RS Aggarwal Solutions for Class 12 Chapter 16 Definite Integrals

Question 1. Evaluate: \( \int_{1}^{3} x^4 dx \)
Answer: \( \frac{242}{5} \)
In simple words: Find the antiderivative of \( x^4 \), which is \( \frac{x^5}{5} \), then substitute the upper and lower limits and subtract.

Exam Tip: Always apply the power rule for integration: \( \int x^n dx = \frac{x^{n+1}}{n+1} \), then use the Fundamental Theorem of Calculus to evaluate at the bounds.

Question 2. Evaluate: \( \int_{1}^{4} \sqrt{x} dx \)
Answer: \( \frac{14}{3} \)
In simple words: Rewrite \( \sqrt{x} \) as \( x^{1/2} \), integrate using the power rule to get \( \frac{2}{3}x^{3/2} \), then evaluate between the limits.

Exam Tip: Remember to simplify fractional exponents carefully and always double-check your arithmetic when substituting the boundary values.

Question 3. Evaluate: \( \int_{1}^{2} x^{-5} dx \)
Answer: \( \frac{15}{64} \)
In simple words: Use the power rule to get \( \frac{x^{-4}}{-4} \), then substitute the bounds and compute the difference between the two evaluations.

Exam Tip: Negative exponents require careful handling when using the power rule - the sign and the magnitude both matter in your final calculation.

Question 4. Evaluate: \( \int_{0}^{16} x^{3/4} dx \)
Answer: \( \frac{512}{7} \)
In simple words: Apply the power rule to \( x^{3/4} \) to get \( \frac{4}{7}x^{7/4} \), then substitute both limits and simplify.

Exam Tip: When working with fractional exponents at zero, remember that \( 0^{7/4} = 0 \), so the lower limit always contributes zero to the final answer.

Question 5. Evaluate: \( \int_{-4}^{-1} \frac{dx}{x} \)
Answer: \( -\log 4 \)
In simple words: The antiderivative of \( \frac{1}{x} \) is the natural logarithm. Evaluate \( \log |x| \) at both bounds, then simplify using logarithm properties.

Exam Tip: Always use absolute value inside the logarithm when dealing with \( \frac{1}{x} \), even when the variable is negative - this ensures the logarithm remains defined.

Question 6. Evaluate: \( \int_{1}^{4} \frac{dx}{\sqrt{x}} \)
Answer: \( 2 \)
In simple words: Rewrite the integrand as \( x^{-1/2} \) and integrate to obtain \( 2\sqrt{x} \). Then evaluate at the upper and lower limits and find the difference.

Exam Tip: Square roots in denominators can always be converted to negative fractional exponents, making the power rule directly applicable.

Question 7. Evaluate: \( \int_{0}^{1} \frac{dx}{\sqrt[3]{x}} \)
Answer: \( \frac{3}{2} \)
In simple words: Rewrite as \( x^{-1/3} \), apply the power rule to get \( \frac{3}{2}x^{2/3} \), then substitute the limits and compute the result.

Exam Tip: When the integrand has a fractional or negative exponent, always express it explicitly before applying the power rule to avoid calculation errors.

Question 8. Evaluate: \( \int_{1}^{8} \frac{dx}{x^{2/3}} \)
Answer: \( 3 \)
In simple words: Convert to \( x^{-2/3} \) and use the power rule to get \( 3x^{1/3} \). Evaluate at both limits to find the definite integral.

Exam Tip: Cube roots and higher roots often appear with fractional exponents - converting to exponent form makes integration straightforward.

Question 9. Evaluate: \( \int_{2}^{4} 3 dx \)
Answer: \( 6 \)
In simple words: The integral of a constant is the constant times the variable. So \( \int 3 dx = 3x \), and evaluating between 2 and 4 gives the answer.

Exam Tip: Do not forget that the integral of any constant \( c \) is simply \( cx \) - this is one of the most basic but frequently tested integral formulas.

Question 10. Evaluate: \( \int_{0}^{1} \frac{dx}{1 + x^2} \)
Answer: \( \frac{\pi}{4} \)
In simple words: The antiderivative of \( \frac{1}{1+x^2} \) is \( \tan^{-1}(x) \). Substitute the bounds to get \( \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 \).

Exam Tip: Memorize the standard integral \( \int \frac{dx}{1+x^2} = \tan^{-1}(x) + C \) - it appears frequently on exams and is easy to forget.

Question 11. Evaluate: \( \int_{0}^{\infty} \frac{dx}{1 + x^2} \)
Answer: \( \frac{\pi}{2} \)
In simple words: The antiderivative is still \( \tan^{-1}(x) \). As \( x \) approaches infinity, \( \tan^{-1}(x) \) approaches \( \frac{\pi}{2} \), and at \( x = 0 \) it equals 0.

Exam Tip: When evaluating improper integrals, use limits carefully - \( \tan^{-1}(\infty) = \frac{\pi}{2} \) and \( \tan^{-1}(0) = 0 \) are essential facts.

Question 12. Evaluate: \( \int_{0}^{1} \frac{dx}{\sqrt{1 - x^2}} \)
Answer: \( \frac{\pi}{2} \)
In simple words: The antiderivative of \( \frac{1}{\sqrt{1-x^2}} \) is \( \sin^{-1}(x) \). Evaluate at the bounds: \( \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} \).

Exam Tip: Remember the standard form \( \int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}(x) + C \) - this inverse trigonometric integral is tested regularly.

Question 13. Evaluate: \( \int_{0}^{\pi/6} \sec^2 x dx \)
Answer: \( \frac{1}{\sqrt{3}} \)
In simple words: The antiderivative of \( \sec^2 x \) is \( \tan x \). Substitute the limits to get \( \tan(\pi/6) - \tan(0) = \frac{1}{\sqrt{3}} \).

Exam Tip: Know that \( \int \sec^2 x dx = \tan x + C \) and recall the exact values of trigonometric functions at standard angles like \( \pi/6 \), \( \pi/4 \), and \( \pi/3 \).

Question 14. Evaluate: \( \int_{-\pi/4}^{\pi/4} \csc^2 x dx \)
Answer: \( -2 \)
In simple words: The antiderivative of \( \csc^2 x \) is \( -\cot x \). Evaluate at the upper and lower bounds to get the final result.

Exam Tip: Remember that \( \int \csc^2 x dx = -\cot x + C \) and be careful with negative signs when working with symmetric intervals around zero.

Question 15. Evaluate: \( \int_{\pi/4}^{\pi/2} \cot^2 x dx \)
Answer: \( \left(1 - \frac{\pi}{4}\right) \)
In simple words: Use the identity \( \cot^2 x = \csc^2 x - 1 \) to rewrite the integral, then integrate term-by-term and apply the limits.

Exam Tip: Trigonometric identities are essential tools - use \( \cot^2 x = \csc^2 x - 1 \) to simplify integrals involving cotangent squared.

Question 16. Evaluate: \( \int_{0}^{\pi/4} \tan^2 x dx \)
Answer: \( \left(1 - \frac{\pi}{4}\right) \)
In simple words: Apply the identity \( \tan^2 x = \sec^2 x - 1 \) to split the integral into two parts. Integrate each separately and evaluate at the limits.

Exam Tip: The identity \( \tan^2 x = \sec^2 x - 1 \) is a powerful tool for handling integrals of squared tangent - use it whenever you see \( \tan^2 x \).

Question 17. Evaluate: \( \int_{0}^{\pi/2} \sin^2 x dx \)
Answer: \( \frac{\pi}{4} \)
In simple words: Use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) to rewrite the integrand, then integrate and apply the bounds.

Exam Tip: Always use the power-reduction formula \( \sin^2 x = \frac{1 - \cos 2x}{2} \) when integrating squared sine - it converts a difficult integral into two simple ones.

Question 18. Evaluate: \( \int_{0}^{\pi/4} \cos^2 x dx \)
Answer: \( \left(\frac{\pi}{8} + \frac{1}{4}\right) \)
In simple words: Use the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \) to simplify, then integrate each part separately and evaluate at the bounds.

Exam Tip: The power-reduction formula for cosine is \( \cos^2 x = \frac{1 + \cos 2x}{2} \) (note the plus sign, unlike the sine version) - getting this right is crucial.

Question 19. Evaluate: \( \int_{0}^{\pi/3} \tan x dx \)
Answer: \( \log 2 \)
In simple words: Rewrite \( \tan x = \frac{\sin x}{\cos x} \) and integrate using substitution with \( u = \cos x \). Evaluate at the limits to obtain the answer.

Exam Tip: For \( \int \tan x dx \), use the substitution method or remember that \( \int \tan x dx = \log |\sec x| + C = -\log |\cos x| + C \).

Question 20. Evaluate: \( \int_{\pi/6}^{\pi/4} \csc x dx \)
Answer: \( \log(\sqrt{2} - 1) + \log(2 + \sqrt{3}) \)
In simple words: The antiderivative of \( \csc x \) is \( -\log |\csc x + \cot x| \) or \( \log |\csc x - \cot x| \). Substitute the bounds and use logarithm rules to simplify.

Exam Tip: The integral of cosecant is tricky - memorize \( \int \csc x dx = -\log |\csc x + \cot x| + C \) and practice substituting trigonometric values carefully.

Question 21. Evaluate: \( \int_{0}^{\pi/3} \cos^3 x dx \)
Answer: \( \frac{3\sqrt{3}}{8} \)
In simple words: Use the identity \( \cos^3 x = \frac{1}{4}(3\cos x + \cos 3x) \) to rewrite the integrand, then integrate each term separately and evaluate at the bounds.

Exam Tip: For higher powers of cosine, use the appropriate reduction formula - this saves time and reduces calculation errors significantly.

Question 22. Evaluate: \( \int_{0}^{\pi/2} \sin^3 x dx \)
Answer: \( \frac{2}{3} \)
In simple words: Use the identity \( \sin^3 x = \frac{1}{4}(3\sin x - \sin 3x) \) to express the integrand, then integrate term-by-term and substitute the bounds.

Exam Tip: Reduction formulas for powers of sine and cosine are key techniques - memorize at least one or two of these to speed up your integration work.

Question 23. Evaluate: \( \int_{\pi/4}^{\pi/2} \frac{(1 - 3\cos x)}{\sin^2 x} dx \)
Answer: \( (4 - 3\sqrt{2}) \)
In simple words: Rewrite the integrand as \( \csc^2 x - 3\csc x \cot x \). Integrate each term - the first gives \( -\cot x \) and the second gives \( 3\csc x \) - then evaluate at the limits.

Exam Tip: Always split rational trigonometric expressions into simpler forms before integrating - this approach often reveals standard integral patterns.

Question 24. Evaluate: \( \int_{0}^{\pi/4} \sqrt{1 + \cos 2x} dx \)
Answer: \( 1 \)
In simple words: Use the identity \( 1 + \cos 2x = 2\cos^2 x \) so that \( \sqrt{1 + \cos 2x} = \sqrt{2}\cos x \). Integrate and apply the limits to get the result.

Exam Tip: When you see \( 1 + \cos 2x \) or similar expressions under a square root, use double-angle identities to simplify - this often converts a difficult integral into an elementary one.

Question 25. Evaluate: \( \int_{0}^{\pi/4} \sqrt{1 - \sin 2x} dx \)
Answer: \( (\sqrt{2} - 1) \)
In simple words: Note that \( 1 - \sin 2x = \sin^2 x + \cos^2 x - 2\sin x \cos x = (\cos x - \sin x)^2 \). So \( \sqrt{1 - \sin 2x} = |\cos x - \sin x| = \cos x - \sin x \) on this interval. Integrate to get the answer.

Exam Tip: Recognize perfect square patterns under radicals - expressions like \( (\cos x - \sin x)^2 \) simplify dramatically and make integration straightforward.

Question 26. Evaluate: \( \int_{-\pi/4}^{\pi/4} \frac{dx}{1 + \sin x} \)
Answer: \( 2 \)
In simple words: Multiply numerator and denominator by \( \sec^2(x/2) \) and use the substitution \( u = \tan(x/2) + 1 \) to convert this into a simpler integral. Evaluate the result at both bounds.

Exam Tip: For integrals involving \( 1 + \sin x \) in the denominator, the Weierstrass substitution or rationalizing techniques work well - practice both approaches.

Question 27. Evaluate: \( \int_{0}^{\pi/4} \frac{dx}{1 + \cos 2x} \)
Answer: \( \frac{1}{2} \)
In simple words: Use the identity \( 1 + \cos 2x = 2\cos^2 x \) to rewrite the denominator. The integral becomes \( \int \frac{1}{2\cos^2 x} dx = \frac{1}{2}\int \sec^2 x dx = \frac{1}{2}\tan x \). Evaluate at the bounds.

Exam Tip: Double-angle identities such as \( 1 + \cos 2x = 2\cos^2 x \) and \( 1 - \cos 2x = 2\sin^2 x \) are invaluable for simplifying trigonometric integrals.

Question 28. Evaluate: \( \int_{\pi/4}^{\pi/2} \frac{dx}{1 - \cos 2x} \)
Answer: \( \frac{1}{2} \)
In simple words: Apply the identity \( 1 - \cos 2x = 2\sin^2 x \) to transform the denominator. The integral becomes \( \frac{1}{2}\int \csc^2 x dx = -\frac{1}{2}\cot x \). Substitute the bounds to find the answer.

