}\right) dx = e^x \tan^{-1} x - \int \frac{e^x}{1+x^2} dx + \int \frac{e^x}{1+x^2} dx = e^x \tan^{-1} x + c \)

where \( c \) is the integrating constant.

Exam Tip: When a function and its derivative both appear in the integrand, construct the parts decomposition so their integrals cancel. This pattern yields immediate answers without additional computation.

Question 61. Evaluate the following integrals: \( \int e^x\left(\cot x + \log \sin x\right) dx \)
Answer: Observe that the derivative of \( \log \sin x \) is \( \frac{\cos x}{\sin x} = \cot x \). Split:

\( \int e^x\left(\cot x + \log \sin x\right) dx = \int e^x \cot x \, dx + \int e^x \log \sin x \, dx \)

For the second integral, apply integration by parts with \( f_1(x) = \log \sin x \) and \( f_2(x) = e^x \):

\( \int e^x \log \sin x \, dx = \log \sin x \int e^x dx - \int \left[\frac{d}{dx}(\log \sin x) \int e^x dx\right] dx = e^x \log \sin x - \int e^x \cot x \, dx \)

Combining:

\( \int e^x\left(\cot x + \log \sin x\right) dx = \int e^x \cot x \, dx + e^x \log \sin x - \int e^x \cot x \, dx = e^x \log \sin x + c \)

where \( c \) is the integrating constant.

Exam Tip: Logarithmic functions paired with their derivative forms integrate cleanly through parts - the derivative term cancels with the other addend, leaving only the logarithmic part times the exponential.

Question 62. Evaluate the following integrals: \( \int e^x\left(\tan x - \log \cos x\right) dx \)
Answer: Note that the derivative of \( \log \cos x \) is \( \frac{-\sin x}{\cos x} = -\tan x \), so \( \frac{d}{dx}(-\log \cos x) = \tan x \). Split the integral:

\( \int e^x\left(\tan x - \log \cos x\right) dx = \int e^x \tan x \, dx - \int e^x \log \cos x \, dx \)

For the second integral, apply integration by parts with \( f_1(x) = \log \cos x \) and \( f_2(x) = e^x \):

\( \int e^x \log \cos x \, dx = \log \cos x \int e^x dx - \int \left[\frac{d}{dx}(\log \cos x) \int e^x dx\right] dx = e^x \log \cos x + \int e^x \tan x \, dx \)

Substituting back:

\( \int e^x\left(\tan x - \log \cos x\right) dx = \int e^x \tan x \, dx - e^x \log \cos x - \int e^x \tan x \, dx = -e^x \log \cos x + c \)

\( = e^x \log \sec x + c \), where \( c \) is the integrating constant.

Exam Tip: Recognize when a logarithm's derivative (with appropriate sign adjustment) matches another term in the integrand - this signals that parts will produce a cancellation and yield a clean final form.

Question 63. Evaluate the following integrals: \( \int e^x\left[\sec x + \log(\sec x + \tan x)\right] dx \)
Answer: Observe that the derivative of \( \log(\sec x + \tan x) \) is \( \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \sec x \). Split:

\( \int e^x\left[\sec x + \log(\sec x + \tan x)\right] dx = \int e^x \sec x \, dx + \int e^x \log(\sec x + \tan x) \, dx \)

For the second part, use integration by parts with \( f_1(x) = \log(\sec x + \tan x) \) and \( f_2(x) = e^x \):

\( \int e^x \log(\sec x + \tan x) \, dx = e^x \log(\sec x + \tan x) - \int e^x \sec x \, dx \)

Combining:

\( \int e^x\left[\sec x + \log(\sec x + \tan x)\right] dx = \int e^x \sec x \, dx + e^x \log(\sec x + \tan x) - \int e^x \sec x \, dx \)

\( = e^x \log(\sec x + \tan x) + c \), where \( c \) is the integrating constant.

Exam Tip: Complex logarithmic arguments often have elegant derivatives - always compute the derivative before proceeding with integration by parts to check if cancellation will occur.

Question 64. Evaluate the following integrals: \( \int e^x\left(\sec^2 x + \tan x\right) dx \)
Answer: Observe that \( \frac{d}{dx}(\tan x) = \sec^2 x \). Split:

\( \int e^x\left(\sec^2 x + \tan x\right) dx = \int e^x \sec^2 x \, dx + \int e^x \tan x \, dx \)

Apply integration by parts to the second integral with \( f_1(x) = \tan x \) and \( f_2(x) = e^x \):

\( \int e^x \tan x \, dx = \tan x \int e^x dx - \int \left[\frac{d}{dx}(\tan x) \int e^x dx\right] dx = e^x \tan x - \int e^x \sec^2 x \, dx \)

Combining:

\( \int e^x\left(\sec^2 x + \tan x\right) dx = \int e^x \sec^2 x \, dx + e^x \tan x - \int e^x \sec^2 x \, dx = e^x \tan x + c \)

where \( c \) is the integrating constant.

Exam Tip: When a function appears alongside its derivative, the parts method creates an immediate cancellation. Identify these complementary pairs before integrating to save time and reduce error risk.

Question 65. Evaluate the following integrals: \( \int e^x\left(\frac{\sin x \cos x - 1}{\sin^2 x}\right) dx \)
Answer: Rewrite the integrand by splitting the fraction:

\( \int e^x\left(\frac{\sin x \cos x - 1}{\sin^2 x}\right) dx = \int e^x(\cot x - \csc^2 x) dx = \int e^x \cot x \, dx - \int e^x \csc^2 x \, dx \)

For the first integral, apply integration by parts with \( f_1(x) = \cot x \) and \( f_2(x) = e^x \):

\( \int e^x \cot x \, dx = \cot x \int e^x dx - \int \left[\frac{d}{dx}(\cot x) \int e^x dx\right] dx = e^x \cot x + \int e^x \csc^2 x \, dx \)

Combining both integrals:

\( \int e^x\left(\frac{\sin x \cos x - 1}{\sin^2 x}\right) dx = e^x \cot x + \int e^x \csc^2 x \, dx - \int e^x \csc^2 x \, dx = e^x \cot x + c \)

where \( c \) is the integrating constant.

Exam Tip: Decompose complex fractions into recognizable trigonometric parts - once simplified, look for function-derivative pairs that enable rapid cancellation through integration by parts.

Question 66. Evaluate the following integrals: \( \int e^x\left(\frac{\cos x + \sin x}{\cos^2 x}\right) dx \)
Answer: Split the fraction:

\( \int e^x\left(\frac{\cos x + \sin x}{\cos^2 x}\right) dx = \int e^x(\sec x + \tan x) dx = \int e^x \sec x \, dx + \int e^x \tan x \, dx \)

For the second integral, use integration by parts with \( f_1(x) = \tan x \) and \( f_2(x) = e^x \):

\( \int e^x \tan x \, dx = \tan x \int e^x dx - \int \left[\frac{d}{dx}(\tan x) \int e^x dx\right] dx = e^x \tan x - \int e^x \sec^2 x \, dx \)

From the definition of the original problem, \( \sec x \) is related by: \( \frac{d}{dx}(\sec x) = \sec x \tan x \). Through a similar process:

\( \int e^x \sec x \, dx = e^x \sec x - \int e^x \sec x \tan x \, dx \)

Carefully combining and recognizing the pattern yields:

\( \int e^x\left(\frac{\cos x + \sin x}{\cos^2 x}\right) dx = e^x \sec x + c \), where \( c \) is the integrating constant.

Exam Tip: Rational trigonometric expressions in exponential integrands often simplify to basic trig plus their derivatives - always attempt to factor or decompose the denominator to expose these hidden structures.

Question 67. Evaluate the following integrals: \( \int e^x\left(\frac{2 - \sin 2x}{1 - \cos 2x}\right) dx \)
Answer: Simplify the integrand using trigonometric identities. Note that \( 1 - \cos 2x = 2\sin^2 x \) and \( \sin 2x = 2\sin x \cos x \):

\( \frac{2 - \sin 2x}{1 - \cos 2x} = \frac{2 - 2\sin x \cos x}{2\sin^2 x} = \frac{1 - \sin x \cos x}{\sin^2 x} = \csc^2 x - \cot x \)

Thus: \( \int e^x(\csc^2 x - \cot x) dx = \int e^x \csc^2 x \, dx - \int e^x \cot x \, dx \)

Using integration by parts on the second integral with \( f_1(x) = \cot x \) and \( f_2(x) = e^x \):

\( \int e^x \cot x \, dx = e^x \cot x + \int e^x \csc^2 x \, dx \)

Combining:

\( \int e^x\left(\frac{2 - \sin 2x}{1 - \cos 2x}\right) dx = \int e^x \csc^2 x \, dx - e^x \cot x - \int e^x \csc^2 x \, dx = -e^x \cot x + c \)

where \( c \) is the integrating constant.

Exam Tip: Always apply double-angle and other identities to simplify complex trigonometric expressions before integrating. The resulting form often reveals a function-derivative pair crucial for rapid solution.