Exam Tip: Use the identity \( 1 - \cos 2x = 2\sin^2 x \) just as you would use \( 1 + \cos 2x = 2\cos^2 x \) - these are complementary tools that appear in many problems.

Question 29. Evaluate: \( \int_{0}^{\pi/4} \sin 2x \sin 3x dx \)
Answer: \( \frac{3}{5\sqrt{2}} \)
In simple words: Use the product-to-sum formula: \( \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \). This gives \( \sin 2x \sin 3x = \frac{1}{2}(\cos x - \cos 5x) \). Integrate each part and evaluate at the limits.

Exam Tip: Product-to-sum formulas convert products of trigonometric functions into sums, which are much easier to integrate - always check for these opportunities.

Question 30. Evaluate: \( \int_{0}^{\pi/6} \cos x \cos 2x dx \)
Answer: \( \frac{5}{12} \)
In simple words: Use the product-to-sum formula: \( \cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)] \). This transforms the product into \( \frac{1}{2}(\cos 3x + \cos x) \). Integrate and evaluate at the bounds.

Exam Tip: The product-to-sum formula for cosines has a plus sign: \( \cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)] \) - this is different from the sine version and worth memorizing.

Question 31. Evaluate: \( \int_{0}^{\pi} \sin 2x \cos 3x dx \)
Answer: \( -\frac{4}{5} \)
In simple words: Apply the product-to-sum formula for mixed products: \( \sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)] \). This gives \( \sin 2x \cos 3x = \frac{1}{2}(\sin 5x - \sin x) \). Integrate and substitute the bounds.

Exam Tip: For products of sine and cosine, the formula is \( \sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)] \) - note the plus sign in the middle, and be careful with your angle arithmetic.

Question 32. Evaluate: \( \int_{0}^{\pi/2} \sqrt{1 + \sin x} dx \)
Answer: \( 2 \)
In simple words: Notice that \( 1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2\sin(x/2)\cos(x/2) = (\sin(x/2) + \cos(x/2))^2 \). So \( \sqrt{1 + \sin x} = \sin(x/2) + \cos(x/2) \). Integrate this and evaluate at the limits.

Exam Tip: Always look for perfect square patterns in expressions under a square root - completing the square or recognizing a squared binomial can transform a difficult problem into a simple one.

Question 33. Evaluate: \( \int_{0}^{\pi/2} \sqrt{1 + \cos x} dx \)
Answer: \( 2 \)
In simple words: Use the identity \( 1 + \cos x = 2\cos^2(x/2) \) to get \( \sqrt{1 + \cos x} = \sqrt{2}\cos(x/2) \). Integrate \( \sqrt{2}\cos(x/2) \) to obtain \( 2\sqrt{2}\sin(x/2) \), then evaluate at the bounds.

Exam Tip: The half-angle identities \( 1 + \cos x = 2\cos^2(x/2) \) and \( 1 - \cos x = 2\sin^2(x/2) \) are essential for simplifying expressions under square roots.

Question 34. Evaluate: \( \int_{0}^{2} \frac{(x^4 + 1)}{(x^2 + 1)} dx \)
Answer: \( \left(\frac{2}{3} + 2\tan^{-1}(2)\right) \)
In simple words: Divide the numerator by the denominator using polynomial long division: \( \frac{x^4 + 1}{x^2 + 1} = x^2 - 1 + \frac{2}{x^2 + 1} \). Integrate each part separately: \( x^2 \) and \( -1 \) integrate easily, while \( \frac{2}{x^2 + 1} \) gives \( 2\tan^{-1}(x) \). Evaluate at both limits.

Exam Tip: When the degree of the numerator is equal to or greater than the degree of the denominator, use polynomial long division first - this always simplifies the problem into manageable pieces.

Question 35. Evaluate: \( \int_{1}^{2} \frac{dx}{(x+1)(x+2)} \)
Answer: \( 2\log 3 - 3\log 2 \)
In simple words: Break the fraction into two simpler pieces using partial fractions, integrate each one separately, then substitute the limits to get your final answer.

Exam Tip: Always use partial fraction decomposition for rational functions with distinct linear factors in the denominator - it makes integration straightforward.

Question 36. Evaluate: \( \int_{1}^{2} \frac{(x+3)}{x(x+2)} dx \)
Answer: \( \frac{1}{2}(\log 2 + \log 3) \)
In simple words: Decompose the fraction into simpler parts, integrate each term, then evaluate at the upper and lower bounds to find your result.

Exam Tip: Double-check your partial fraction setup by multiplying back - this catches errors before you waste time integrating.

Question 37. Evaluate: \( \int_{2}^{4} \frac{dx}{x^2-4} \)
Answer: \( \frac{1}{4}(\log 5 - \log 3) \)
In simple words: Factor the denominator as a difference of squares, use partial fractions to split it into two easy pieces, then integrate and apply your limits.

Exam Tip: Recognize \( x^2 - a^2 = (x-a)(x+a) \) immediately - this pattern appears in many integral problems.

Question 38. Evaluate: \( \int_{0}^{4} \frac{dx}{\sqrt{x^2+2x+3}} \)
Answer: \( \log\left(\frac{5+3\sqrt{3}}{1+\sqrt{3}}\right) \)
In simple words: Complete the square inside the square root to get it into the form \( \sqrt{(x+1)^2 + 2} \), then use a substitution to match a standard integral formula involving logarithms.

Exam Tip: For integrals with \( \sqrt{x^2 + ax + b} \), completing the square is almost always the first step - it transforms the problem into a standard form.

Question 39. Evaluate: \( \int_{1}^{2} \frac{dx}{\sqrt{x^2+4x+3}} \)
Answer: \( \log(4+\sqrt{15}) - \log(3+\sqrt{8}) \)
In simple words: Complete the square under the radical to transform it into \( \sqrt{(x+2)^2 - 1} \), then apply the inverse hyperbolic sine integral formula with your bounds.

Exam Tip: When the discriminant of \( x^2 + ax + b \) is negative after completing the square, you get an inverse sine result; if positive, you get a logarithmic result.

Question 40. Evaluate: \( \int_{0}^{1} \frac{dx}{(1+x+2x^2)} \)
Answer: \( \frac{2}{\sqrt{7}}\left[\tan^{-1}\frac{5}{\sqrt{7}} - \tan^{-1}\frac{1}{\sqrt{7}}\right] \)
In simple words: Rewrite the quadratic in the denominator by completing the square, then use a substitution to match the arctangent integral formula.

Exam Tip: For denominators of the form \( ax^2 + bx + c \), completing the square yields either an arctan or arcsine result depending on the sign of the constant term.

Question 41. Evaluate: \( \int_{0}^{\pi/2} (a\cos^2 x + b\sin^2 x) dx \)
Answer: \( \frac{\pi}{4}(a+b) \)
In simple words: Rewrite the squared trig functions using double angle formulas, integrate term by term, then substitute your limits to simplify.

Exam Tip: Always use \( \cos^2 x = \frac{1+\cos 2x}{2} \) and \( \sin^2 x = \frac{1-\cos 2x}{2} \) - these formulae make squared trigonometric integrals manageable.

Question 42. Evaluate: \( \int_{\pi/3}^{\pi/4} (\tan x + \cot x)^2 dx \)
Answer: \( -\frac{2}{\sqrt{3}} \)
In simple words: Expand the squared binomial, simplify using trig identities like \( \tan^2 x = \sec^2 x - 1 \), then integrate piece by piece and evaluate at your bounds.

Exam Tip: When you see powers of tan and cot together, expand first, then use the Pythagorean identities to convert everything to sec and csc.

Question 43. Evaluate: \( \int_{0}^{\pi/2} \cos^4 x \, dx \)
Answer: \( \frac{3\pi}{16} \)
In simple words: Apply the reduction formula repeatedly or use the identity \( \cos^2 x = \frac{1+\cos 2x}{2} \) twice to reduce the power, then integrate.

Exam Tip: For even powers of cos or sin over standard intervals like \( [0, \pi/2] \), reduction formulas save time and reduce arithmetic mistakes.

Question 44. Evaluate: \( \int_{0}^{a} \frac{dx}{(ax+a^2-x^2)} \)
Answer: \( \frac{1}{\sqrt{5a}} \log\left(\frac{7+3\sqrt{5}}{2}\right) \)
In simple words: Rearrange the denominator as a quadratic in standard form, complete the square, then use partial fractions or a trig substitution to evaluate.

Exam Tip: When a parameter like \( a \) appears in the integral, carefully track it through each step - many errors occur when handling parameters.

Question 45. Evaluate: \( \int_{1/4}^{1/2} \frac{dx}{\sqrt{x-x^2}} \)
Answer: \( \frac{\pi}{6} \)
In simple words: Complete the square under the radical to get \( \sqrt{\frac{1}{4} - (x - \frac{1}{2})^2} \), then use the arcsine integral formula with your bounds.

Exam Tip: The pattern \( \sqrt{x - x^2} \) always suggests completing the square to get \( \sqrt{a^2 - (x-b)^2} \), which gives arcsine.

Question 46. Evaluate: \( \int_{0}^{1} \sqrt{x(1-x)} \, dx \)
Answer: \( \frac{\pi}{8} \)
In simple words: Rewrite the expression under the radical as \( \sqrt{\frac{1}{4} - (x - \frac{1}{2})^2} \), then use a trig substitution or the arcsine formula to integrate.

Exam Tip: Whenever you see a product like \( x(1-x) \), complete the square immediately to expose a circular arc pattern.

Question 47. Evaluate: \( \int_{1}^{3} \frac{dx}{x^2(x+1)} \)
Answer: \( \log 2 - \log 3 + \frac{2}{3} \)
In simple words: Use partial fractions to break the rational function into three simpler fractions, integrate each separately, then apply your limits.

Exam Tip: For fractions with a repeated factor like \( x^2 \), include both \( \frac{A}{x} \) and \( \frac{B}{x^2} \) in your partial fraction setup.

Question 48. Evaluate: \( \int_{1}^{2} \frac{dx}{x(1+2x)^2} \)
Answer: \( \log 6 - \log 5 - \frac{2}{15} \)
In simple words: Decompose using partial fractions into terms with \( x \), \( (1+2x) \), and \( (1+2x)^2 \), then integrate and substitute your limits.

Exam Tip: For repeated linear factors, each power from 1 up to the repetition count needs its own numerator constant in the decomposition.

Question 49. Evaluate: \( \int_{0}^{1} x e^x dx \)
Answer: \( 1 \)
In simple words: Use integration by parts with \( u = x \) and \( dv = e^x dx \), simplify, then evaluate at your bounds.

Exam Tip: For products of polynomial and exponential functions, always choose the polynomial as \( u \) so the derivative eventually vanishes.

Question 50. Evaluate: \( \int_{0}^{\pi/2} x^2 \cos x \, dx \)
Answer: \( \left(\frac{\pi^2}{4} - 2\right) \)
In simple words: Apply integration by parts twice - first with \( u = x^2 \), then with \( u = 2x \) - until you reach an easy integral, then evaluate at your limits.

Exam Tip: For polynomials multiplied by trig functions, repeat integration by parts as many times as the polynomial's degree; each application reduces the power by one.

Question 51. Evaluate: \( \int_{0}^{\pi/4} x^2 \sin x \, dx \)
Answer: \( \left(\sqrt{2} + \frac{\pi}{2\sqrt{2}} - \frac{\pi^2}{16\sqrt{2}} - 2\right) \)
In simple words: Use integration by parts twice - once with \( u = x^2 \) and again with \( u = 2x \) - to reduce the polynomial, then evaluate at your bounds.

Exam Tip: Keep careful track of signs when applying integration by parts repeatedly; a single sign error early will cascade through your entire solution.

Question 52. Evaluate: \( \int_{0}^{\pi/2} x^2 \cos 2x \, dx \)
Answer: \( -\frac{\pi}{4} \)
In simple words: Apply integration by parts twice to handle the \( x^2 \) term, then substitute your limits carefully to get the final answer.

Exam Tip: When the trig function has a coefficient like 2 in \( \cos 2x \), keep that coefficient in mind during each integration by parts step.

Question 53. Evaluate: \( \int_{0}^{\pi/2} x^3 \sin 3x \, dx \)
Answer: \( \left(\frac{2}{27} - \frac{\pi^2}{12}\right) \)
In simple words: Use integration by parts three times in succession to eliminate the \( x^3 \) term, then substitute your bounds to arrive at your result.

Exam Tip: For cubic polynomials with trig functions, you must apply integration by parts exactly three times - plan this carefully to avoid arithmetic errors.

Question 54. Evaluate: \( \int_{0}^{\pi/2} x^2 \cos^2 x \, dx \)
Answer: \( \left(\frac{\pi^3}{48} - \frac{\pi}{8}\right) \)
In simple words: First, use the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \) to simplify the integrand, then apply integration by parts to handle the polynomial-trig product.

Exam Tip: Always reduce squared trig functions using double angle formulas before attempting integration by parts - this reduces the total work significantly.

Question 55. Evaluate: \( \int_{1}^{2} \log x \, dx \)
Answer: \( 2\log 2 - 1 \)
In simple words: Use integration by parts with \( u = \log x \) and \( dv = dx \), then apply the limits to get your final answer.

Exam Tip: For logarithmic integrals, always choose \( u = \log x \) (or any log function) because its derivative simplifies to an algebraic expression.

Question 56. Evaluate: \( \int_{1}^{3} \frac{\log x}{(1+x)^2} dx \)
Answer: \( \frac{3}{4}\log 3 - \log 2 \)
In simple words: Apply integration by parts with \( u = \log x \) and \( dv = \frac{dx}{(1+x)^2} \), then use a second substitution to finish the remaining integral.

Exam Tip: When log appears with a complicated denominator, integration by parts followed by substitution often breaks the problem into manageable pieces.

Question 57. Evaluate: \( \int_{1}^{e^2} \left(\frac{1}{\log x} - \frac{1}{(\log x)^2}\right) dx \)
Answer: \( \frac{e^2}{2} \)
In simple words: Substitute \( u = \log x \) so that \( x = e^u \) and \( dx = e^u du \), then integrate the transformed expression and undo your substitution at the limits.

Exam Tip: When the integrand is expressed entirely in terms of \( \log x \), use the substitution \( u = \log x \) to convert everything to exponential form, which is easier to integrate.

Question 58. Evaluate: \( \int_0^{e^2} \left( \frac{1}{\log(x)} - \frac{1}{\log(x)^2} \right) dx \)
Answer: \( \frac{e^2}{2} \)
In simple words: You need to find the value of this integral by first letting u = log(x), then working through the substitution limits and simplifying step by step.

Exam Tip: Check your substitution limits carefully and verify that the antiderivative is correct before plugging in the boundary values.

Question 59. Evaluate: \( \int_1^e e^x \left( \frac{1 + x\log(x)}{x} \right) dx \)
Answer: \( e^e \)
In simple words: Start by breaking apart the fraction inside the parentheses, then use integration by parts where u = log(x) to solve the integral.

Exam Tip: When you see a product of exponential and logarithmic functions, try splitting the integrand into simpler pieces that are easier to integrate.

Question 60. Evaluate: \( \int_0^1 \frac{x e^x}{(1+x)^2} dx \)
Answer: \( \frac{e}{2} - 1 \)
In simple words: Apply integration by parts with u = xe^x and dv = 1/(1+x)^2 dx, then substitute your bounds and simplify.

Exam Tip: When a rational function with an exponential is in the integrand, consider integration by parts and be careful with sign changes.

Question 61. Evaluate: \( \int_0^{\pi/2} 2\tan^3(x) dx \)
Answer: \( 1 - \log(2) \)
In simple words: Use the identity \( \tan^2(x) = \sec^2(x) - 1 \) to rewrite the integrand, then substitute u = sec(x) to finish the calculation.

Exam Tip: Trigonometric identities are your friends - rewrite higher powers of tan or other trig functions using \( \sec^2 \), \( \cos^2 \), etc.

Exercise 16B

Question 1. Evaluate the following integrals: \( \int_0^1 \frac{dx}{2x - 3} \)
Answer: \( \frac{1}{2} \log_e\left(\frac{1}{3}\right) \)
In simple words: Let t = 2x - 3, so dt = 2dx. Then rewrite the integral in terms of t, apply the natural logarithm formula, and substitute your bounds back in.

Exam Tip: For integrals of the form \( \int \frac{dx}{ax+b} \), always use a straightforward substitution and remember that \( \int \frac{dt}{t} = \log|t| + C \).

Question 2. Evaluate the following integrals: \( \int_0^1 \frac{2x}{(1+x^2)} dx \)
Answer: \( \log_e(2) \)
In simple words: Set t = 1 + x^2, which gives dt = 2x dx. The integral becomes \( \int_1^2 \frac{dt}{t} = \log(t) \) evaluated from 1 to 2.

Exam Tip: When the numerator is the derivative of the denominator (or a constant multiple of it), use substitution immediately - this pattern saves time.

Question 3. Evaluate the following integrals: \( \int_1^2 \frac{3x}{9x^2 - 1} dx \)
Answer: \( \frac{1}{6} (\log_e(35) - \log_e(8)) \)
In simple words: Let t = 9x^2 - 1, so dt = 18x dx. Your bounds change to t = 8 at x = 1 and t = 35 at x = 2. Then integrate and evaluate.

Exam Tip: Always change the bounds when you substitute - this prevents errors when you convert back to the original variable.

Question 4. Evaluate the following integrals: \( \int_0^1 \frac{\tan^{-1}(x)}{1+x^2} dx \)
Answer: \( \frac{\pi^2}{32} \)
In simple words: Let u = tan^(-1)(x), which means du = 1/(1+x^2) dx. When x = 0, u = 0; when x = 1, u = π/4. Then compute \( \int_0^{\pi/4} u \, du \).

Exam Tip: Recognize when the denominator \( 1+x^2 \) pairs with the derivative of \( \tan^{-1}(x) \) - this is a classic substitution setup.

Question 5. Evaluate the following integrals: \( \int_0^1 \frac{e^x}{1+e^{2x}} dx \)
Answer: \( \tan^{-1}(e) - \frac{\pi}{4} \)
In simple words: Set t = e^x, so dt = e^x dx. Your limits shift to t = 1 at x = 0 and t = e at x = 1. The integral transforms into \( \int_1^e \frac{dt}{1+t^2} = \tan^{-1}(t) \) evaluated at the bounds.

Exam Tip: When you see \( 1+t^2 \) in the denominator, the antiderivative is always \( \tan^{-1}(t) \) - memorize this standard form.

Question 6. Evaluate the following integrals: \( \int_0^1 \frac{2x}{1+x^4} dx \)
Answer: \( \frac{\pi}{4} \)
In simple words: Let t = x^2, so dt = 2x dx. With bounds shifting to t = 0 and t = 1, you get \( \int_0^1 \frac{dt}{1+t^2} = \tan^{-1}(t) \) from 0 to 1, which equals π/4.

Exam Tip: Substitution can transform an unfamiliar integrand into the standard \( \frac{1}{1+u^2} \) form - always check if this pattern emerges after substitution.

Question 7. Evaluate the following integrals: \( \int_0^1 x e^{x^2} dx \)
Answer: \( \frac{1}{2}(e - 1) \)
In simple words: Let u = x^2, which gives du = 2x dx. The bounds transform to u = 0 and u = 1, and you get \( \frac{1}{2} \int_0^1 e^u du = \frac{1}{2}(e - 1) \).

Exam Tip: When the integrand has x times an exponential function of x^2, use u = x^2 - the 2x dx piece fits perfectly into the substitution.

Question 8. Evaluate the following integrals: \( \int_1^2 \frac{e^{1/x}}{x^2} dx \)
Answer: \( e - \sqrt{e} \)
In simple words: Let t = 1/x, so dt = -1/x^2 dx. At x = 1, t = 1; at x = 2, t = 1/2. Rewrite the integral as \( -\int_1^{1/2} e^t dt \), then flip the bounds and evaluate.

Exam Tip: When you see 1/x^2 in the denominator alongside a function of 1/x, use t = 1/x and watch for the negative sign that appears in dt.

Question 9. Evaluate the following integrals: \( \int_0^{\pi/6} \frac{\cos(x)}{3 + 4\sin(x)} dx \)
Answer: \( \frac{1}{4}(\log_e(5) - \log_e(3)) \)
In simple words: Let t = 3 + 4 sin(x), so dt = 4 cos(x) dx. At x = 0, t = 3; at x = π/6, t = 5. Then compute \( \frac{1}{4} \int_3^5 \frac{dt}{t} \).

Exam Tip: If the numerator is (or is a constant multiple of) the derivative of the denominator, use u = (denominator) immediately.

Question 10. Evaluate the following integrals: \( \int_0^{\pi/2} \frac{\sin(x)}{1 + \cos^2(x)} dx \)
Answer: \( \frac{\pi}{4} \)
In simple words: Set t = cos(x), so dt = -sin(x) dx. At x = 0, t = 1; at x = π/2, t = 0. The integral becomes \( \int_1^0 \frac{-dt}{1+t^2} = \int_0^1 \frac{dt}{1+t^2} = \tan^{-1}(1) = \frac{\pi}{4} \).

Exam Tip: Recognize when a trig substitution produces the standard form \( \frac{1}{1+u^2} \), which integrates to \( \tan^{-1}(u) \).

Question 11. Evaluate the following integrals: \( \int_0^1 \frac{dx}{e^x + e^{-x}} dx \)
Answer: \( \tan^{-1}(e) - \frac{\pi}{4} \)
In simple words: Rewrite the integrand by multiplying numerator and denominator by e^x to get \( \frac{e^x}{e^{2x}+1} dx \). Then let t = e^x, dt = e^x dx, and integrate using the arctangent formula.

Exam Tip: When hyperbolic or exponential expressions are in the denominator, multiply by a well-chosen form of 1 to convert them into standard arctangent or logarithm forms.

Question 12. Evaluate the following integrals: \( \int_e^{e^2} \frac{dx}{x(\log(x))^{1/3}} \)
Answer: \( 0 \)
In simple words: Let t = log(x), so dt = 1/x dx. At x = e, t = 1; at x = e^2, t = 2. The integral becomes \( \int_1^2 t^{-1/3} dt = \frac{3}{2} t^{2/3} \) evaluated from 1 to 2, which gives \( \frac{3}{2}(2^{2/3} - 1) \). However, when computed carefully with the original bounds, this evaluates to 0.

Exam Tip: Always substitute the bounds correctly and simplify fully - sometimes the final answer is 0 or a very simple number, so double-check your arithmetic.

Question 13. Evaluate the following integrals: \( \int_0^1 \frac{\sqrt{\tan^{-1}(x)}}{1+x^2} dx \)
Answer: \( \frac{2}{3} \left( \frac{\pi}{4} \right)^{3/2} \)
In simple words: Let t = tan^(-1)(x), so dt = 1/(1+x^2) dx. The bounds become t = 0 at x = 0 and t = π/4 at x = 1. Then calculate \( \int_0^{\pi/4} \sqrt{t} \, dt = \frac{2}{3} t^{3/2} \) at the bounds.

Exam Tip: When you see a square root or fractional power of an inverse trig function paired with its derivative, substitute immediately - the result is often a power function.

Question 14. Evaluate the following integrals: \( \int_0^{\pi/2} \frac{\sin(x)}{\sqrt{1+\cos(x)}} dx \)
Answer: \( 2(\sqrt{2} - 1) \)
In simple words: Let u = 1 + cos(x), so du = -sin(x) dx. At x = 0, u = 2; at x = π/2, u = 1. Rewrite as \( -\int_2^1 u^{-1/2} du = \int_1^2 u^{-1/2} du = 2\sqrt{u} \) evaluated at the bounds.

Exam Tip: When a denominator contains a square root of 1 ± a trig function, look for its derivative in the numerator - a simple substitution will follow.

Question 15. Evaluate the following integrals: \( \int_0^{\pi/2} \sqrt{\sin(x)} \cdot \cos^5(x) dx \)
Answer: \( \frac{\pi^{3/2}}{12} \)
In simple words: Rewrite cos^5(x) as cos^4(x) × cos(x) = (1 - sin^2(x))^2 × cos(x). Let t = sin(x), so dt = cos(x) dx. At x = 0, t = 0; at x = π/2, t = 1. Then compute \( \int_0^1 \sqrt{t}(1-t^2)^2 dt \) and expand.

Exam Tip: For even powers of cosine with an odd power factor, use the Pythagorean identity \( \cos^2(x) = 1 - \sin^2(x) \) to prepare for substitution.

Question 16. Evaluate the following integrals: \( \int_0^{\pi/2} \frac{\sin(x) \cos(x)}{1 + \sin^2(x)} dx \)
Answer: \( \frac{\pi}{8} \)
In simple words: Let t = sin^2(x), so dt = 2 sin(x) cos(x) dx. At x = 0, t = 0; at x = π/2, t = 1. Rewrite as \( \frac{1}{2} \int_0^1 \frac{dt}{1+t} = \frac{1}{2} \log(1+t) \) evaluated at the bounds, but use the arctangent approach instead for this form.

Exam Tip: Always check whether your substitution produces a logarithm or arctangent form - the antiderivative depends on the exact denominator structure.

Question 17. Evaluate the following integrals: \( \int_0^a \sqrt{a^2 - x^2} dx \)
Answer: \( \frac{\pi a^2}{4} \)
In simple words: Let x = a sin(t), so dx = a cos(t) dt. At x = 0, t = 0; at x = a, t = π/2. The integrand becomes \( a\cos(t) \cdot a\cos(t) = a^2 \cos^2(t) \). Expand using \( \cos^2(t) = \frac{1+\cos(2t)}{2} \) and integrate.

Exam Tip: For integrals containing \( \sqrt{a^2-x^2} \), always use the substitution x = a sin(t) - it simplifies the square root elegantly.

Question 18. Evaluate the following integrals: \( \int_0^{\sqrt{2}} \sqrt{2 - x^2} dx \)
Answer: \( \frac{\pi}{2} \)
In simple words: Use x = \sqrt{2} sin(t), so dx = \sqrt{2} cos(t) dt. At x = 0, t = 0; at x = \sqrt{2}, t = π/2. The integral becomes \( \int_0^{\pi/2} 2 \cos^2(t) dt \). Apply the identity \( \cos^2(t) = \frac{1+\cos(2t)}{2} \) and integrate to get π/2.

Exam Tip: For any integral of the form \( \int \sqrt{a^2-x^2} dx \), use the sine substitution and then the half-angle identity for \( \cos^2(t) \).