Question 68. Evaluate the following integrals: \( \int e^x\left(\frac{1 + \sin x}{1 + \cos x}\right) dx \)
Answer: Using half-angle identities, note that \( 1 + \sin x = \left(1 + \tan \frac{x}{2}\right) \cdot \frac{2}{1 + \tan^2(x/2)} \) and \( 1 + \cos x = \frac{2}{1 + \tan^2(x/2)} \).

This simplifies to: \( \frac{1 + \sin x}{1 + \cos x} = \frac{\left(1 + \tan \frac{x}{2}\right)^2}{2} \)

Thus: \( \int e^x\left(\frac{1 + \sin x}{1 + \cos x}\right) dx = \int e^x \cdot \frac{(1 + \tan \frac{x}{2})^2}{2} dx = \int \frac{e^x(1 + \tan^2 \frac{x}{2} + 2\tan \frac{x}{2})}{2} dx \)

\( = \int \frac{e^x(\sec^2 \frac{x}{2} + 2\tan \frac{x}{2})}{2} dx = \int \frac{e^x \sec^2 \frac{x}{2}}{2} dx + \int e^x \tan \frac{x}{2} dx \)

For the second integral, use integration by parts with \( f_1(x) = \tan \frac{x}{2} \) and \( f_2(x) = e^x \):

\( \int e^x \tan \frac{x}{2} dx = e^x \tan \frac{x}{2} - \int \frac{e^x \sec^2 \frac{x}{2}}{2} dx \)

Combining:

\( \int e^x\left(\frac{1 + \sin x}{1 + \cos x}\right) dx = e^x \tan \frac{x}{2} + c \), where \( c \) is the integrating constant.

Exam Tip: Half-angle substitutions are powerful for trigonometric expressions involving sums in numerator and denominator. After simplification, the resulting function-derivative pattern often yields an immediate cancellation.

Question 69. Evaluate the following integrals: \( \int e^x\left(\frac{\sin 4x - 1}{1 - \cos 4x}\right) dx \)
Answer: Using the identity \( 1 - \cos 4x = 2\sin^2 2x \):

\( \frac{\sin 4x - 1}{1 - \cos 4x} = \frac{2\sin 2x \cos 2x - 1}{2\sin^2 2x} \)

Rewrite as: \( \int e^x(\csc^2 2x - \cot 2x) dx = \int e^x \csc^2 2x \, dx - \int e^x \cot 2x \, dx \)

For the second integral, apply integration by parts with \( f_1(x) = \cot 2x \) and \( f_2(x) = e^x \):

\( \int e^x \cot 2x \, dx = e^x \cot 2x + 2\int e^x \csc^2 2x \, dx \)

Substituting back carefully and accounting for the coefficient from the derivative:

\( \int e^x\left(\frac{\sin 4x - 1}{1 - \cos 4x}\right) dx = -e^x \cot 2x + c \), where \( c \) is the integrating constant.

Exam Tip: When derivatives involve coefficient multipliers (e.g., \( \frac{d}{dx}\cot 2x = -2\csc^2 2x \)), track these carefully through the parts formula. The final answer's coefficient structure will reflect this multiplicative factor.

Question 69. Evaluate the following integrals:
\( \int \frac{e^{x}\left[\sqrt{1-x^{2}} \sin^{-1} x+1\right]}{\sqrt{1-x^{2}}} dx \)
Answer: \( e^{x} \sin^{-1} x + c \), where c is the integrating constant
In simple words: When you integrate this expression using integration by parts, the inverse sine term remains, and all other terms cancel out to give a clean result.

Exam Tip: Watch for expressions where the derivative of the inverse trigonometric function simplifies the other terms - this is a key pattern in integration by parts problems.

Question 70. Evaluate the following integrals:
\( \int e^{x} \left(\frac{1+x \log x}{x}\right) dx \)
Answer: \( e^{x} \log x + c \), where c is the integrating constant
In simple words: Split the fraction into two parts, then use integration by parts on the logarithmic term - everything simplifies to leave just the exponential times the logarithm.

Exam Tip: Always split rational expressions with multiple terms in the numerator before choosing which term to use as your first function in integration by parts.

Question 71. Evaluate the following integrals:
\( \int e^{x} \cdot \frac{x}{(1+x)^{2}} dx \)
Answer: \( \frac{e^{x}}{1+x} + c \), where c is the integrating constant
In simple words: Use partial fractions to decompose the rational part, then apply integration by parts - this breaks down a complex fraction into manageable pieces.

Exam Tip: Partial fractions combined with integration by parts is powerful for rational functions multiplied by exponentials - always check if your denominator factors.

Question 72. Evaluate the following integrals:
\( \int e^{x} \cdot \frac{x-1}{(x+1)^{3}} dx \)
Answer: \( \frac{e^{x}}{(x+1)^{2}} + c \), where c is the integrating constant
In simple words: Express the fraction as a sum of simpler fractions, then pair each with the exponential using integration by parts - repeated application of the formula handles higher powers.

Exam Tip: When a denominator has a high power like \( (x+1)^{3} \), set up partial fractions with separate terms for each power level, then integrate each piece separately.

Question 73. Evaluate the following integrals:
\( \int e^{x} \cdot \frac{2-x}{(1-x)^{2}} dx \)
Answer: \( \frac{e^{x}}{1-x} + c \), where c is the integrating constant
In simple words: Break apart the numerator over the denominator using partial fractions, then integrate each fraction multiplied by the exponential term - the integrals simplify nicely.

Exam Tip: Notice that this is similar to Question 72 but with opposite signs in the binomial - the same partial fractions method works, just watch your arithmetic carefully.

Question 74. Evaluate the following integrals:
\( \int e^{x} \cdot \frac{x-3}{(x-1)^{3}} dx \)
Answer: \( \frac{e^{x}}{(x-1)^{2}} + c \), where c is the integrating constant
In simple words: Decompose the rational expression into partial fractions with terms for the linear and quadratic powers in the denominator, then combine with integration by parts.

Exam Tip: The power in the final answer is always one less than the highest power in your partial fraction decomposition - use this as a quick check of your work.

Question 75. Evaluate the following integrals:
\( \int e^{3x} \left(\frac{3x-1}{9x^{2}}\right) dx \)
Answer: \( \frac{e^{3x}}{9x} + c \), where c is the integrating constant
In simple words: Factor out and separate the terms in the numerator so you have two simpler fractions, then use integration by parts with the exponential \( e^{3x} \).

Exam Tip: When the exponential base is not \( e^{x} \) but rather \( e^{ax} \), remember that the derivative introduces a factor of a - account for this carefully in your integration by parts setup.

Question 76. Evaluate the following integrals:
\( \int \frac{(x+1)}{(x+2)^{2}} e^{x} dx \)
Answer: \( \frac{e^{x}}{x+2} + c \), where c is the integrating constant
In simple words: Use partial fractions to split the rational part into linear and quadratic denominators, then integrate each piece with the exponential using integration by parts.

Exam Tip: For integrals of the form \( \int \frac{P(x)}{(x+a)^{n}} e^{x} dx \), always reduce the power of the denominator by 1 from what you see in the final answer.

Question 77. Evaluate the following integrals:
\( \int \frac{x e^{2x}}{(1+2x)^{2}} dx \)
Answer: \( \frac{e^{2x}}{4(2x+1)} + c \), where c is the integrating constant
In simple words: Decompose the fraction using partial fractions, multiply each term by the exponential, then apply integration by parts - the exponential \( e^{2x} \) requires adjusting for the factor 2.

Exam Tip: When exponential base is \( e^{2x} \) or higher multiples, be extra careful about the chain rule adjustments that appear in your integration by parts formula.

Question 78. Evaluate the following integrals:
\( \int e^{2x} \left(\frac{2x-1}{4x^{2}}\right) dx \)
Answer: \( \frac{e^{2x}}{4x} + c \), where c is the integrating constant
In simple words: Rewrite the fraction as a sum of two terms divided by \( 4x^{2} \), then multiply each by \( e^{2x} \) and integrate using the standard integration by parts method.

Exam Tip: Notice the coefficients: if you have \( e^{2x} \) in the integral and your final answer has \( \frac{e^{2x}}{4x} \), check that the 4 matches the 4 in the denominator of your original fraction.

Question 79. Evaluate the following integrals:
\( \int e^{x} \left(\log x + \frac{1}{x^{2}}\right) dx \)
Answer: \( e^{x} \left(\log x - \frac{1}{x}\right) + c \), where c is the integrating constant
In simple words: Integrate the logarithm term using integration by parts - apply the formula with \( \log x \) as the first function. The \( \frac{1}{x^{2}} \) term combines with the result to form a perfect answer.

Exam Tip: When a sum appears inside the integral with one term being logarithmic, always use integration by parts on that logarithmic part first - the other terms often work out to cancel intermediate steps.

Question 80. Evaluate the following integrals:
\( \int \frac{\log x}{(1+\log x)^{2}} dx \)
Answer: \( \frac{x}{1+\log x} + c \), where c is the integrating constant
In simple words: Use partial fractions on the denominator after recognizing that \( 1 + \log x \) appears in a squared form - then apply integration by parts with \( \frac{1}{1+\log x} \) as the first function.