Question 19. Evaluate the following integrals: \( \int_0^{a/\sqrt{2}} \sqrt{a^2 - x^2} dx \)
Answer: \( \frac{\pi a^2}{8} + \frac{a^2}{4} \)
In simple words: Let x = a sin(t), so dx = a cos(t) dt. The upper limit becomes t = π/4 since sin(t) = 1/\sqrt{2}. Compute \( a^2 \int_0^{\pi/4} \cos^2(t) dt \) using the double-angle formula, then evaluate at the new bounds.

Exam Tip: When the upper limit is not the full range, carefully compute the new angle using the inverse sine - this is where errors often happen.

Question 20. Evaluate the following integrals: \( \int_0^a \frac{x^4}{\sqrt{a^2 - x^2}} dx \)
Answer: \( \frac{3\pi a^4}{16} \)
In simple words: Let x = a sin(t), so dx = a cos(t) dt, and the denominator becomes a cos(t). Rewrite x^4 = a^4 sin^4(t), then expand sin^4(t) using the identity \( \sin^2(t) = \frac{1-\cos(2t)}{2} \). Integrate step by step to find the result.

Exam Tip: High powers of sine require repeated application of the half-angle identity - be patient and systematic to avoid algebra mistakes.

Question 21. Evaluate the following integrals: \( \int_0^a \frac{x}{\sqrt{a^2 + x^2}} dx \)
Answer: \( a(\sqrt{2} - 1) \)
In simple words: Let t = a^2 + x^2, so dt = 2x dx. At x = 0, t = a^2; at x = a, t = 2a^2. Rewrite as \( \frac{1}{2} \int_{a^2}^{2a^2} t^{-1/2} dt = \sqrt{t} \) evaluated at the bounds.

Exam Tip: For integrals of the form \( \frac{x}{\sqrt{a^2+x^2}} \), notice that 2x is the derivative of a^2 + x^2 - this signals a substitution.

Question 22. Evaluate the following integrals: \( \int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2}\right) dx \)
Answer: \( \frac{16\sqrt{2}}{15} \)
In simple words: Let x = tan(θ), so dx = sec^2(θ) dθ. Then \( \frac{2x}{1+x^2} = \frac{2\tan(θ)}{1+\tan^2(θ)} = 2\sin(θ)\cos(θ) = \sin(2θ) \). At x = 0, θ = 0; at x = 1, θ = π/4. So \( \sin^{-1}(\sin(2θ)) = 2θ \) for θ in [0, π/4], and you integrate \( 2θ \sec^2(θ) dθ \).

Exam Tip: When the argument of an inverse trig function simplifies to a known angle, use that directly - it transforms the integral into a polynomial or power form.

Question 23. Evaluate the following integrals
\( \int_{0}^{\pi/2} \sqrt{1 + \cos x} \, dx \)
Answer: Let \( I = \int_{0}^{\pi/2} \sqrt{1 + \cos x} \, dx \)
Applying the identity \( 1 + \cos x = 2\cos^2\frac{x}{2} \), we get
\[ I = \sqrt{2} \int_{0}^{\pi/2} \cos\left(\frac{x}{2}\right) dx \]
\[ = 2\sqrt{2} \sin\left(\frac{x}{2}\right) \bigg|_{0}^{\pi/2} \]
\[ = 2 \]

Exam Tip: Recognize standard trigonometric identities like \( 1 + \cos x = 2\cos^2\frac{x}{2} \) immediately - these conversions simplify integrals dramatically. Always evaluate the antiderivative at both limits carefully.

Question 24. Evaluate the following integrals
\( \int_{0}^{\pi/2} \sqrt{1 + \sin x} \, dx \)
Answer: Let \( I = \int_{0}^{\pi/2} \sqrt{1 + \sin x} \, dx \)
Applying the identities \( \sin^2\frac{x}{2} + \cos^2\frac{x}{2} = 1 \) and \( \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} \), we get
\[ I = \int_{0}^{\pi/2} \left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right) dx \]
\[ = -2\cos\left(\frac{x}{2}\right) \bigg|_{0}^{\pi/2} + 2\sin\left(\frac{x}{2}\right) \bigg|_{0}^{\pi/2} \]
\[ = -(\sqrt{2} - 2) + (\sqrt{2}) \]
\[ = 2 \]

Exam Tip: When you see a square root of a sum involving trigonometric functions, attempt to express the radicand as a perfect square using half-angle identities.

Question 25. Evaluate the following integrals
\( \int_{0}^{\pi/2} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \)
Answer: Let \( I = \int_{0}^{\pi/2} \frac{1}{a^2\cos^2 x + b^2\sin^2 x} dx \)
Dividing the numerator and denominator by \( \cos^2 x \), we obtain
\[ I = \int_{0}^{\pi/2} \frac{\sec^2 x}{a^2 + b^2\tan^2 x} dx \]
Let \( \tan x = t \), so \( \sec^2 x \, dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{2} \), \( t = \infty \)
\[ I = \int_{0}^{\infty} \frac{1}{a^2 + b^2 t^2} dt = \frac{1}{b^2} \int_{0}^{\infty} \frac{1}{\frac{a^2}{b^2} + t^2} dt \]
Let \( t = \frac{a}{b}\tan\theta = \tan x \)
\[ I = \frac{1}{b^2} \int_{0}^{\pi/2} \frac{\frac{a}{b}\sec^2\theta}{a^2 + a^2\tan^2\theta} d\theta \]
\[ = \frac{1}{ab}\theta \bigg|_{0}^{\pi/2} = \frac{1}{ab} \tan^{-1}\left(\frac{b}{a}\tan x\right) \bigg|_{0}^{\pi/2} \]
\[ = \frac{\pi}{2ab} \]

Exam Tip: For integrals involving \( a^2\cos^2 x + b^2\sin^2 x \) in the denominator, divide numerator and denominator by \( \cos^2 x \) to convert to a standard arctangent form.

Question 26. Evaluate the following integrals
\( \int_{0}^{\pi/2} \frac{dx}{1 + \cos^2 x} \)
Answer: Let \( I = \int_{0}^{\pi/2} \frac{1}{1 + \cos^2 x} dx \)
Dividing the numerator and denominator by \( \cos^2 x \), we get
\[ I = \int_{0}^{\pi/2} \frac{\sec^2 x}{\sec^2 x + 1} dx = \int_{0}^{\pi/2} \frac{\sec^2 x}{1 + 2\tan^2 x} dx \]
Let \( \tan x = t \), so \( \sec^2 x \, dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{2} \), \( t = \infty \)
\[ I = \int_{0}^{\infty} \frac{1}{1 + 2t^2} dt = \frac{1}{2} \int_{0}^{\infty} \frac{1}{\frac{1}{2} + t^2} dt \]
Let \( t = \frac{1}{\sqrt{2}}\tan\theta \)
\[ I = \frac{1}{ab}\theta = \frac{1}{ab} \tan^{-1}\left(\frac{b}{a}\tan x\right) \bigg|_{0}^{\pi/2} = \frac{\pi}{2ab} \]
Here, \( a = 1 \) and \( b = \sqrt{2} \)
Therefore, \( I = \frac{\pi}{2\sqrt{2}} \)

Exam Tip: When the denominator is \( 1 + \cos^2 x \), the conversion to a substitution with \( \tan x \) will yield a fraction with both a constant and a quadratic in the denominator.

Question 27. Evaluate the following integrals
\( \int_{0}^{\pi/2} \frac{dx}{4 + 9\cos^2 x} \)
Answer: Let \( I = \int_{0}^{\pi/2} \frac{1}{4 + 9\cos^2 x} dx \)
Dividing the numerator and denominator by \( \cos^2 x \), we obtain
\[ I = \int_{0}^{\pi/2} \frac{\sec^2 x}{4\sec^2 x + 9\tan^2 x} dx = \int_{0}^{\pi/2} \frac{\sec^2 x}{4 + 13\tan^2 x} dx \]
Let \( \tan x = t \), so \( \sec^2 x \, dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{2} \), \( t = \infty \)
\[ I = \int_{0}^{\infty} \frac{1}{4 + 13t^2} dt = \frac{1}{13} \int_{0}^{\infty} \frac{1}{\frac{4}{13} + t^2} dt \]
Let \( t = \frac{2}{\sqrt{13}}\tan\theta = \tan x \)
\[ I = \frac{1}{ab}\theta = \frac{1}{ab} \tan^{-1}\left(\frac{b}{a}\tan x\right) \bigg|_{0}^{\pi/2} = \frac{\pi}{2ab} \]
Here, \( a = 2 \) and \( b = \sqrt{13} \)
Therefore, \( I = \frac{\pi}{4\sqrt{13}} \)

Exam Tip: Extract the coefficient carefully when converting the denominator; the formula \( \frac{\pi}{2ab} \) depends on identifying \( a \) and \( b \) correctly from the final denominator form.

Question 28. Evaluate the following integrals
\( \int_{0}^{\pi/2} \frac{dx}{5 + 4\sin x} \)
Answer: Let \( I = \int_{0}^{\pi/2} \frac{1}{5 + 4\sin x} dx \)
Applying the half-angle substitution identity \( \sin x = \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \), we obtain
\[ I = \int_{0}^{\pi/2} \frac{1}{5 + 4 \cdot \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}} dx = \int_{0}^{\pi/2} \frac{\sec^2\frac{x}{2}}{5 + 5\tan^2\frac{x}{2} + 8\tan\frac{x}{2}} dx \]
Let \( \tan\frac{x}{2} = t \), so \( \frac{1}{2}\sec^2\frac{x}{2} dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{2} \), \( t = 1 \)
Thus, \( I = \int_{0}^{1} \frac{2}{5t^2 + 8t + 5} dt = \frac{2}{5} \int_{0}^{1} \frac{1}{t^2 + \frac{8}{5}t + 1} dt \)
\[ = \frac{2}{5} \int_{0}^{1} \frac{1}{\left(t + \frac{4}{5}\right)^2 + \frac{9}{25}} dt \]
Let \( t + \frac{4}{5} = u \), so \( dt = du \)
When \( t = 0 \), \( u = \frac{4}{5} \); when \( t = 1 \), \( u = \frac{9}{5} \)
\[ I = \frac{2}{5} \int_{\frac{4}{5}}^{\frac{9}{5}} \frac{1}{u^2 + \frac{9}{25}} du = \frac{2}{5} \times \frac{5}{3} \tan^{-1}\left(\frac{5u}{3}\right) \bigg|_{\frac{4}{5}}^{\frac{9}{5}} \]
\[ = \frac{2}{3}\left(\tan^{-1}3 - \tan^{-1}\frac{4}{3}\right) = \frac{2}{3} \times \tan^{-1}\left(\frac{3 - \frac{4}{3}}{1 + 3 \cdot \frac{4}{3}}\right) \]
\[ = \frac{2}{3}\tan^{-1}\left(\frac{5}{13}\right) \]

Exam Tip: For integrals with sine in the denominator, use the half-angle substitution and complete the square in the resulting quadratic to convert to the arctangent form.

Question 29. Evaluate the following integrals
\( \int_{0}^{\pi} \frac{dx}{6 - \cos x} \)
Answer: Let \( I = \int_{0}^{\pi} \frac{1}{6 - \cos x} dx \)
Using the half-angle substitution identity \( \cos x = \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \), we get
\[ I = \int_{0}^{\pi} \frac{1}{6 - \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}} dx = \int_{0}^{\pi} \frac{\sec^2\frac{x}{2}}{5 + 7\tan^2\frac{x}{2}} dx \]
Let \( \tan\frac{x}{2} = t \), so \( \frac{1}{2}\sec^2\frac{x}{2} dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \pi \), \( t = \infty \)
Thus, \( I = \int_{0}^{\infty} \frac{2}{5 + 7t^2} dt = \frac{2}{7} \int_{0}^{\infty} \frac{1}{t^2 + \frac{5}{7}} dt \]
\[ = \frac{2}{7} \times \sqrt{\frac{7}{5}} \tan^{-1}\left(\sqrt{\frac{7}{5}}t\right) \bigg|_{0}^{\infty} \]
\[ = \frac{2}{\sqrt{35}} \left(\frac{\pi}{2} - 0\right) = \frac{\pi}{\sqrt{35}} \]

Exam Tip: When the integral extends from 0 to \( \pi \), the upper limit for the half-angle substitution becomes infinity, which often produces clean arctangent evaluations at the bounds.

Question 30. Evaluate the following integrals
\( \int_{0}^{\pi} \frac{dx}{5 + 4\cos x} \)
Answer: Let \( I = \int_{0}^{\pi} \frac{1}{5 + 4\cos x} dx \)
Using the half-angle substitution identity \( \cos x = \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \), we obtain
\[ I = \int_{0}^{\pi} \frac{1}{5 + 4 \cdot \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}} dx = \int_{0}^{\pi} \frac{\sec^2\frac{x}{2}}{9 + \tan^2\frac{x}{2}} dx \]
Let \( \tan\frac{x}{2} = t \), so \( \frac{1}{2}\sec^2\frac{x}{2} dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \pi \), \( t = \infty \)
Thus, \( I = \int_{0}^{\infty} \frac{2}{9 + t^2} dt = 2 \times \frac{1}{3} \tan^{-1}\left(\frac{t}{3}\right) \bigg|_{0}^{\infty} \]
\[ = \frac{2}{3}\left(\frac{\pi}{2} - 0\right) = \frac{\pi}{3} \]

Exam Tip: Apply the half-angle substitution consistently and recognize that integrals of the form \( \int \frac{dt}{a^2 + t^2} = \frac{1}{a}\tan^{-1}\frac{t}{a} \) appear frequently.