Exam Tip: This problem shows that not every integral has an exponential \( e^{x} \) - sometimes the answer still simplifies beautifully without it if you choose the right integration by parts pair.

Question 81. Evaluate the following integrals:
\( \int \{sin(\log x) + \cos(\log x)\} dx \)
Answer: \( x \sin(\log x) + c \), where c is the integrating constant
In simple words: Use integration by parts with \( \sin(\log x) \) as your first function and 1 as your second - the cosine term that appears cancels with the second part of the sum.

Exam Tip: Whenever you see \( \sin(\log x) + \cos(\log x) \) together, expect them to cancel during integration by parts - this is a sign you chose the right approach.

Question 82. Evaluate the following integrals:
\( \int \left[\frac{1}{\log x} - \frac{1}{(\log x)^{2}}\right] dx \)
Answer: \( \frac{x}{\log x} + c \), where c is the integrating constant
In simple words: Apply integration by parts to the first term using \( \frac{1}{\log x} \) as your first function - the second term then becomes the perfect follow-up that cancels the extra integral that appears.

Exam Tip: The presence of both \( \frac{1}{\log x} \) and \( \frac{1}{(\log x)^{2}} \) in the same integral is a red flag that they are designed to work together through integration by parts.

Question 83. Evaluate the following integrals:
\( \int \left[\log(\log x) + \frac{1}{(\log x)^{2}}\right] dx \)
Answer: \( x \left[\log(\log x) - \frac{1}{\log x}\right] + c \), where c is the integrating constant
In simple words: Use integration by parts on the logarithm of a logarithm - this requires careful chaining of the derivative formula, and the second term contributes to give the final answer.

Exam Tip: Compositions like \( \log(\log x) \) require the chain rule when you differentiate - track your steps carefully to avoid losing a factor during integration by parts.

Question 84. Evaluate the following integrals:
\( \int \frac{\sin^{-1}\sqrt{x} - \cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x}} dx \)
Answer: \( \frac{2}{\pi}\left[\sqrt{x - x^{2}} + x(\sin^{-1}\sqrt{x} - \cos^{-1}\sqrt{x}) + \sin^{-1}\sqrt{1-x}\right] + c \), where c is the integrating constant
In simple words: Use the identity that \( \sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x} = \frac{\pi}{2} \) to simplify the denominator, then integrate each of the two resulting terms separately using integration by parts with substitution.

Exam Tip: Inverse trigonometric identities are essential here - always simplify the denominator first before diving into complicated integration by parts on the numerator.

Question 85. Evaluate the following integrals:
\( \int 5^{5^{5^{x}}} \cdot 5^{5^{x}} \cdot 5^{x} dx \)
Answer: \( \frac{5^{5^{5^{x}}}}{(\log 5)^{3}} + c \), where c is the integrating constant
In simple words: Use substitution twice - first replace \( 5^{x} \) with a variable, then replace the new exponential with another - this layers the chain rule and builds up the power structure step by step.

Exam Tip: Nested exponentials like \( 5^{5^{x}} \) always require layered substitution - count how many exponentials are stacked to know how many substitution steps you need.

Question 86. Evaluate the following integrals:
\( \int e^{2x} \tan x dx \)
Answer: \( \frac{1}{2} e^{2x} \tan x + c \), where c is the integrating constant
In simple words: Use integration by parts with \( \tan x \) as your first function - when you apply the method and simplify using trigonometric identities, the repeating integrals cancel out leaving a clean result.

Exam Tip: Integrals of trigonometric functions times exponentials often involve cyclic integration by parts - if you get the same integral on both sides, solve algebraically rather than continuing to integrate.

Question 87. Evaluate the following integrals:
\( \int e^{2x} \left(\frac{1+\sin 2x}{1+\cos 2x}\right) dx \)
Answer: \( \frac{1}{2} e^{2x} \tan x + c \), where c is the integrating constant
In simple words: Rewrite the trigonometric fraction using double angle formulas - recognizing that \( 1 + \cos 2x = 2\cos^{2}x \) and \( 1 + \sin 2x = (\cos x + \sin x)^{2} \) transforms this into a tangent expression.

Exam Tip: When a problem gives a complex trigonometric ratio like \( \frac{1+\sin 2x}{1+\cos 2x} \), always try double angle identities first - they often simplify into recognizable forms.

Question 88. Evaluate the following integrals:
\( \int e^{2x} \left(\frac{1-\sin 2x}{1-\cos 2x}\right) dx \)
Answer: \( -\frac{1}{2} e^{2x} \cot x + c \), where c is the integrating constant
In simple words: Apply double angle identities to rewrite the fraction - \( 1 - \cos 2x = 2\sin^{2}x \) and \( 1 - \sin 2x = (\cos x - \sin x)^{2} \) simplifies to a cotangent form times the exponential.

Exam Tip: Notice the pattern - Question 87 has a plus sign giving tangent, Question 88 has a minus sign giving negative cotangent - double angle identities reveal this structure.

Question 1. Mark (√) against the correct answer: \( \int x e^x dx = ? \)
(A) \( e^x (1 - x) + C \)
(B) \( e^x (x - 1) + C \)
(C) \( e^x (x - 1) + C \)
(D) none of these
Answer: (C) \( e^x (x - 1) + C \)
In simple words: When you integrate \( x e^x \) using integration by parts, you get \( e^x \) multiplied by \( (x - 1) \), plus the constant.

Exam Tip: Always apply the integration by parts formula: \( \int f(x) g(x) dx = f(x) \int g(x) dx - \int [f'(x) \int g(x) dx] dx \). Choose the first function wisely - algebraic or logarithmic functions go first.

Question 2. Mark (√) against the correct answer: \( \int x e^{2x} dx = ? \)
(A) \( \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C \)
(B) \( \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C \)
(C) \( 2x e^{2x} + 4e^{2x} + C \)
(D) none of these
Answer: (B) \( \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C \)
In simple words: Using integration by parts where \( x \) is the first function and \( e^{2x} \) is the second, the answer separates into two terms - one with the original variable and one with just the exponential.

Exam Tip: When the exponent has a coefficient (like 2x), remember to divide by that coefficient when integrating the exponential function.

Question 3. Mark (√) against the correct answer: \( \int x \cos 2x dx = ? \)
(A) \( \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C \)
(B) \( \frac{1}{2} x \sin 2x - \frac{1}{4} \cos 2x + C \)
(C) \( 2x \sin 2x + 4 \cos 2x + C \)
(D) none of these
Answer: (A) \( \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C \)
In simple words: When integrating a product of \( x \) and a trigonometric function, apply parts - the \( x \) multiplies the integrated trig function, then subtract the integral of the derivative of \( x \) times the integrated trig.

Exam Tip: For \( \int x \cos kx dx \), use substitution \( t = 2x \) to simplify coefficients early, making the arithmetic cleaner.

Question 4. Mark (√) against the correct answer: \( \int x \sec^2 x dx = ? \)
(A) \( x \tan x - \log |\cos x| + C \)
(B) \( x \tan x + \log |\cos x| + C \)
(C) \( x \tan x + \log |\sec x| + C \)
(D) none of these
Answer: (B) \( x \tan x + \log |\cos x| + C \)
In simple words: The integral of \( x \sec^2 x \) yields \( x \tan x \) as the main term plus an additional logarithmic term involving the cosine function.

Exam Tip: Remember that the derivative of \( \tan x \) is \( \sec^2 x \), making it straightforward to identify which trig function to integrate first in the parts formula.

Question 5. Mark (√) against the correct answer: \( \int x \sin 2x dx = ? \)
(A) \( \frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x + C \)
(B) \( -\frac{1}{2} x \cos 2x - \frac{1}{4} \sin 2x + C \)
(C) \( -\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x + C \)
(D) none of these
Answer: (C) \( -\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x + C \)
In simple words: When integrating \( x \sin 2x \), the answer involves \( x \) multiplied by a negative cosine term, plus a positive sine term with a smaller coefficient.

Exam Tip: Watch the signs carefully - the cosine term appears with a negative sign because the integral of \( \sin 2x \) is \( -\cos 2x \) divided by 2.

Question 6. Mark (√) against the correct answer: \( \int x \log x dx = ? \)
(A) \( x \log x + \frac{1}{2} x^2 + C \)
(B) \( \frac{1}{2} x^2 \log x + \frac{1}{4} x^2 + C \)
(C) \( \frac{1}{2} x^2 \log x - \frac{1}{4} x^2 + C \)
(D) none of these
Answer: (C) \( \frac{1}{2} x^2 \log x - \frac{1}{4} x^2 + C \)
In simple words: The integral of \( x \log x \) gives the logarithm multiplied by half of \( x^2 \), minus a quarter of \( x^2 \).

Exam Tip: Logarithmic functions should always be chosen as the first function in integration by parts since their derivative becomes simpler.

Question 7. Mark (√) against the correct answer: \( \int x \cosec^2 x dx = ? \)
(A) \( x \cot x - \log |\sin x| + C \)
(B) \( -\cot x + \log |\sin x| + C \)
(C) \( x \tan x - \log |\sec x| + C \)
(D) none of these
Answer: (D) none of these
In simple words: The integral gives a term with \( x \) times the negative cotangent, combined with the natural logarithm of the sine function's absolute value.

Exam Tip: The derivative of \( \cot x \) is \( -\cosec^2 x \), so watch for negative signs when applying integration by parts.

Question 8. Mark (√) against the correct answer: \( \int x \sin x \cos x dx = ? \)
(A) \( -\frac{1}{4} x \sin 2x + \frac{1}{8} \cos 2x + C \)
(B) \( \frac{1}{4} x \cos 2x - \frac{1}{8} \sin 2x + C \)
(C) \( \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C \)
(D) \( -\frac{1}{4} x \cos 2x + \frac{1}{8} \sin 2x + C \)
Answer: (D) \( -\frac{1}{4} x \cos 2x + \frac{1}{8} \sin 2x + C \)
In simple words: First use the double angle identity \( 2 \sin x \cos x = \sin 2x \), then apply integration by parts to handle the resulting product of \( x \) and a trigonometric function.

Exam Tip: Always look for trigonometric identities to simplify products before integrating - this reduces the complexity of the calculation.

Question 9. Mark (√) against the correct answer: \( \int x \cos^2 x dx = ? \)
(A) \( \frac{x^2}{4} - \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + C \)
(B) \( \frac{x^2}{4} + \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + C \)
(C) \( \frac{x^2}{4} + \frac{x \sin 2x}{4} - \frac{\cos 2x}{8} + C \)
(D) none of these
Answer: (D) \( \frac{x^2}{4} + \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + C \)
In simple words: Use the half-angle formula \( \cos^2 x = \frac{1 + \cos 2x}{2} \) to rewrite the integrand, then split into two simpler integrals and apply parts to the second piece.

Exam Tip: Power-reduction formulas are essential for integrals containing squared trigonometric functions - they transform the problem into manageable pieces.

Question 10. Mark (√) against the correct answer: \( \int \frac{\log x}{x^2} dx = ? \)
(A) \( -\frac{1}{x} (\log x + 1) + C \)
(B) \( \frac{1}{x} (\log x - 1) + C \)
(C) \( \frac{1}{x} (\log x + 1) + C \)
(D) none of these
Answer: (A) \( -\frac{1}{x} (\log x + 1) + C \)
In simple words: Set \( \log x \) as the first function and \( x^{-2} \) as the second, then use integration by parts to obtain a product involving the reciprocal.

Exam Tip: Rewrite \( \frac{1}{x^2} \) as \( x^{-2} \) to clarify the form for integration, making the power rule application straightforward.

Question 11. Mark (√) against the correct answer: \( \int \log x dx = ? \)
(A) \( \frac{1}{x} + C \)
(B) \( \frac{1}{2} (\log x)^2 + C \)
(C) \( x (\log x + 1) + C \)
(D) \( x (\log x - 1) + C \)
Answer: (D) \( x (\log x - 1) + C \)
In simple words: Treat \( \log x \) as the first function and 1 as the second, apply parts to get \( x \log x \) minus the integral of \( x \) times the derivative of \( \log x \).

Exam Tip: When a single logarithmic function is the entire integrand, always use parts with 1 as the multiplier - this avoids repeated integration.

Question 12. Mark (√) against the correct answer: \( \int \log_{10} x dx = ? \)
(A) \( \frac{1}{x} \log_e 10 + C \)
(B) \( \frac{1}{x} \log_{10} e + C \)
(C) \( x (\log x - 1) \log_e 10 + C \)
(D) \( x(\log x - 1) \log_{10} e + C \)
Answer: (D) \( x(\log x - 1) \log_{10} e + C \)
In simple words: Convert \( \log_{10} x \) to \( \frac{\log x}{\log 10} \) using the change-of-base formula, then integrate using parts with the constant coefficient factored out.

Exam Tip: Always apply the change-of-base formula to convert non-natural logarithms into natural logarithms before integration - this keeps the formula standard.

Question 13. Mark (√) against the correct answer: \( \int (\log x)^2 dx = ? \)
(A) \( \frac{2 \log x}{x} + C \)
(B) \( \frac{1}{3} (\log x)^3 + C \)
(C) \( x (\log x)^2 - 2x \log x + 2x + C \)
(D) \( x (\log x)^2 + 2x \log x - 2x + C \)
Answer: (C) \( x (\log x)^2 - 2x \log x + 2x + C \)
In simple words: Apply integration by parts twice - first with \( (\log x)^2 \) and then with \( \log x \) - to progressively reduce the power of the logarithm.

Exam Tip: For higher powers of logarithms, repeated application of integration by parts is necessary; track each step to avoid losing intermediate results.

Question 14. Mark (√) against the correct answer: \( \int e^{\sqrt{x}} dx = ? \)
(A) \( e^{\sqrt{x}} + \sqrt{x} + C \)
(B) \( \frac{1}{2} e^{\sqrt{x}} (\sqrt{x} + 1) + C \)
(C) \( 2e^{\sqrt{x}} (\sqrt{x} - 1) + C \)
(D) none of these
Answer: (C) \( 2e^{\sqrt{x}} (\sqrt{x} - 1) + C \)
In simple words: Substitute \( \sqrt{x} = t \), then \( dx = 2t dt \), and apply integration by parts to the product \( t e^t \) that emerges.

Exam Tip: Substitution first simplifies nested functions; once substituted, use integration by parts systematically on the new variable.

Question 15. Mark (√) against the correct answer: \( \int \cos \sqrt{x} dx = ? \)
(A) \( \sin \sqrt{x} + \cos \sqrt{x} + C \)
(B) \( \frac{1}{2} (\sqrt{x} \sin \sqrt{x} - \cos \sqrt{x}) + C \)
(C) \( 2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + C \)
(D) none of these
Answer: (C) \( 2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + C \)
In simple words: Substitute \( \sqrt{x} = t \) so that \( dx = 2t dt \), then apply integration by parts to \( 2t \cos t \).

Exam Tip: Always express the result in terms of the original variable - substitute back \( t = \sqrt{x} \) before writing the final answer.

Question 16. Mark (√) against the correct answer: \( \int \cos(\log x) dx = ? \)
(A) \( \frac{x}{2} [\cos (\log x) - \sin (\log x)] + C \)
(B) \( \frac{x}{2} [\cos (\log x) + \sin (\log x)] + C \)
(C) \( 2x [\cos (\log x) + \sin (\log x)] + C \)
(D) \( 2x [\cos (\log x) - \sin (\log x)] + C \)
Answer: (B) \( \frac{x}{2} [\cos (\log x) + \sin (\log x)] + C \)
In simple words: Use integration by parts with \( \cos(\log x) \) as the first function; this requires applying parts twice due to the cyclic nature of trigonometric integrals.

Exam Tip: When integrals cycle back (the integral reappears), collect like terms and solve for the integral algebraically - this avoids infinite recursion.

Question 17. Mark (√) against the correct answer: \( \int \sec^3 x dx = ? \)
(A) \( \frac{1}{2} \{\sec x \tan x - \log |\sec x + \tan x|\} + C \)
(B) \( \frac{1}{2} \{\sec x \tan x + \log |\sec x + \tan x|\} + C \)
(C) \( 2\{\sec x \tan x + \log |\sec x + \tan x|\} + C \)
(D) none of these
Answer: (B) \( \frac{1}{2} \{\sec x \tan x + \log |\sec x + \tan x|\} + C \)
In simple words: Apply integration by parts with \( \sec x \) and \( \sec^2 x \), then recognize a cyclic integral and solve it algebraically to isolate the original integral.

Exam Tip: For \( \sec^3 x \), the reduction formula or cyclic method works best - traditional parts alone leads to complexity that is resolved by algebraic manipulation.

Question 18. Mark (√) against the correct answer in each of the following: \( \int \left[\frac{1}{(\log x)} - \frac{1}{(\log x)^2}\right] dx = ? \)
(a) x log x + C
(b) \( \frac{x}{\log x} + C \)
(c) \( x + \frac{1}{\log x} + C \)
(d) none of these
Answer: (b) \( \frac{x}{\log x} + C \)
In simple words: When you integrate this expression, substitute t = log x to simplify it. After working through the integration by parts and substitution, you get x divided by log x, plus a constant.

Exam Tip: Always recognize when a substitution can reduce a complex integral into a standard form - here, u-substitution with log x is key to finding the answer quickly.

Question 19. Mark (√) against the correct answer in each of the following: \( \int 2x^2 e^{x^3} dx = ? \)
(a) \( e^{x^2} (x^2 - 1) + C \)
(b) \( e^{x^2} (x^2 + 1) + C \)
(c) \( e^{x^2} (x + 1) + C \)
(d) none of these
Answer: (b) \( e^{x^2} (x^2 + 1) + C \)
In simple words: Let t = x³, then dt = 3x² dx. Rearrange to get x² dx = dt/3. The integral becomes easier and when you work it out, you get e to the power x³, multiplied by (x³ + 1), all plus a constant.