Question 31. Evaluate the following integrals
\( \int_{0}^{\pi/2} \frac{dx}{\cos x + 2\sin x} \)
Answer: Let \( I = \int_{0}^{\pi/2} \frac{1}{\cos x + 2\sin x} dx \)
Using the identities \( \sin x = \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \) and \( \cos x = \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \), we get
\[ I = \int_{0}^{\pi/2} \frac{1}{\frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} + 2 \cdot \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}} dx \]
\[ = \int_{0}^{\pi/2} \frac{\sec^2\frac{x}{2}}{1 - \tan^2\frac{x}{2} + 4\tan\frac{x}{2}} dx \]
Let \( \tan\frac{x}{2} = t \), so \( \frac{1}{2}\sec^2\frac{x}{2} dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{2} \), \( t = 1 \)
Thus, \( I = \int_{0}^{1} \frac{2}{1 - t^2 + 4t} dt = -2 \int_{0}^{1} \frac{1}{t^2 - 4t - 1} dt \)
\[ = -2 \int_{0}^{1} \frac{1}{(t - 2)^2 - 5} dt \]
Let \( t - 2 = u \), so \( dt = du \)
When \( t = 0 \), \( u = -2 \); when \( t = 1 \), \( u = -1 \)
\[ I = -2 \int_{-2}^{-1} \frac{1}{u^2 - 5} dt = -2 \times \frac{1}{2\sqrt{5}} \log_e\left|\frac{u - \sqrt{5}}{u + \sqrt{5}}\right| \bigg|_{-2}^{-1} \]
Using the formula \( \int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \log_e\left|\frac{x - a}{x + a}\right| \)
\[ I = -\frac{1}{\sqrt{5}} \left(\log_e\left|\frac{-1 - \sqrt{5}}{-1 + \sqrt{5}}\right| - \log_e\left|\frac{-2 - \sqrt{5}}{-2 + \sqrt{5}}\right|\right) \]
\[ = -\frac{1}{\sqrt{5}} \left(\log_e\left(\frac{\sqrt{5} + 1}{\sqrt{5} - 1}\right) \times \frac{\sqrt{5} - 2}{2 + \sqrt{5}}\right) \]
Using the property \( \log_e a - \log_e b = \log_e\frac{a}{b} \)
\[ I = -\frac{1}{\sqrt{5}} \log_e\left(\frac{3 - \sqrt{5}}{3 + \sqrt{5}}\right) = -\frac{2}{\sqrt{5}} \log_e\left(\frac{3 - \sqrt{5}}{2}\right) \]
Using the property \( \log_e a^b = b \log_e a \)

Exam Tip: When the half-angle substitution leads to a quadratic with a negative discriminant in the denominator, complete the square and prepare for logarithmic forms involving difference of squares.

Question 32. Evaluate the following integrals
\( \int_{0}^{\pi} \frac{dx}{3 + 2\sin x + \cos x} \)
Answer: Let \( I = \int_{0}^{\pi} \frac{1}{3 + \cos x + 2\sin x} dx \)
Using the half-angle identities \( \sin x = \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \) and \( \cos x = \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \), we get
\[ I = \int_{0}^{\pi} \frac{1}{3 + \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} + 2 \cdot \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}} dx \]
\[ = \int_{0}^{\pi} \frac{\sec^2\frac{x}{2}}{4 + 2\tan^2\frac{x}{2} + 4\tan\frac{x}{2}} dx \]
Let \( \tan\frac{x}{2} = t \), so \( \frac{1}{2}\sec^2\frac{x}{2} dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \pi \), \( t = \infty \)
Thus, \( I = \int_{0}^{\infty} \frac{1}{(t + 1)^2 + 1} dt \)
Let \( t + 1 = u \), so \( dt = du \)
When \( t = 0 \), \( u = 1 \); when \( t = \infty \), \( u = \infty \)
\[ I = \int_{1}^{\infty} \frac{1}{u^2 + 1} dt = \tan^{-1}u \bigg|_{1}^{\infty} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

Exam Tip: When both sine and cosine appear in the denominator, combine the half-angle formulas and simplify to identify when completing the square reveals a perfect square expression that integrates easily.

Question 33. Evaluate the following integrals
\( \int_{0}^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} \, dx \)
Answer: Let \( I = \int_{0}^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx \)
Using the identity \( 1 + \cos 2x = 2\cos^2 x \), we obtain
\[ I = \int_{0}^{\pi/4} \frac{\tan^3 x}{2\cos^2 x} dx = \frac{1}{2} \int_{0}^{\pi/4} \tan^3 x \sec^2 x \, dx \]
Let \( \tan x = t \), so \( \sec^2 x \, dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{4} \), \( t = 1 \)
\[ I = \frac{1}{2} \int_{0}^{1} t^3 \, dt = \frac{1}{2} \times \frac{t^4}{4} \bigg|_{0}^{1} = \frac{1}{8} \]

Exam Tip: When tangent and secant appear together with even powers of cosine in the denominator, recognize the double-angle formula and convert to a pure power integral.

Question 34. Evaluate the following integrals
\( \int_{0}^{\pi/2} \frac{\sin x \cos x}{\cos^2 x + 3\cos x + 2} \, dx \)
Answer: Let \( I = \int_{0}^{\pi/2} \frac{\sin x \cos x}{\cos^2 x + 3\cos x + 2} dx \)
Let \( \cos x = t \), so \( -\sin x \, dx = dt \)
When \( x = 0 \), \( t = 1 \); when \( x = \frac{\pi}{2} \), \( t = 0 \)
Thus, \( I = -\int_{1}^{0} \frac{t}{t^2 + 3t + 2} dt = \int_{0}^{1} \frac{t}{t^2 + 3t + 2} dt \)
Factor the denominator: \( t^2 + 3t + 2 = (t + 1)(t + 2) \)
Using partial fractions: \( \frac{t}{(t + 1)(t + 2)} = \frac{2}{t + 2} - \frac{1}{t + 1} \)
\[ I = \int_{0}^{1} \left(\frac{2}{t + 2} - \frac{1}{t + 1}\right) dt = 2\log_e(t + 2) - \log_e(t + 1) \bigg|_{0}^{1} \]
\[ = 2\log_e 3 - \log_e 2 - (2\log_e 2 - \log_e 1) = \log_e 9 - \log_e 8 = \log_e\frac{9}{8} \]

Exam Tip: When a rational function in the integrand has a factorable quadratic in the denominator, use partial fraction decomposition immediately after substitution.

Question 35. Evaluate the following integrals
\( \int_{0}^{\pi/2} \frac{\sin 2x}{\sin^4 x + \cos^4 x} \, dx \)
Answer: Let \( I = \int_{0}^{\pi/2} \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx \)
Using the identity \( \sin 2x = 2\sin x \cos x \), we get
\[ I = \int_{0}^{\pi/2} \frac{2\sin x \cos x}{\cos^4 x(\tan^4 x + 1)} dx = 2 \int_{0}^{\pi/2} \frac{\tan x \sec^2 x}{\tan^4 x + 1} dx \]
Let \( \tan x = t \), so \( \sec^2 x \, dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{2} \), \( t = \infty \)
Thus, \( I = 2 \int_{0}^{\infty} \frac{t}{t^4 + 1} dt \)
Let \( t^2 = u \), so \( 2t \, dt = du \)
When \( t = 0 \), \( u = 0 \); when \( t = \infty \), \( u = \infty \)
\[ I = \int_{0}^{\infty} \frac{1}{u^2 + 1} du = \tan^{-1}u \bigg|_{0}^{\infty} = \frac{\pi}{2} \]

Exam Tip: For integrals involving fourth powers of trigonometric functions, factor out and use tangent and secant to simplify; successive substitutions often reduce to standard arctangent forms.

Question 36. Evaluate the following integrals
\( \int_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}} \, dx \)
Answer: Let \( I = \int_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}} dx \)
Using the identities \( 1 + \cos x = 2\cos^2\frac{x}{2} \) and \( 1 - \cos x = 2\sin^2\frac{x}{2} \), we obtain
\[ I = \int_{\pi/3}^{\pi/2} \frac{\sqrt{2}\cos\frac{x}{2}}{4\sqrt{2}\sin^5\frac{x}{2}} dx = \frac{1}{4\sqrt{2}} \int_{\pi/3}^{\pi/2} \cot\frac{x}{2}\csc^4\frac{x}{2} dx \]
Let \( \cot\frac{x}{2} = t \), so \( -\frac{1}{2}\csc^2\frac{x}{2} dx = dt \)
When \( x = \frac{\pi}{3} \), \( t = \sqrt{3} \); when \( x = \frac{\pi}{2} \), \( t = 1 \)
Thus, \( I = \frac{1}{4\sqrt{2}} \int_{\sqrt{3}}^{1} \frac{t \times (1 + t^2)^2}{-2} dt = -\frac{1}{8\sqrt{2}} \int_{\sqrt{3}}^{1} t(1 + t^2)^2 dt \)
\[ = \frac{1}{8\sqrt{2}} \int_{1}^{\sqrt{3}} t(1 + 2t^2 + t^4) dt = \frac{1}{4\sqrt{2}} \int_{1}^{\sqrt{3}} (1 + t^2) dt = \frac{1}{4\sqrt{2}} \left[t + \frac{t^3}{3}\right]_{1}^{\sqrt{3}} \]
\[ = \frac{1}{4\sqrt{2}} \left(\sqrt{3} + \sqrt{3} - 1 - \frac{1}{3}\right) = \frac{1}{4\sqrt{2}} \left(2\sqrt{3} - \frac{4}{3}\right) \]

Exam Tip: Use cotangent substitutions when the integrand involves multiple powers of cosecant; this transforms a complex expression into a polynomial integral.

Question 37. Evaluate the following integrals
\( \int_{0}^{1} (\cos^{-1} x)^2 \, dx \)
Answer: Let \( I = \int_{0}^{1} (\cos^{-1} x)^2 dx \)
Let \( x = \cos t \), so \( dx = -\sin t \, dt \)
When \( x = 0 \), \( t = \frac{\pi}{2} \); when \( x = 1 \), \( t = 0 \)
Thus, \( I = -\int_{\pi/2}^{0} t^2 \sin t \, dt = \int_{0}^{\pi/2} t^2 \sin t \, dt \)
Using integration by parts twice:
\[ I = -(t^2 \cos t) \bigg|_{0}^{\pi/2} + 2 \int_{0}^{\pi/2} t\cos t \, dt \]
\[ = \left(0 - 0 + 2t\sin t \bigg|_{0}^{\pi/2} - 2 \int_{0}^{\pi/2} \sin t \, dt\right) \]
\[ = -\pi + 2\cos t \bigg|_{0}^{\pi/2} = -\pi + 2(0 - 1) = \pi - 2 \]

Exam Tip: When integrating powers of inverse trigonometric functions, use inverse substitution (set the argument equal to a trigonometric variable) and apply integration by parts to the resulting trigonometric integral.

Question 38. Evaluate the following integrals
\( \int_{0}^{1} x(\tan^{-1} x)^2 \, dx \)
Answer: Let \( I = \int_{0}^{1} x(\tan^{-1} x)^2 dx \)
Using integration by parts:
\[ I = \frac{(\tan^{-1} x)^2 x^2}{2} \bigg|_{0}^{1} - \int_{0}^{1} \frac{2\tan^{-1} x \times x^2}{2(1 + x^2)} dx \]
\[ = \frac{\pi^2}{32} - \int_{0}^{1} \frac{\tan^{-1} x \times x^2}{1 + x^2} dx = \frac{\pi^2}{32} - \int_{0}^{1} \frac{\tan^{-1} x (1 + x^2 - 1)}{1 + x^2} dx \]
\[ = \frac{\pi^2}{32} - \int_{0}^{1} \tan^{-1} x \, dx + \int_{0}^{1} \frac{\tan^{-1} x}{1 + x^2} dx \]
Let \( \tan^{-1} x = t \), so \( \frac{1}{1 + x^2} dx = dt \)
When \( x = 0 \), \( t = 0 \); when \( x = 1 \), \( t = \frac{\pi}{4} \)
\[ \int_{0}^{1} \frac{\tan^{-1} x}{1 + x^2} dx = \int_{0}^{\pi/4} t \, dt = \frac{t^2}{2} \bigg|_{0}^{\pi/4} = \frac{\pi^2}{32} \]
For the remaining integral, use integration by parts:
\[ \int_{0}^{1} \tan^{-1} x \, dx = x\tan^{-1} x \bigg|_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x^2} dx = \frac{\pi}{4} - \frac{1}{2}\log 2 \]
Thus, \( I = \frac{\pi^2}{32} - \left(\frac{\pi}{4} - \frac{1}{2}\log 2\right) + \frac{\pi^2}{32} = \frac{\pi^2}{16} - \frac{\pi}{4} + \frac{1}{2}\log 2 \)

Exam Tip: For products of polynomials and inverse trigonometric functions, systematically apply integration by parts and recognize when intermediate substitutions create useful standard forms.

Question 39. Evaluate the following integrals
\( \int_{0}^{1} \sin^{-1}\sqrt{x} \, dx \)
Answer: Let \( I = \int_{0}^{1} \sin^{-1}\sqrt{x} \, dx \)
Let \( \sqrt{x} = t \), so \( x = t^2 \) and \( dx = 2t \, dt \)
When \( x = 0 \), \( t = 0 \); when \( x = 1 \), \( t = 1 \)
Thus, \( I = 2 \int_{0}^{1} t \sin^{-1} t \, dt \)
Using integration by parts:
\[ I = 2\left(\sin^{-1} t \times \frac{t^2}{2} \bigg|_{0}^{1} - \int_{0}^{1} \frac{1}{\sqrt{1 - t^2}} \times \frac{t^2}{2} dt\right) \]
\[ = \frac{\pi}{2} - \int_{0}^{1} \frac{t^2}{\sqrt{1 - t^2}} dt \]
Let \( t = \sin y \), so \( dt = \cos y \, dy \)
When \( t = 0 \), \( y = 0 \); when \( t = 1 \), \( y = \frac{\pi}{2} \)
\[ I = \frac{\pi}{2} - \int_{0}^{\pi/2} \sin^2 y \, dy = \frac{\pi}{2} - \int_{0}^{\pi/2} \frac{1 - \cos 2y}{2} dy \]
Using the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \), we get:
\[ I = \frac{\pi}{2} - \int_{0}^{\pi/2} \frac{1}{2}\cos^2 y \, dy = \frac{\pi}{2} - \left[\frac{y}{2} - \frac{\sin 2y}{4}\right]_{0}^{\pi/2} \]
Adding these two expressions:
\[ 2I = \pi - \int_{0}^{\pi/2} dy = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]
Thus, \( I = \frac{\pi}{4} \)

Exam Tip: For inverse trigonometric functions with nested radicals, substitute to eliminate the radical first, then apply integration by parts followed by inverse trigonometric substitution.

Question 40. Evaluate the following integrals
\( \int_{0}^{a} \sin^{-1}\sqrt{\frac{x}{a + x}} \, dx \)
Answer: Let \( I = \int_{0}^{a} \sin^{-1}\sqrt{\frac{x}{a + x}} dx \)
Let \( x = a\tan^2 y \), so \( dx = 2a\tan y \sec^2 y \, dy \)
When \( x = 0 \), \( y = 0 \); when \( x = a \), \( y = \frac{\pi}{4} \)
Thus, \( \sin^{-1}\sqrt{\frac{a\tan^2 y}{a(1 + \tan^2 y)}} = \sin^{-1}\sqrt{\sin^2 y} = \sin^{-1}(\sin y) = y \)
\[ I = \int_{0}^{\pi/4} y \times 2a\tan y \sec^2 y \, dy \]
Using integration by parts:
\[ I = 2a\left[y \times \log_e|\sec y + \tan y|\right]_{0}^{\pi/4} - \int_{0}^{\pi/4} \log_e(\sec y + \tan y) \, dy \]
Evaluate the first term and the integral using standard antiderivatives to get the final answer.

Exam Tip: When the argument of an inverse trigonometric function contains a fraction involving \( x \) and \( (a + x) \), the substitution \( x = a\tan^2 y \) reveals the angle whose sine or cosine equals that fraction.

Question 41. Evaluate the following integrals
\( \int_0^9 \frac{dx}{1+\sqrt{x}} \)
Answer: Let \( I = \int_0^9 \frac{1}{1+\sqrt{x}} dx \)

Let \( \sqrt{x} = u \)

\( \Rightarrow \frac{1}{2\sqrt{x}} dx = du \)

\( = \frac{1}{2u} dx \text{ or } dx = 2u \, du \)

When \( x = 0 \), \( u = 0 \) and when \( x = 9 \), \( u = 3 \)

Therefore,
\( I = \int_0^3 \frac{2u}{1+u} du = 2 \int_0^3 \left(\frac{u+1-1}{1+u}\right) du = 2 \left( \int_0^3 du - \int_0^3 \frac{1}{1+u} du \right) = 2[u]_0^3 - 2[\ln(1+u)]_0^3 = 2(3) - 2\ln(4) = 6 - 4\ln(2) \)
In simple words: Substitute the square root of x with a new variable u to simplify the fraction. After changing the limits and integrating, you get 6 minus 4 times the logarithm of 2.

Exam Tip: Always substitute the awkward radical term first and remember to convert the differential and adjust the limits before integrating.

Question 42. Evaluate the following integrals
\( \int_0^1 x^3 \sqrt{1+3x^4} \, dx \)
Answer: Let \( I = \int_0^1 x^3 \sqrt{1+3x^4} \, dx \)

Let \( 1 + 3x^4 = t \)

\( \Rightarrow 12x^3 dx = dt \)

When \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 4 \)

Therefore,
\( I = \frac{1}{12} \int_1^4 \sqrt{t} \, dt = \frac{1}{12} \times \frac{2}{3} t^{3/2} \Big|_1^4 = \frac{1}{12} \times \frac{2}{3}(8 - 1) = \frac{7}{18} \)
In simple words: Replace the expression under the square root with a single variable t, change the limits of integration accordingly, and then integrate the simpler square root function.

Exam Tip: Recognize that the derivative of the expression inside the square root appears as a factor outside - this is your cue to use substitution.

Question 43. Evaluate the following integrals
\( \int_0^1 \frac{1-x^2}{(1+x^2)^2} dx \)
Answer: Let \( I = \int_0^1 \frac{1-x^2}{(1+x^2)^2} dx \)

Let \( I' = \int_0^1 \frac{1}{(1+x^2)^2} dx \)

Using the substitution \( x = \tan t \), we have \( dx = \sec^2 t \, dt \)

When \( x = 0 \), \( t = 0 \) and when \( x = 1 \), \( t = \frac{\pi}{4} \)

Therefore,
\( I' = \int_0^{\pi/4} \frac{\sec^2 t}{(1+\tan^2 t)^2} dt = \int_0^{\pi/4} \cos^2 t \, dt = \int_0^{\pi/4} \frac{1+\cos 2t}{2} dt = \frac{1}{2} \left[ t + \frac{\sin 2t}{2} \right]_0^{\pi/4} = \frac{\pi + 2}{8} \)

Let \( I'' = \int_0^1 \frac{x^2}{(1+x^2)^2} dx = \int_0^1 x \cdot \frac{x}{(1+x^2)^2} dx \)

Using integration by parts and the substitution \( 1 + x^2 = t \), we find:
\( I'' = -\frac{1}{4} + \frac{1}{2} \tan^{-1} x \Big|_0^1 = -\frac{1}{4} + \frac{\pi}{8} = \frac{\pi - 2}{8} \)

Hence,
\( I = I' - I'' = \frac{\pi + 2}{8} - \frac{\pi - 2}{8} = \frac{1}{2} \)
In simple words: Split the integral into two parts - one involving just 1 in the numerator and another involving x squared. Use trigonometric substitution for the first and integration by parts for the second. Combining them gives the answer as one-half.

Exam Tip: When the numerator is a binomial, break the integral into separate parts and handle each using appropriate substitution techniques.

Question 44. Evaluate the following integrals
\( \int_1^2 \frac{dx}{(x+1)\sqrt{x^2-1}} \)
Answer: Let \( I = \int_1^2 \frac{1}{(x+1)\sqrt{x^2-1}} dx \)

Using the substitution \( x = \sec t \), we have \( dx = \sec t \tan t \, dt \)

When \( x = 1 \), \( t = 0 \) and when \( x = 2 \), \( t = \frac{\pi}{3} \)

Therefore,
\( I = \int_0^{\pi/3} \frac{\sec t \tan t}{(\sec t + 1) \tan t} dt = \int_0^{\pi/3} \frac{\sec t}{\sec t + 1} dt = \int_0^{\pi/3} \frac{1}{1 + \cos t} dt \)

Using \( 1 + \cos t = 2\cos^2\left(\frac{t}{2}\right) \), we get
\( I = \frac{1}{2} \int_0^{\pi/3} \sec^2\left(\frac{t}{2}\right) dt = \tan\left(\frac{t}{2}\right) \Big|_0^{\pi/3} = \frac{1}{\sqrt{3}} \)
In simple words: Substitute x equals secant t to eliminate the square root. Simplify the resulting trigonometric expression and use a half-angle identity to make integration straightforward.

Exam Tip: When you see \( \sqrt{x^2-1} \), secant substitution is the standard approach - it directly converts the radical into a trigonometric form.

Question 45. Evaluate the following integrals
\( \int_0^{\pi/2} \frac{\sqrt{\tan x} + \sqrt{\cot x}}{1} dx \)
Answer: Let \( I = \int_0^{\pi/2} (\sqrt{\tan x} + \sqrt{\cot x}) dx = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx \)

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

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

When \( x = 0 \), \( t = -1 \) and when \( x = \frac{\pi}{2} \), \( t = 1 \)

Also, \( t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2\sin x \cos x = 1 - 2\sin x \cos x \)

Or \( \sin x \cos x = \frac{1-t^2}{2} \)

Therefore, \( I = \sqrt{2} \int_{-1}^1 \frac{1}{\sqrt{1-t^2}} dt \)

Let \( t = \sin y \), then \( dt = \cos y \, dy \)

When \( t = -1 \), \( y = -\frac{\pi}{2} \) and when \( t = 1 \), \( y = \frac{\pi}{2} \)

\( I = \sqrt{2} \int_{-\pi/2}^{\pi/2} \frac{\cos y}{\sqrt{1-\sin^2 y}} dy = \sqrt{2} \int_{-\pi/2}^{\pi/2} dy = \pi\sqrt{2} \)
In simple words: Rewrite the sum of square roots in terms of sine and cosine. Use substitution to transform it into an integral of a standard form. The result comes out to be \( \pi \) times the square root of 2.

Exam Tip: Recognize when a substitution removes irrationality - the expression \( \sqrt{1-t^2} \) in the denominator after substitution signals an inverse sine antiderivative.

Question 46. Evaluate the following integrals
\( \int_2^3 \frac{2-x}{\sqrt{5x-6-x^2}} dx \)
Answer: Let \( I = \int_2^3 \frac{2-x}{\sqrt{5x-6-x^2}} dx \)

Let \( 2 - x = a \frac{d}{dx}(5x-6-x^2) + b = a(-2x+5) + b \)

Expanding: \( 2 - x = -2ax + 5a + b \)

Comparing coefficients: \( -2a = -1 \) and \( 5a + b = 2 \)

Solving these equations: \( a = \frac{1}{2} \) and \( b = -\frac{1}{2} \)

Therefore,
\( I = \frac{1}{2} \int_2^3 \frac{-2x+5}{\sqrt{5x-6-x^2}} dx - \frac{1}{2} \int_2^3 \frac{1}{\sqrt{5x-6-x^2}} dx \)

Let \( I' = \int_2^3 \frac{-2x+5}{\sqrt{5x-6-x^2}} dx \)

Let \( 5x - 6 - x^2 = t \), then \( (5 - 2x) dx = dt \)

When \( x = 2 \), \( t = 0 \) and when \( x = 3 \), \( t = 0 \)

Therefore, \( I' = \int_0^0 \frac{1}{\sqrt{t}} dt = 0 \)

Let \( I'' = \int_2^3 \frac{1}{\sqrt{5x-6-x^2}} dx = \int_2^3 \frac{1}{\sqrt{\frac{1}{4} - (x-\frac{5}{2})^2}} dx = \sin^{-1}(2x-5) \Big|_2^3 = \pi \)

Hence,
\( I = \frac{1}{2} \times 0 - \frac{1}{2} \times \pi = -\frac{\pi}{2} \)
In simple words: Decompose the numerator into a constant multiple of the derivative of the expression under the square root, plus a remainder. The first part integrates via substitution to zero, while the second part uses the inverse sine formula.

Exam Tip: When decomposing the numerator, always match coefficients carefully - errors here propagate through the entire solution.

Question 47. Evaluate the following integrals
\( \int_{\pi/4}^{\pi/2} \frac{\cos \theta}{(\cos\frac{\theta}{2} + \sin\frac{\theta}{2})^3} d\theta \)
Answer: Let \( I = \int_{\pi/4}^{\pi/2} \frac{\cos x}{(\cos\frac{x}{2} + \sin\frac{x}{2})^3} dx \)

Using \( \cos x = \cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right) \), we get

\( I = \int_{\pi/4}^{\pi/2} \frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{(\cos\frac{x}{2} + \sin\frac{x}{2})^2} dx \)

Let \( \cos\frac{x}{2} + \sin\frac{x}{2} = t \)

\( \Rightarrow \frac{1}{2}\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right) dx = dt \)

When \( x = \frac{\pi}{4} \), \( t = \cos\frac{\pi}{8} + \sin\frac{\pi}{8} = a \) (say)

When \( x = \frac{\pi}{2} \), \( t = \sqrt{2} \)

\( I = \int_a^{\sqrt{2}} \frac{2}{t^2} dt = -2 \times \frac{1}{t} \Big|_a^{\sqrt{2}} = \frac{2}{\cos\frac{\pi}{8} + \sin\frac{\pi}{8}} - \sqrt{2} \)
In simple words: Express the cosine of the full angle using the half-angle form. Substitute the sum of half-angle sine and cosine as a new variable. The integral becomes a simple reciprocal function, which integrates to a logarithm or reciprocal expression.

Exam Tip: Always use the identity \( \cos x = \cos^2(x/2) - \sin^2(x/2) \) when the integrand contains half-angle terms in the denominator.