Exam Tip: Watch for exponential functions paired with polynomial terms - substitution with the exponent is usually the fastest path to simplification.

Question 20. Mark (√) against the correct answer in each of the following: \( \int (x 2^x) dx = ? \)
(a) \( \frac{2^x}{(\log 2)} (x + \log 2) + C \)
(b) \( \frac{2^x}{(\log 2)^2} (x + \log 2 - 1) + C \)
(c) \( \frac{x \cdot 2^x}{(\log 2)} - \frac{2^x}{(\log 2)^2} + C \)
(d) none of these
Answer: (d) none of these
In simple words: Using integration by parts with x as the first function and 2ˣ as the second, you simplify the integral step by step. When you gather all the terms together, the final answer does not match any of the first three choices.

Exam Tip: Always check the constant term and the coefficient structure carefully - a small difference in grouping or signs can make an option incorrect.

Question 21. Mark (√) against the correct answer in each of the following: \( \int x \cot^2 x \, dx = ? \)
(a) \( -x \cot x + \frac{x^2}{2} + \log |\sin x| + C \)
(b) \( -x \cot x - \frac{x^2}{2} + \log |\sin x| + C \)
(c) \( -x \cot x + \frac{x^2}{2} - \log |\sin x| + C \)
(d) none of these
Answer: (b) \( -x \cot x - \frac{x^2}{2} + \log |\sin x| + C \)
In simple words: Apply integration by parts twice. Rewrite cot² x as (cosec² x - 1) to make the integral manageable. Work through the steps carefully and collect all terms - you'll get negative x times cot x, minus x² divided by 2, plus the logarithm of the absolute value of sin x, plus a constant.

Exam Tip: When working with trigonometric integrals, breaking down identities like cot² x = cosec² x - 1 early saves time and reduces errors.

Question 22. Mark (√) against the correct answer in each of the following: \( \int \sin \sqrt{x} \, dx = ? \)
(a) \( -\sqrt{x} \cos \sqrt{x} + C \)
(b) \( -\sqrt{x} \cos \sqrt{x} - 2 \sin \sqrt{x} - C \)
(c) \( -2\sqrt{x} \cos \sqrt{x} + 2 \sin \sqrt{x} - C \)
(d) none of these
Answer: (c) \( -2\sqrt{x} \cos \sqrt{x} + 2 \sin \sqrt{x} - C \)
In simple words: Let t = √x, so x = t², and dx = 2t dt. The integral becomes 2 times the integral of t times sin t. Use integration by parts - first function is t, second is sin t. Complete the working and substitute back to get the answer in terms of √x.

Exam Tip: With nested functions like sin √x, always set up a substitution to remove the square root first - this transforms the integral into a standard form.

Question 23. Mark (√) against the correct answer in each of the following: \( \int e^{\sin x} \sin 2x \, dx = ? \)
(a) (2 sin x) e^(sin x) + C
(b) (2 cos x) e^(sin x) + C
(c) 2e^(sin x) (sin x + 1) + C
(d) 2e^(sin x) (sin x - 1) + C
Answer: (d) 2e^(sin x) (sin x - 1) + C
In simple words: Rewrite sin 2x as 2 sin x cos x. Let t = sin x, so dt = cos x dx. The integral becomes 2 times the integral of t times e to the power t, dt. Apply integration by parts - t is the first function and e^t is the second. Work through both steps and substitute back to finish.

Exam Tip: Recognize that sin 2x = 2 sin x cos x immediately - this clever rewrite often reveals a hidden substitution opportunity.

Question 24. Mark (√) against the correct answer in each of the following: \( \int \frac{\sin^{-1} x}{(1-x^2)^{3/2}} dx = ? \)
(a) \( \frac{\sin^{-1} x}{\sqrt{1-x^2}} - \frac{1}{2} \log |1-x^2| + C \)
(b) \( x \sin^{-1} x + \frac{1}{2} \log |1-x^2| + C \)
(c) \( \frac{\sin^{-1} x}{\sqrt{1-x^2}} + \frac{1}{2} \log |1-x^2| + C \)
(d) none of these
Answer: (c) \( \frac{\sin^{-1} x}{\sqrt{1-x^2}} + \frac{1}{2} \log |1-x^2| + C \)
In simple words: Put sin⁻¹ x = t, which means x = sin t. Then \( \frac{1}{\sqrt{1-x^2}} = \frac{dt}{dx} \). Rewrite the integral using t and dx in terms of dt. Apply integration by parts with t as the first function. After substituting back, collect all terms to arrive at the final form.

Exam Tip: When inverse trigonometric functions appear, substituting the inverse function as a new variable often simplifies complex expressions underneath.

Question 25. Mark (√) against the correct answer in each of the following: \( \int \frac{x \tan^{-1} x}{(1-x^2)^{3/2}} dx = ? \)
(a) \( \frac{\tan^{-1} x}{\sqrt{1+x^2}} - \frac{x}{\sqrt{1+x^2}} + C \)
(b) \( \frac{-\tan^{-1} x}{\sqrt{1+x^2}} + \frac{x}{\sqrt{1+x^2}} + C \)
(c) \( \frac{x \tan^{-1} x}{\sqrt{1+x^2}} + \frac{1}{2} \log \left|\frac{x}{\sqrt{1+x^2}}\right| + C \)
(d) none of these
Answer: (b) \( \frac{-\tan^{-1} x}{\sqrt{1+x^2}} + \frac{x}{\sqrt{1+x^2}} + C \)
In simple words: Let tan⁻¹ x = t and x = tan t. Then dx = sec² t dt. From the substitution, find that 1 + x² = sec² t and other needed identities. The integral reduces to t times sin t, integrated by parts. Once you complete the working, substitute back and simplify to get the final answer.

Exam Tip: Building the trig relationships carefully (like 1 + tan² t = sec² t) during substitution prevents algebra mistakes later.

Question 26. Mark (√) against the correct answer in each of the following: \( \int x \tan^{-1} x \, dx = ? \)
(a) \( \frac{1}{2} \tan^{-1} x + \log(1+x^2) - \frac{1}{2} x + C \)
(b) \( \frac{1}{2} x^2 \tan^{-1} x + \frac{1}{2} x + C \)
(c) \( \frac{1}{2} (1+x^2) \tan^{-1} x - \frac{1}{2} x + C \)
(d) none of these
Answer: (c) \( \frac{1}{2} (1+x^2) \tan^{-1} x - \frac{1}{2} x + C \)
In simple words: Use integration by parts with tan⁻¹ x as the first function and x as the second. This gives you \( \frac{x^2}{2} \tan^{-1} x \) minus the integral of \( \frac{x^2}{2} \) divided by (1 + x²), all dx. Break the second integral into \( \frac{1}{2} \int 1 \, dx \) and \( \frac{1}{2} \int \frac{1}{1+x^2} dx \). Combine everything to get the final form.

Exam Tip: For inverse trig functions, always choose the inverse function as u in integration by parts - it eliminates the inverse function from the remaining integral.

Question 27. Mark (√) against the correct answer in each of the following: \( \int \tan^{-1} \sqrt{x} \, dx = ? \)
(a) \( (x-1) \tan^{-1} \sqrt{x} + \sqrt{x} + C \)
(b) \( (x+1) \tan^{-1} \sqrt{x} - \sqrt{x} + C \)
(c) \( \frac{1}{2}\sqrt{x} \tan^{-1} \sqrt{x} - \frac{1}{2}\sqrt{x} + C \)
(d) none of these
Answer: (b) \( (x+1) \tan^{-1} \sqrt{x} - \sqrt{x} + C \)
In simple words: Let √x = t, so x = t² and dx = 2t dt. The integral becomes 2 times the integral of t times tan⁻¹ t, dt. Apply integration by parts with tan⁻¹ t as the first function and t as the second. The remaining integral simplifies using polynomial division. After completing all steps, substitute back √x for t to get the answer.

Exam Tip: Nested radicals like √x inside inverse functions signal a substitution - remove the radical first to reveal the standard form beneath.

Question 28. Mark (√) against the correct answer in each of the following: \( \int \cos^{-1} x \, dx = ? \)
(a) \( x \cos^{-1} x - \sqrt{1-x^2} + C \)
(b) \( x \cos^{-1} x + \sqrt{1-x^2} + C \)
(c) \( x \sin^{-1} x - \sqrt{1-x^2} + C \)
(d) none of these
Answer: (b) \( x \cos^{-1} x + \sqrt{1-x^2} + C \)
In simple words: Apply integration by parts with cos⁻¹ x as the first function and 1 as the second. This gives you \( x \cos^{-1} x \) plus the integral of \( \frac{x}{\sqrt{1-x^2}} \) dx. For the remaining integral, let u = 1 - x², so du = -2x dx. This integral becomes \( -\frac{1}{2} \int u^{-1/2} du \), which yields \( -\sqrt{1-x^2} \). Combine to get the final result.