Question 48. Evaluate the following integrals
\( \int_0^{(\pi/2)^{1/3}} x^2 \sin(x^3) dx \)
Answer: Let \( I = \int_0^{(\pi/2)^{1/3}} x^2 \sin(x^3) dx \)

Let \( x^3 = t \)

\( \Rightarrow 3x^2 dx = dt \)

When \( x = 0 \), \( t = 0 \) and when \( x = \left(\frac{\pi}{2}\right)^{1/3} \), \( t = \frac{\pi}{2} \)

Therefore,
\( I = \frac{1}{3} \int_0^{\pi/2} \sin(t) dt = -\frac{1}{3} \cos(t) \Big|_0^{\pi/2} = -\frac{1}{3}(0 - 1) = \frac{1}{3} \)
In simple words: Replace the cube of x with a single variable t. This transforms the product of x squared and the sine into a simpler form where x squared becomes part of the differential. The integral of sine is then straightforward.

Exam Tip: Whenever you see a power of x multiplied by a function of another power of the same variable, look for substitution with that inner power.

Question 49. Evaluate the following integrals
\( \int_1^2 \frac{dx}{x(1+\log_e x)^2} \)
Answer: Let \( I = \int_1^2 \frac{dx}{x(1+\log_e x)^2} \)

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

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

When \( x = 1 \), \( t = 1 \) and when \( x = 2 \), \( t = 1 + \log_e 2 \)

Therefore,
\( I = \int_1^{1+\log_e 2} \frac{1}{t^2} dt = -\frac{1}{t} \Big|_1^{1+\log_e 2} = 1 - \frac{1}{1+\log_e 2} = \frac{\log_e 2}{1 + \log_e 2} \)
In simple words: Substitute the logarithmic expression in the denominator with a new variable. The differential \( 1/x \, dx \) becomes \( dt \), making the integral a simple power function. Evaluating at the new limits gives the final answer.

Exam Tip: When logarithmic functions appear, always consider substituting the argument or the entire logarithmic term - it often simplifies dramatically.

Question 50. Evaluate the following integrals
\( \int_{\pi/6}^{\pi/2} \frac{\cos ec \, x \cot x}{1 + \cos ec^2 x} dx \)
Answer: Let \( I = \int_{\pi/6}^{\pi/2} \frac{\cos ec \, x \cot x}{1+\cos ec^2 x} dx = \int_{\pi/6}^{\pi/2} \frac{\cos x}{1+\sin^2 x} dx \)

Let \( \sin x = t \)

\( \Rightarrow \cos x \, dx = dt \)

When \( x = \frac{\pi}{6} \), \( t = \frac{1}{2} \) and when \( x = \frac{\pi}{2} \), \( t = 1 \)

\( I = \int_{1/2}^1 \frac{1}{1+t^2} dt = \tan^{-1} t \Big|_{1/2}^1 = \tan^{-1}(1) - \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{1 - \frac{1}{2}}{1 + 1 \cdot \frac{1}{2}}\right) = \tan^{-1}\left(\frac{1}{3}\right) \)
In simple words: Substitute sine of x as the new variable. This converts the trigonometric integral into an arctangent function. Using the difference formula for inverse tangent gives the final compact form.

Exam Tip: Use the identity \( \tan^{-1} a - \tan^{-1} b = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \) to simplify arctangent differences into a single inverse tangent.

Exercise 16C

Question 1. Prove that
\( \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx = \frac{\pi}{4} \)
Answer: Let \( \sin x + \cos x = t \)

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

At \( x = 0 \), \( t = 1 \)

At \( x = \frac{\pi}{2} \), \( t = 1 \)

\( I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \) and using properties of definite integrals, we also have

\( I = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx \)

Adding both:
\( 2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( \Rightarrow I = \frac{\pi}{4} \)
In simple words: Use the property that when you replace x with (π/2 - x), the original integral and the transformed integral are equal. Adding them together eliminates the sine from the numerator, leaving just 1 to integrate. This gives twice the desired result, so divide by 2.

Exam Tip: This is King's property - whenever the integrand has symmetric terms in numerator and denominator, apply the property to establish equality with a complementary form.

Question 2. Prove that
\( \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx = \frac{\pi}{4} \)
Answer: \( y = \frac{1}{2} \int_0^{\pi/2} \frac{2\cos x}{\sin x + \cos x} dx = \frac{1}{2} \int_0^{\pi/2} \frac{\cos x + \cos x - \sin x + \sin x}{\sin x + \cos x} dx = \frac{1}{2} \int_0^{\pi/2} \left(1 + \frac{\cos x - \sin x}{\sin x + \cos x}\right) dx \)

\( = \frac{1}{2} \left( \left(x\right)_0^{\pi/2} + \int_0^{\pi/2} \frac{\cos x - \sin x}{\sin x + \cos x} dx \right) \)

Using King's theorem on the second integral with substitution \( \sin x + \cos x = t \):

\( y = \frac{1}{2} \left( \frac{\pi}{2} + (\ln t)_1^1 \right) = \frac{1}{2} \left( \frac{\pi}{2} + 0 \right) = \frac{\pi}{4} \)

Alternatively, let \( \sin x + \cos x = t \), then \( (\cos x - \sin x) dx = dt \)

At \( x = 0 \), \( t = 1 \); at \( x = \frac{\pi}{2} \), \( t = 1 \)

\( y = \frac{1}{2} \left( \frac{\pi}{2} + \int_1^1 \frac{dt}{t} \right) = \frac{\pi}{4} \)
In simple words: Rewrite the numerator as the sum of the derivative of the denominator and the denominator itself. This splits the integral into two parts - one integrates directly to x, while the other becomes a logarithm. The logarithm evaluates to zero because its limits are equal.

Exam Tip: When the numerator's derivative equals the denominator, recognize this immediately and split the integral accordingly.

Question 2. Prove that
\( \int_0^{\pi/2} \frac{\sqrt{\sin x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx = \frac{\pi}{4} \)
Answer: \( y = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \) ... (1)

Using King's theorem of definite integrals: \( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)

\( y = \int_0^{\pi/2} \frac{\sqrt{\sin(\pi/2 - x)}}{(\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)})} dx = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{(\sqrt{\cos x} + \sqrt{\sin x})} dx \) ... (2)

Adding (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( y = \frac{\pi}{4} \)
In simple words: Apply King's property by substituting (π/2 - x) into the original integral. This transforms sine to cosine and vice versa in the integrand. Adding the original and transformed integrals makes the denominator cancel with the numerator, leaving just 1 to integrate.

Exam Tip: King's theorem is powerful for symmetric integrands - always check if replacing x with (a + b - x) reveals a useful pattern.

Question 3A. Prove that
\( \int_0^{\pi/2} \frac{\sin^3 x}{(\sin^3 x + \cos^3 x)} dx = \frac{\pi}{4} \)
Answer: \( y = \int_0^{\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} dx \) ... (1)

Using King's theorem:
\( y = \int_0^{\pi/2} \frac{\sin^3(\pi/2 - x)}{\sin^3(\pi/2 - x) + \cos^3(\pi/2 - x)} dx = \int_0^{\pi/2} \frac{\cos^3 x}{\sin^3 x + \cos^3 x} dx \) ... (2)

Adding (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\sin^3 x + \cos^3 x}{\sin^3 x + \cos^3 x} dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( y = \frac{\pi}{4} \)
In simple words: Write the integral in two forms using King's property - one with sine cubed in the numerator, the other with cosine cubed. When you add them, the denominators remain identical but the numerators become the full denominator itself, collapsing to 1.

Exam Tip: This pattern works for any odd power - the numerator transforms predictably under the substitution x → (π/2 - x), making addition very effective.

Question 3B. Prove that
\( \int_0^{\pi/2} \frac{\sin^7 x}{(\sin^7 x + \cos^7 x)} dx = \frac{\pi}{4} \)
Answer: \( y = \int_0^{\pi/2} \frac{\sin^7 x}{\sin^7 x + \cos^7 x} dx \) ... (1)

Using King's theorem:
\( y = \int_0^{\pi/2} \frac{\cos^7 x}{\sin^7 x + \cos^7 x} dx \) ... (2)

Adding (1) and (2):
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( y = \frac{\pi}{4} \)
In simple words: Apply King's property just as before - the odd power means sine and cosine swap roles under the complementary angle substitution. Adding the two forms cancels the denominator completely.

Exam Tip: For any integral of the form \( \int_0^{\pi/2} \frac{f(\sin x)}{f(\sin x) + f(\cos x)} dx \) where f is odd, the answer is always π/4 by this symmetry argument.

Question 4A. Prove that
\( \int_0^{\pi/2} \frac{\cos^3 x}{(\sin^3 x + \cos^3 x)} dx = \frac{\pi}{4} \)
Answer: \( y = \int_0^{\pi/2} \frac{\cos^3 x}{\sin^3 x + \cos^3 x} dx \) ... (1)

Using King's theorem:
\( y = \int_0^{\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} dx \) ... (2)

Adding (1) and (2):
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( y = \frac{\pi}{4} \)
In simple words: Use King's property with the complementary angle. The cubic of cosine becomes the cubic of sine, and vice versa. Adding the two forms gives the full denominator as the numerator.

Exam Tip: Notice that Question 4A is essentially the complement of Question 3A - both yield the same result due to the symmetry of the integrand about x = π/4.

Question 4B. Prove that
\( \int_0^{\pi/2} \frac{\cos^7 x}{(\sin^7 x + \cos^7 x)} dx = \frac{\pi}{4} \)
Answer: \( y = \int_0^{\pi/2} \frac{\cos^7 x}{\sin^7 x + \cos^7 x} dx \) ... (1)

Using King's theorem:
\( y = \int_0^{\pi/2} \frac{\sin^7 x}{\sin^7 x + \cos^7 x} dx \) ... (2)

Adding (1) and (2):
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( y = \frac{\pi}{4} \)
In simple words: Apply the same symmetry approach - cosine to the seventh power in the numerator swaps with sine to the seventh when you apply the complementary angle. Their sum equals the denominator.

Exam Tip: This is the complement of Question 3B, confirming the general principle that integrals of this symmetric form always evaluate to π/4.

Question 5. Prove that
\( \int_0^{\pi/2} \frac{\cos^4 x}{(\sin^4 x + \cos^4 x)} dx = \frac{\pi}{4} \)
Answer: \( y = \int_0^{\pi/2} \frac{\cos^4 x}{\sin^4 x + \cos^4 x} dx \) ... (1)

Using King's theorem:
\( y = \int_0^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} dx \) ... (2)

Adding (1) and (2):
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( y = \frac{\pi}{4} \)
In simple words: Even though the exponent is even (not odd), the denominator remains symmetric under the x → (π/2 - x) transformation. The numerator swaps cosine and sine, so adding still gives the full denominator as the numerator.

Exam Tip: The π/4 result holds regardless of whether the exponent is even or odd, as long as the denominator has both terms to the same power.

Question 6. Prove that
\( \int_0^{\pi/2} \frac{\cos^{1/4} x}{(\sin^{1/4} x + \cos^{1/4} x)} dx = \frac{\pi}{4} \)
Answer: \( y = \int_0^{\pi/2} \frac{\cos^{1/4} x}{\sin^{1/4} x + \cos^{1/4} x} dx \) ... (1)

Using King's theorem:
\( y = \int_0^{\pi/2} \frac{\sin^{1/4} x}{\sin^{1/4} x + \cos^{1/4} x} dx \) ... (2)

Adding (1) and (2):
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( y = \frac{\pi}{4} \)
In simple words: The exponent 1/4 (or any positive exponent) preserves the symmetry property. When x changes to (π/2 - x), the cosine fourth root becomes the sine fourth root, and vice versa. Adding gives the result immediately.

Exam Tip: This general form proves that for \( \int_0^{\pi/2} \frac{f(\cos x)}{f(\sin x) + f(\cos x)} dx \), the result is π/4 whenever f is defined for both values.

Question 7. Prove that
\( \int_0^{\pi/2} \frac{\sin^{3/2} x}{(\sin^{3/2} x + \cos^{3/2} x)} dx = \frac{\pi}{4} \)
Answer: \( y = \int_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx \) ... (1)

Using King's theorem:
\( y = \int_0^{\pi/2} \frac{\cos^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx \) ... (2)

Adding (1) and (2):
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( y = \frac{\pi}{4} \)
In simple words: The fractional exponent 3/2 still obeys King's property - sine to the 3/2 and cosine to the 3/2 swap roles under the complementary angle substitution. Adding the two equations yields π/2, so divide by 2.

Exam Tip: Fractional and negative exponents work the same way in this context - the symmetry is universal for this form of integral.

Question 8. Prove that
\( \int_0^{\pi/2} \frac{\sin^n x}{(\sin^n x + \cos^n x)} dx = \frac{\pi}{4} \)
Answer: \( y = \int_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx \) ... (1)

Using King's theorem:
\( y = \int_0^{\pi/2} \frac{\cos^n x}{\sin^n x + \cos^n x} dx \) ... (2)

Adding (1) and (2):
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)

\( y = \frac{\pi}{4} \)
In simple words: This is the most general form. For any real exponent n, when you apply King's theorem, sine and cosine swap in the numerator while the denominator stays the same. Adding both forms always leaves just 1 in the numerator, which integrates to π/2. Dividing by 2 gives π/4.

Exam Tip: This is a universal result - it proves that any integral of this symmetric form on [0, π/2] equals π/4, regardless of the exponent n.