Exam Tip: The integral of cos⁻¹ x is closely related to sin⁻¹ x - recognize the pattern and watch the sign carefully when combining terms.

Question 29. Mark (√) against the correct answer in each of the following: \( \int \tan^{-1} x \, dx = ? \)
(a) \( x \tan^{-1} x + \frac{1}{2} \log |1+x^2| + C \)
(b) \( x \tan^{-1} x - \frac{1}{2} \log |1+x^2| + C \)
(c) \( -x \tan^{-1} x + \frac{1}{2} \log |1+x^2| + C \)
(d) none of these
Answer: (b) \( x \tan^{-1} x - \frac{1}{2} \log |1+x^2| + C \)
In simple words: Use integration by parts with tan⁻¹ x as the first function and 1 as the second. You get \( x \tan^{-1} x \) minus the integral of \( \frac{x}{1+x^2} dx \). Recognize that the remaining integral is the derivative pattern for a logarithm - it equals \( \frac{1}{2} \log(1+x^2) \). Subtract this from the first part to arrive at the answer.

Exam Tip: Once you apply integration by parts, always check if the remaining integral fits a standard logarithmic form - \( \frac{f'(x)}{f(x)} dx = \log |f(x)| + C \).

Question 30. Mark (√) against the correct answer in each of the following: \( \int \sec^{-1} x \, dx = ? \)
(a) \( x \sec^{-1} x + \log |x + \sqrt{x^2-1}| + C \)
(b) \( x \sec^{-1} x - \log |x + \sqrt{x^2-1}| + C \)
(c) \( x \sec^{-1} x + \log |x - \sqrt{x^2-1}| + C \)
(d) none of these
Answer: (b) \( x \sec^{-1} x - \log |x + \sqrt{x^2-1}| + C \)
In simple words: Apply integration by parts with sec⁻¹ x as the first function and 1 as the second. This gives you \( x \sec^{-1} x \) minus the integral of \( \frac{x}{\sqrt{x^2-1}} dx \). For the remaining integral, substitute u = x² - 1, so du = 2x dx. This becomes \( \frac{1}{2} \int u^{-1/2} du = \sqrt{x^2-1} \). Now combine: the result is \( x \sec^{-1} x \) minus the integral of \( \frac{1}{\sqrt{x^2-1}} dx \), which gives the logarithmic form shown.

Exam Tip: The integral of sec⁻¹ x involves the logarithmic derivative of x + √(x² - 1) - memorizing this relationship saves calculation time.

Question 31. Mark (√) against the correct answer in each of the following: \( \int \sin^{-1} (3x - 4x^3) \, dx = ? \)
(a) \( 3 \left[ x \sin^{-1} x + \sqrt{1-x^2} \right] + C \)
(b) \( 3 \left[ x \sin^{-1} x - \sqrt{1-x^2} \right] + C \)
(c) \( \frac{3x^2}{2} + C \)
(d) none of these
Answer: (a) \( 3 \left[ x \sin^{-1} x + \sqrt{1-x^2} \right] + C \)
In simple words: Let x = sin θ, so dx = cos θ dθ and \( \sqrt{1-x^2} = \cos \theta \). Recognize that 3x - 4x³ = 3 sin θ - 4 sin³ θ = sin 3θ (this is the triple-angle formula for sine). So the integral becomes 3θ times cos θ, integrated with respect to θ. Apply integration by parts, then substitute back x = sin θ and θ = sin⁻¹ x to get the final form.

Exam Tip: Always check if an expression inside an inverse function matches a known trig identity (like the triple-angle formula) - this can transform the entire integral into something much simpler.

Question 32. Mark (√) against the correct answer in each of the following: \( \int \sin^{-1}\left(\frac{2x}{1+x^2}\right) dx = ? \)
(a) 2x tan⁻¹ x + log |1 + x²| + C
(b) 2x tan⁻¹ x - log |1 + x²| + C
(c) 2x sin⁻¹ x + log |1 + x²| + C
(d) None of the options
Answer: (b) 2x tan⁻¹ x - log |1 + x²| + C
In simple words: When you substitute x = tan θ and apply integration by parts twice, the inverse sine simplifies to tan⁻¹ x. The final answer includes a linear term in tan⁻¹ x and a logarithmic correction term.

Exam Tip: Recognize the substitution x = tan θ when you see expressions with (1 + x²) in the denominator - this transforms inverse trigonometric functions and simplifies the integral dramatically.

Question 33. Mark (√) against the correct answer in each of the following: \( \int \tan^{-1}\sqrt{\frac{1-x}{1+x}} dx = ? \)
(a) \( \frac{1}{2}x\left(\cos^{-1}x\right) + \frac{1}{2}\sqrt{1-x^2} + C \)
(b) \( \frac{1}{2}x\left(\sin^{-1}x\right) + \frac{1}{2}\sqrt{1-x^2} + C \)
(c) \( \frac{1}{2}x\left(\cos^{-1}x\right) - \frac{1}{2}\sqrt{1-x^2} + C \)
(d) None of the options
Answer: (c) \( \frac{1}{2}x\left(\cos^{-1}x\right) - \frac{1}{2}\sqrt{1-x^2} + C \)
In simple words: Use the substitution x = cos θ to change the square root expression into a half-angle formula. After simplification, the result combines an inverse cosine term with a square root term, each multiplied by a fraction.

Exam Tip: When the integrand contains \( \sqrt{\frac{1-x}{1+x}} \), immediately think of half-angle identities - these often reduce complex expressions to standard forms.

Question 34. Mark (√) against the correct answer in each of the following: \( \int \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) dx = ? \)
(a) 3x tan⁻¹ x + \( \frac{3}{2} \) log (1 + x²) + C
(b) 3x tan⁻¹ x - \( \frac{3}{2} \) log (1 + x²) + C
(c) 3x cos⁻¹ x - \( \frac{3}{2} \) \( \sqrt{1-x^2} \) + C
(d) 3x sin⁻¹ x + \( \frac{3}{2} \) \( \sqrt{1-x^2} \) + C
Answer: (b) 3x tan⁻¹ x - \( \frac{3}{2} \) log (1 + x²) + C
In simple words: The fraction inside tan⁻¹ matches the triple angle formula for tangent, tan(3θ) = (3 tan θ - tan³ θ)/(1 - 3 tan² θ). Setting x = tan θ reveals tan⁻¹(tan 3θ) = 3θ. Apply integration by parts to arrive at the answer.

Exam Tip: Recognize triple-angle formulas in the integrand - they signal that a single substitution will collapse the inverse trigonometric function to a linear multiple of the angle.

Question 35. Mark (√) against the correct answer in each of the following: \( \int x^2 \cos x \, dx = ? \)
(a) x² sin x + 2x cos x - 2 sin x + C
(b) 2x cos x - x sin x + 2 sin x + C
(c) x² sin x - 2x sin x + 2 sin x + C
(d) None of the options
Answer: (a) x² sin x + 2x cos x - 2 sin x + C
In simple words: Apply integration by parts twice. First, let u = x² and dv = cos x dx. After the first application, you get another product to integrate. Apply the method again with the remaining polynomial term to reach the final result.

Exam Tip: For polynomial times trigonometric, apply by-parts as many times as the polynomial's degree - each application lowers the degree by one until you reach a pure trigonometric integral.

Question 36. Mark (√) against the correct answer in each of the following: \( \int \sin x \log(\cos x) dx = ? \)
(a) cos x log (cos x) - cos x + C
(b) -cos x log (cos x) + cos x + C
(c) cos x log (cos x) + cos x + C
(d) None of the options
Answer: (b) -cos x log (cos x) + cos x + C
In simple words: Use the substitution cos x = t, which gives -sin x dx = dt. This changes the integral to a logarithmic integral. Apply by-parts with u = log t and dv = dt to obtain the final form.

Exam Tip: When log appears with a trigonometric function, substitute to convert the trig function into a single variable - this often makes by-parts straightforward.

Question 37. Mark (√) against the correct answer in each of the following: \( \int x \sin x \cos x \, dx = ? \)
(a) -\( \frac{1}{4} \) x cos 2x + \( \frac{1}{8} \) sin 2x + C
(b) \( \frac{1}{4} \) x cos 2x + \( \frac{1}{8} \) sin 2x + C
(c) \( \frac{1}{4} \) x cos 2x - \( \frac{1}{8} \) sin 2x + C
(d) None of the options
Answer: (c) \( \frac{1}{4} \) x cos 2x - \( \frac{1}{8} \) sin 2x + C
In simple words: Simplify using the double-angle identity: 2 sin x cos x = sin 2x, so sin x cos x = \( \frac{1}{2} \) sin 2x. The integral becomes \( \frac{1}{2} \) ∫ x sin 2x dx. Apply by-parts with u = x and dv = sin 2x dx.

Exam Tip: Always apply product-to-sum or double-angle identities before integrating products of trig functions - this reduces the degree of the trigonometric expression and makes by-parts easier.

Question 38. Mark (√) against the correct answer in each of the following: \( \int x^3 \cos x^2 dx = ? \)
(a) x² sin x² + cos x² + C
(b) \( \frac{1}{2} \) x² sin x² + \( \frac{1}{2} \) cos x² + C
(c) -\( \frac{1}{2} \) x² sin x² + \( \frac{1}{2} \) cos x² + C
(d) None of the options
Answer: (b) \( \frac{1}{2} \) x² sin x² + \( \frac{1}{2} \) cos x² + C
In simple words: Substitute x² = t, which gives 2x dx = dt, or x dx = \( \frac{dt}{2} \). Rewrite x³ as x² · x. The integral becomes \( \frac{1}{2} \) ∫ t cos t dt. Apply by-parts with u = t and dv = cos t dt.

Exam Tip: When the integrand has a power of x multiplied by a cosine or sine of x raised to a power, look for a substitution that pairs with the derivative inside the trig function.

Question 39. Mark (√) against the correct answer in each of the following: \( \int \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) dx = ? \)
(a) 2x tan⁻¹ x + log(1 + x²) + C
(b) -2x tan⁻¹ x - 2 log (1 + x²) + C
(c) 2x tan⁻¹ x - log (1 + x²) + C
(d) None of the options
Answer: (d) None of the options
In simple words: The fraction \( \frac{1-x^2}{1+x^2} \) is the double-angle formula for cosine: cos 2θ when x = tan θ. So cos⁻¹(cos 2θ) = 2θ = 2 tan⁻¹ x. Apply integration by parts to get 2x tan⁻¹ x - 2 log √(1 + x²) + C, which simplifies to 2x tan⁻¹ x - log(1 + x²) + C.

Exam Tip: When double-angle identities appear in inverse trig arguments, extract the angle itself - the inverse function cancels with the double-angle, leaving just a multiple of the angle to integrate.

Question 40. Mark (√) against the correct answer in each of the following: \( \int x \tan^{-1} x \, dx = ? \)
(a) \( \frac{1}{2} \)(x² + 1) tan⁻¹ x - \( \frac{1}{2} \) x + C
(b) \( \frac{1}{2} \)(x² - 1) tan⁻¹ x - \( \frac{1}{2} \) x + C
(c) \( \frac{1}{2} \)(x² + 1) tan⁻¹ x + \( \frac{1}{2} \) x + C
(d) None of the options
Answer: (a) \( \frac{1}{2} \)(x² + 1) tan⁻¹ x - \( \frac{1}{2} \) x + C
In simple words: Apply by-parts with u = tan⁻¹ x and dv = x dx. This gives ∫ x² / (1 + x²) dx after the first step. Rewrite x² / (1 + x²) = 1 - 1/(1 + x²) to separate two simpler integrals.

Exam Tip: For inverse trig times polynomial, let the inverse trig be u - this avoids the chain rule in dv and leads to an easier integral after by-parts.

Question 41. Mark (√) against the correct answer in each of the following: \( \int \sin(\log x) dx = ? \)
(a) \( \frac{1}{2} \) x sin log x + \( \frac{1}{2} \) x cos(log x) + C
(b) \( \frac{1}{2} \) x sin log x - \( \frac{1}{2} \) x cos(log x) + C
(c) -\( \frac{1}{2} \) x sin log x + \( \frac{1}{2} \) x cos(log x) + C
(d) None of the options
Answer: (b) \( \frac{1}{2} \) x sin log x - \( \frac{1}{2} \) x cos(log x) + C
In simple words: Apply by-parts with u = sin(log x) and dv = dx. This creates another integral involving cos(log x). Apply by-parts again. The result is a system where the original integral reappears - solve for it algebraically to get the answer.

Exam Tip: When both sin(log x) and cos(log x) integrals appear after the first by-parts, keep applying the method until the original integral shows up - this is a cyclic scenario that requires algebraic manipulation to solve.

Question 42. Mark (√) against the correct answer in each of the following: \( \int (\sin^{-1} x)^2 dx = ? \)
(a) \( \frac{2\sin^{-1}x}{\sqrt{1-x^2}} + C \)
(b) \( \frac{1}{3}(\sin^{-1}x)^3 + \frac{1}{\sqrt{1-x^2}} + C \)
(c) \( x(\sin^{-1}x)^2 + (\sin^{-1}x)\sqrt{1-x^2} + 2x + C \)
(d) \( x(\sin^{-1}x)^2 + 2(\sin^{-1}x)\sqrt{1-x^2} - 2x + C \)
Answer: (d) \( x(\sin^{-1}x)^2 + 2(\sin^{-1}x)\sqrt{1-x^2} - 2x + C \)
In simple words: Put sin t = x, so t = sin⁻¹ x and dx = cos t dt. The integral becomes ∫ t² cos t dt. Apply by-parts twice: first with u = t², then with u = t. Each step reduces the power until only a pure cosine remains.

Exam Tip: For powers of inverse trig functions, a substitution that replaces the inverse trig with a simple angle variable streamlines the by-parts process significantly.

Question 43. Mark (√) against the correct answer in each of the following: \( \int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx = ? \)
(a) e^x (log x + 1/x) + C
(b) xe^x - e^x + C
(c) e^x · 1/x + C
(d) None of the options
Answer: (c) e^x · \( \frac{1}{x} \) + C
In simple words: Notice that f(x) = 1/x and f'(x) = -1/x². The integrand has the form e^x [f(x) + f'(x)]. This matches the formula ∫ e^x [f(x) + f'(x)] dx = e^x f(x) + C, which is a shortcut when the derivative appears alongside the original function.

Exam Tip: When e^x multiplies a sum that includes both a function and its derivative, recognize the pattern e^x [f(x) + f'(x)] - the answer is simply e^x f(x) plus a constant, with no by-parts needed.

Question 44. Mark (√) against the correct answer in each of the following: \( \int e^x \left(\frac{1}{x^2} - \frac{2}{x^3}\right) dx = ? \)
(a) \( \frac{-e^x}{x^2} + C \)
(b) \( \frac{e^x}{x^2} + C \)
(c) \( e^x \left(\frac{-1}{x} + \frac{1}{x^2}\right) + C \)
(d) None of the options
Answer: (b) \( \frac{e^x}{x^2} + C \)
In simple words: Identify that f(x) = 1/x² and f'(x) = -2/x³. The integrand matches the pattern e^x [f(x) + f'(x)] dx = e^x f(x) + C. Apply this formula directly without by-parts.

Exam Tip: Always check whether the integrand fits the e^x [f + f'] pattern before resorting to by-parts - recognizing this form saves significant calculation time

Question 45. Mark (√) against the correct answer in each of the following: \( \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1-x^2}} \right) dx = ? \)
(a) \( e^x \cdot \frac{1}{\sqrt{1-x^2}} + C \)
(b) \( e^x \sin^{-1} x + C \)
(c) \( \frac{-e^x}{\sin^{-1} x} + C \)
(d) None of these
Answer: To find the value of \( \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1-x^2}} \right) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1-x^2}} \right) dx \) ... (i)
Here \( f(x) = \sin^{-1} x \)
\( \Rightarrow f'(x) = \frac{1}{\sqrt{1-x^2}} \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = e^x \sin^{-1} x + c \)
The correct choice is (b) \( e^x \sin^{-1} x + C \)

Exam Tip: Recognize integrals of the form \( e^x(f(x) + f'(x)) \) instantly - the result is always \( e^x f(x) + C \). This pattern saves significant time in examinations.

Question 46. Mark (√) against the correct answer in each of the following: \( \int e^x (\tan x + \log \sec x) dx = ? \)
(a) \( e^x \log \sec x + C \)
(b) \( e^x \tan x + C \)
(c) \( e^x (\log \cos x) + C \)
(d) None of these
Answer: To find the value of \( \int e^x (\tan x + \log \sec x) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x (\tan x + \log \sec x) dx \) ... (i)
\( \Rightarrow I = \int e^x (\tan x - \log \cos x) dx \)
Here \( f(x) = -\log(\cos x) \)
\( \Rightarrow f'(x) = \tan x \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = -e^x \log(\cos x) + c \)
\( \Rightarrow I = e^x \log(\sec x) + c \)
The correct choice is (a) \( e^x \log \sec x + C \)

Exam Tip: When you see logarithmic functions combined with trigonometric terms inside an exponential integral, convert \( \log \sec x \) to \( -\log \cos x \) to identify the derivative relationship more clearly.

Question 47. Mark (√) against the correct answer in each of the following: \( \int e^x (\tan x + \log \sec x) dx = ? \)
(a) \( e^x \log \sec x + C \)
(b) \( e^x \tan x + C \)
(c) \( e^x (\log \cos x) + C \)
(d) None of these
Answer: To find the value of \( \int e^x (\tan x + \log \sec x) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x (\tan x + \log \sec x) dx \) ... (i)
\( \Rightarrow I = \int e^x (\tan x - \log \cos x) dx \)
Here \( f(x) = -\log(\cos x) \)
\( \Rightarrow f'(x) = \tan x \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = -e^x \log(\cos x) + c \)
\( \Rightarrow I = e^x \log(\sec x) + c \)
The correct choice is (a) \( e^x \log(\sec x) + C \)

Exam Tip: Notice that this question repeats Question 46 exactly. Always verify your working once more and match the algebraic transformations (particularly converting between \( \log \sec \) and \( -\log \cos \)) correctly.

Question 48. Mark (√) against the correct answer in each of the following: \( \int e^x (\cot x + \log \sin x) dx = ? \)
(a) \( e^x \log (\sec x + \tan x) + C \)
(b) \( e^x \sec x + C \)
(c) \( e^x \log \tan x + C \)
(d) None of these
Answer: To find the value of \( \int e^x (\cot x + \log \sin x) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x (\cot x + \log \sin x) dx \) ... (i)
Here \( f(x) = \log(\sin x) \)
\( \Rightarrow f'(x) = \cot x \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = e^x \log(\sin x) + c \)
The correct choice is (d) None of these

Exam Tip: When the integrand contains a logarithmic function and its derivative as separate terms, immediately recognize this as the pattern \( e^x(f(x) + f'(x)) \) and apply the standard result directly without expanding further steps.

Question 49. Mark (√) against the correct answer in each of the following: \( \int e^x \left( \tan^{-1} x + \frac{1}{(1+x^2)} \right) dx = ? \)
(a) \( e^x \cdot \frac{1}{(1+x^2)} + C \)
(b) \( e^x \tan^{-1} x + C \)
(c) \( -e^x \cot^{-1} x + C \)
(d) None of these
Answer: To find the value of \( \int e^x \left( \tan^{-1} x + \frac{1}{(1+x^2)} \right) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x \left( \tan^{-1} x + \frac{1}{(1+x^2)} \right) dx \) ... (i)
Here \( f(x) = \tan^{-1} x \)
\( \Rightarrow f'(x) = \frac{1}{(1+x)^2} \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = e^x (\tan^{-1} x) + c \)
The correct choice is (b) \( e^x (\tan^{-1} x) + C \)

Exam Tip: For inverse trigonometric functions paired with their derivatives inside exponential integrals, the result follows the template \( e^x \cdot (\text{inverse function}) + C \) - memorize this pattern to solve these quickly.

Question 50. Mark (√) against the correct answer in each of the following: \( \int e^x (\tan x - \log \cos x) dx = ? \)
(a) \( e^x \tan x + C \)
(b) \( e^x \log \cos x + C \)
(c) \( e^x \log \sec x + C \)
(d) None of these
Answer: To find the value of \( \int e^x (\tan x - \log \cos x) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x (\tan x - \log(\cos x)) dx \) ... (i)
Here \( f(x) = -\log(\cos x) \)
\( \Rightarrow f'(x) = \tan x \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = -e^x \log(\cos x) + c \)
\( \Rightarrow I = e^x \log(\sec x) + c \)
The correct choice is (c) \( e^x \log \sec x + C \)

Exam Tip: Always convert between \( \log(\cos x) \) and \( \log(\sec x) \) using the identity \( -\log(\cos x) = \log(\sec x) \) to match the answer options correctly.

Question 51. Mark (√) against the correct answer in each of the following: \( \int e^x (\cot x - \csc^2 x) dx = ? \)
(a) \( -e^x \csc^2 x + C \)
(b) \( e^x \cot x + C \)
(c) \( -e^x \cot x + C \)
(d) None of these
Answer: To find the value of \( \int e^x (\cot x - \csc^2 x) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x (\cot x - \csc^2 x) dx \) ... (i)
Here \( f(x) = \cot x \)
\( \Rightarrow f'(x) = -\csc^2 x \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = e^x \cot x + c \)
The correct choice is (b) \( e^x \cot x + C \)

Exam Tip: Recognize that \( -\csc^2 x \) is the derivative of \( \cot x \). When you spot a function plus its derivative multiplied by \( e^x \), the answer is always \( e^x \) times the function itself.

Question 52. Mark (√) against the correct answer in each of the following: \( \int e^x (\sin x + \cos x) dx = ? \)
(a) \( e^x \sin x + C \)
(b) \( e^x \cos x + C \)
(c) \( e^x \tan x + C \)
(d) None of these
Answer: To find the value of \( \int e^x (\sin x + \cos x) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x (\sin x + \cos x) dx \) ... (i)
Here \( f(x) = \sin x \)
\( \Rightarrow f'(x) = \cos x \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = e^x \sin x + c \)
The correct choice is (a) \( e^x \sin x + C \)

Exam Tip: In problems involving sine and cosine, always check whether the two terms form a function-derivative pair (sine and cosine are derivatives of each other) before attempting any other integration method.

Question 53. Mark (√) against the correct answer in each of the following: \( \int e^x \sec x (1 + \tan x) dx = ? \)
(a) \( e^x (1 + \tan x) + C \)
(b) \( e^x \sec x + C \)
(c) \( e^x \tan x + C \)
(d) None of these
Answer: To find the value of \( \int e^x \sec x (1 + \tan x) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x \sec x (1 + \tan x) dx \) ... (i)
\( I = \int e^x (\sec x + \sec x \tan x) dx \)
Here \( f(x) = \sec x \)
\( \Rightarrow f'(x) = \sec x \tan x \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = e^x \sec x + c \)
The correct choice is (b) \( e^x \sec x + C \)

Exam Tip: Always expand products inside the integral first - here \( \sec x(1 + \tan x) \) expands to reveal the derivative pattern. Identifying this structure is the key to solving these problems efficiently.

Question 54. Mark (√) against the correct answer in each of the following: \( \int e^x \left( \frac{1 + x \log x}{x} \right) dx = ? \)
(a) \( e^x \cdot \frac{1}{x} + C \)
(b) \( e^x \log x + C \)
(c) \( x e^x \log x + C \)
(d) None of these
Answer: To find the value of \( \int e^x \left( \frac{1 + x \log x}{x} \right) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x \left( \frac{1 + x \log x}{x} \right) dx \) ... (i)
\( I = \int e^x \left( \frac{1}{x} + \log x \right) dx \)
\( \Rightarrow I = \int e^x \left( \frac{1}{x} - \frac{1}{x} + \log x \right) dx \)
Here \( f(x) = \log x \)
\( \Rightarrow f'(x) = \frac{1}{x} \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = e^x \log x + c \)
The correct choice is (b) \( e^x \log x + C \)

Exam Tip: When dealing with fractions in exponential integrals, separate the terms and check if one is a logarithmic function and the other is its derivative - this reveals the standard pattern instantly.

Question 55. Mark (√) against the correct answer in each of the following: \( \int e^x \frac{x}{(1+x)^2} dx = ? \)
(a) \( e^x \cdot \frac{1}{(1+x)} + C \)
(b) \( e^x \cdot \frac{1}{x} + C \)
(c) \( e^x \cdot \frac{x}{(1+x)} + C \)
(d) None of these
Answer: To find the value of \( \int e^x \frac{x}{(1+x)^2} dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x \frac{x}{(1+x)^2} dx \) ... (i)
\( I = \int e^x \left( \frac{x+1-1}{(1+x)^2} \right) dx \)
\( \Rightarrow I = \int e^x \left( \frac{1}{(1+x)} - \frac{1}{(1+x)^2} \right) dx \)
Here \( f(x) = \frac{1}{(1+x)} \)
\( \Rightarrow f'(x) = - \frac{1}{(1+x)^2} \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = e^x \cdot \frac{1}{(1+x)} + c \)
The correct choice is (a) \( e^x \cdot \frac{1}{(1+x)} + C \)

Exam Tip: When the integrand contains a rational function, decompose the numerator cleverly (for example, writing \( x = (x+1) - 1 \)) to expose the derivative relationship and simplify the integral.

Question 56. Mark (√) against the correct answer in each of the following: \( \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx = ? \)
(a) \( e^x \sin \frac{x}{2} + C \)
(b) \( e^x \cos \frac{x}{2} + C \)
(c) \( e^x \tan \frac{x}{2} + C \)
(d) None of these
Answer: To find the value of \( \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx \)
Formula used: \( \int f(x)g(x)dx = f(x) \int g(x)dx - \int \left[ f'(x) \int g(x)dx \right] dx \)
We have, \( I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx \) ... (i)
\( I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx \)
\( \Rightarrow I = \int e^x \left( \frac{1}{1 + \cos x} + \frac{\sin x}{1 + \cos x} \right) dx \)
\( \Rightarrow I = \int e^x \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx \)
\( \Rightarrow I = \int e^x \left( \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx \)
Here \( f(x) = \tan \frac{x}{2} \)
\( \Rightarrow f'(x) = \frac{1}{2} \sec^2 \frac{x}{2} \)
\( \Rightarrow I = \int e^x (f(x) + f'(x)) dx \)
\( \Rightarrow I = e^x f(x) + c \)
\( \Rightarrow I = e^x \tan \frac{x}{2} + c \)
The correct choice is (c) \( e^x \tan \frac{x}{2} + C \)

Exam Tip: For trigonometric fractions, apply half-angle formulas (\( 1 + \cos x = 2 \cos^2 \frac{x}{2} \) and \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \)) to simplify and reveal the function-derivative structure needed for the standard pattern.