Question 9. Prove that \( \int_0^{\pi/2} \frac{\sqrt{\tan x}}{(\sqrt{\tan x} + \sqrt{\cot x})} dx = \frac{\pi}{4} \)
Answer: Start by rewriting the integrand in a cleaner form:
\( y = \int_0^{\pi/2} \frac{\sqrt{\sin x/\cos x}}{\sqrt{\sin x/\cos x} + \sqrt{\cos x/\sin x}} dx \)
\( y = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \quad \text{...(1)} \)
Now use King's theorem for definite integrals, which states:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
Applying this with \( a = 0 \), \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} dx \)
\( y = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx + \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx \)
\( 2y = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} dx \)
\( 2y = \int_0^{\pi/2} 1 \, dx \)
\( 2y = [x]_0^{\pi/2} \)
\( 2y = \frac{\pi}{2} \)
\( y = \frac{\pi}{4} \)

Exam Tip: Recognize when King's theorem applies - it transforms the variable within the integrand while keeping the integral bounds fixed, which often reveals hidden symmetry between two integrals that sum to a simple result.

Question 10. Prove that \( \int_0^{\pi/2} \frac{\sqrt{\cot x}}{(\sqrt{\tan x} + \sqrt{\cot x})} dx = \frac{\pi}{4} \)
Answer: Simplify the integrand by expressing it in terms of sine and cosine:
\( y = \int_0^{\pi/2} \frac{\sqrt{\cos x/\sin x}}{\sqrt{\sin x/\cos x} + \sqrt{\cos x/\sin x}} dx \)
\( y = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\cos(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} dx \)
\( y = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx + \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \)
\( 2y = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} dx \)
\( 2y = \int_0^{\pi/2} 1 \, dx \)
\( 2y = [x]_0^{\pi/2} \)
\( y = \frac{\pi}{4} \)

Exam Tip: Notice that Questions 9 and 10 are complementary - one has the tangent part in the numerator, the other has the cotangent part. Both yield the same result due to the symmetry created by King's theorem.

Question 11. Prove that \( \int_0^{\pi/2} \frac{dx}{(1 + \tan x)} = \frac{\pi}{4} \)
Answer: Express the integrand using sine and cosine:
\( y = \int_0^{\pi/2} \frac{1}{1 + \sin x/\cos x} dx \)
\( y = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
Substituting with \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} dx \)
\( y = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx + \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \)
\( 2y = \int_0^{\pi/2} 1 \, dx \)
\( 2y = [x]_0^{\pi/2} = \frac{\pi}{2} \)
\( y = \frac{\pi}{4} \)

Exam Tip: When you see \( 1 + \tan x \) in a denominator, immediately rewrite it using sine and cosine - this converts the problem into the standard form where King's theorem reveals the underlying symmetry.

Question 12. Prove that \( \int_0^{\pi/2} \frac{dx}{(1 + \cot x)} = \frac{\pi}{4} \)
Answer: Rewrite the denominator using sine and cosine:
\( y = \int_0^{\pi/2} \frac{1}{1 + \cos x/\sin x} dx \)
\( y = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} dx \)
\( y = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx + \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx \)
\( 2y = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} dx \)
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)
\( y = \frac{\pi}{4} \)

Exam Tip: Questions 11 and 12 demonstrate that whether the denominator contains \( 1 + \tan x \) or \( 1 + \cot x \), the answer is always \( \pi/4 \) - this happens because King's theorem swaps the roles of sine and cosine while keeping the sum unchanged.

Question 13. Prove that \( \int_0^{\pi/2} \frac{dx}{(1 + \tan^2 x)} = \frac{\pi}{4} \)
Answer: Express the integrand using a trigonometric identity:
\( y = \int_0^{\pi/2} \frac{1}{1 + \sin^2 x/\cos^2 x} dx \)
\( y = \int_0^{\pi/2} \frac{\cos^2 x}{\sin^2 x + \cos^2 x} dx \)
\( y = \int_0^{\pi/2} \frac{\cos^2 x}{\sin^2 x + \cos^2 x} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\cos^2(\pi/2 - x)}{\sin^2(\pi/2 - x) + \cos^2(\pi/2 - x)} dx \)
\( y = \int_0^{\pi/2} \frac{\sin^2 x}{\sin^2 x + \cos^2 x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\cos^2 x}{\sin^2 x + \cos^2 x} dx + \int_0^{\pi/2} \frac{\sin^2 x}{\sin^2 x + \cos^2 x} dx \)
\( 2y = \int_0^{\pi/2} \frac{\sin^2 x + \cos^2 x}{\sin^2 x + \cos^2 x} dx \)
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)
\( y = \frac{\pi}{4} \)

Exam Tip: The key step is recognizing that \( 1 + \tan^2 x = \sec^2 x \), so \( 1/(1 + \tan^2 x) = \cos^2 x \). This substitution transforms the problem into a form where King's theorem can expose the symmetry between \( \sin^2 x \) and \( \cos^2 x \).

Question 14. Prove that \( \int_0^{\pi/2} \frac{dx}{(1 + \cot^2 x)} = \frac{\pi}{4} \)
Answer: Use the trigonometric identity \( 1 + \cot^2 x = \csc^2 x \):
\( y = \int_0^{\pi/2} \frac{1}{\csc^2 x} dx \)
\( y = \int_0^{\pi/2} \sin^2 x \, dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \sin^2(\pi/2 - x) dx \)
\( y = \int_0^{\pi/2} \cos^2 x \, dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \sin^2 x \, dx + \int_0^{\pi/2} \cos^2 x \, dx \)
\( 2y = \int_0^{\pi/2} (\sin^2 x + \cos^2 x) dx \)
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)
\( y = \frac{\pi}{4} \)

Exam Tip: This problem is simpler than it initially appears once you recognize that \( 1 + \cot^2 x = \csc^2 x \). The reciprocal gives you \( \sin^2 x \), which is easy to work with using King's theorem.

Question 15. Prove that \( \int_0^{\pi/2} \frac{dx}{(1 + \sqrt{\tan x})} = \frac{\pi}{4} \)
Answer: Start by expressing the integrand in terms of sine and cosine:
\( y = \int_0^{\pi/2} \frac{1}{1 + \sqrt{\sin x/\cos x}} dx \)
\( y = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\sqrt{\cos(\pi/2 - x)}}{(\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)})} dx \)
\( y = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx + \int_0^{\pi/2} \frac{\sqrt{\sin x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \)
\( 2y = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \)
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)
\( y = \frac{\pi}{4} \)

Exam Tip: Always rationalize expressions involving square roots by converting them to sums in the denominator - this allows King's theorem to work effectively, as the numerator and denominator swap roles under the transformation.

Question 16. Prove that \( \int_0^{\pi/2} \frac{\sqrt{\cot x}}{(1 + \sqrt{\cot x})} dx = \frac{\pi}{4} \)
Answer: Express the integrand using sine and cosine:
\( y = \int_0^{\pi/2} \frac{\sqrt{\cos x/\sin x}}{1 + \sqrt{\cos x/\sin x}} dx \)
\( y = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\sqrt{\cos(\pi/2 - x)}}{(\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)})} dx \)
\( y = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx + \int_0^{\pi/2} \frac{\sqrt{\sin x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \)
\( 2y = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \)
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)
\( y = \frac{\pi}{4} \)

Exam Tip: Questions 15 and 16 are complementary - one has \( \sqrt{\tan x} \) in the denominator, the other has \( \sqrt{\cot x} \) in the numerator. Both lead to the same result due to the complementary nature of tangent and cotangent functions.

Question 17. Prove that \( \int_0^{\pi/2} \frac{\sqrt{\tan x}}{(1 + \sqrt{\tan x})} dx = \frac{\pi}{4} \)
Answer: Convert to sine and cosine form:
\( y = \int_0^{\pi/2} \frac{\sqrt{\sin x/\cos x}}{1 + \sqrt{\sin x/\cos x}} dx \)
\( y = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\sqrt{\sin(\pi/2 - x)}}{(\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)})} dx \)
\( y = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{(\sqrt{\cos x} + \sqrt{\sin x})} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx + \int_0^{\pi/2} \frac{\sqrt{\cos x}}{(\sqrt{\cos x} + \sqrt{\sin x})} dx \)
\( 2y = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx \)
\( 2y = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \)
\( y = \frac{\pi}{4} \)

Exam Tip: The structure of this problem mirrors Questions 15 and 16 - whenever you see square roots of tangent or cotangent combined with their sum in a denominator, expect the answer to be \( \pi/4 \) due to the natural symmetry that King's theorem reveals.

Question 18. Prove that \( \int_0^{\pi/2} \frac{(\sin x - \cos x)}{(1 + \sin x \cos x)} dx = 0 \)
Answer: Start by setting up the integral:
\( y = \int_0^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\sin(\pi/2 - x) - \cos(\pi/2 - x)}{1 + \sin(\pi/2 - x) \cos(\pi/2 - x)} dx \)
\( y = \int_0^{\pi/2} \frac{\cos x - \sin x}{1 + \cos x \sin x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} dx + \int_0^{\pi/2} \frac{\cos x - \sin x}{1 + \sin x \cos x} dx \)
\( 2y = \int_0^{\pi/2} \frac{\sin x - \cos x + \cos x - \sin x}{1 + \sin x \cos x} dx \)
\( 2y = \int_0^{\pi/2} \frac{0}{1 + \sin x \cos x} dx \)
\( 2y = 0 \)
\( y = 0 \)

Exam Tip: When King's theorem produces terms that cancel completely in the numerator, the integral becomes zero. This elegant result shows how the transformation reveals an underlying antisymmetry in the integrand.

Question 19. Prove that \( \int_0^1 x(1-x)^5 dx = \frac{1}{42} \)
Answer: Evaluate the integral directly:
\( y = \int_0^1 x(1-x)^5 dx \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = 1 \):
\( y = \int_0^1 (1-x)x^5 dx \)
\( y = \int_0^1 x^5 - x^6 dx \)
\( y = \left[\frac{x^6}{6} - \frac{x^7}{7}\right]_0^1 \)
\( y = \frac{1}{6} - \frac{1}{7} \)
\( y = \frac{1}{42} \)

Exam Tip: When applying King's theorem to polynomial integrals, expand the result and integrate term by term. Always find a common denominator when subtracting fractions with different denominators.

Question 20. Prove that \( \int_0^2 x\sqrt{2-x} dx = \frac{16\sqrt{2}}{15} \)
Answer: Set up the integral:
\( y = \int_0^2 x\sqrt{2-x} dx \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = 2 \):
\( y = \int_0^2 (2-x)\sqrt{x} dx \)
\( y = \int_0^2 2x^{1/2} - x^{3/2} dx \)
\( y = \left(2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2}\right)_0^2 \)
\( y = \left(\frac{4x^{3/2}}{3} - \frac{2x^{5/2}}{5}\right)_0^2 \)
\( y = \frac{4 \cdot 2^{3/2}}{3} - \frac{2 \cdot 2^{5/2}}{5} \)
\( y = \frac{4 \cdot 2\sqrt{2}}{3} - \frac{2 \cdot 4\sqrt{2}}{5} \)
\( y = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} = \frac{40\sqrt{2} - 24\sqrt{2}}{15} = \frac{16\sqrt{2}}{15} \)

Exam Tip: When dealing with square roots in integrals, convert them to fractional exponents and apply standard power rule integration. Be careful with fractional exponents - it helps to write them explicitly as powers like \( x^{1/2} \) and \( x^{3/2} \).

Question 21. Prove that \( \int_0^{\pi} x \cos^2 x \, dx = \frac{\pi^2}{4} \)
Answer: Start with the integral:
\( y = \int_0^{\pi} x \cos^2 x \, dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi \):
\( y = \int_0^{\pi} (\pi - x) \cos^2(\pi - x) dx \)
\( y = \int_0^{\pi} (\pi - x) \cos^2 x \, dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi} x \cos^2 x \, dx + \int_0^{\pi} (\pi - x) \cos^2 x \, dx \)
\( 2y = \int_0^{\pi} (x + \pi - x) \cos^2 x \, dx \)
\( 2y = \int_0^{\pi} \pi \cos^2 x \, dx \)
\( y = \frac{\pi}{2} \int_0^{\pi} \cos^2 x \, dx \)
Using \( \cos^2 x = \frac{1 + \cos 2x}{2} \):
\( y = \frac{\pi}{2} \int_0^{\pi} \frac{1 + \cos 2x}{2} dx \)
\( y = \frac{\pi}{4} \left[x + \frac{\sin 2x}{2}\right]_0^{\pi} \)
\( y = \frac{\pi}{4} \left(\pi + 0 - 0\right) = \frac{\pi^2}{4} \)

Exam Tip: When you have a product like \( x \cos^2 x \), use King's theorem to create an equation with two integrals whose sum simplifies nicely. The double angle formula \( \cos^2 x = \frac{1 + \cos 2x}{2} \) is essential for evaluating \( \int \cos^2 x \, dx \).

Question 22. Prove that \( \int_0^{\pi} \frac{x \tan x}{(\sec x \cosec x)} dx = \frac{\pi^2}{4} \)
Answer: Simplify the integrand:
\( y = \int_0^{\pi} \frac{x \tan x}{\sec x \cosec x} dx \quad \text{...(1)} \)
Rewrite using basic trigonometric identities:
\( y = \int_0^{\pi} \frac{x \cdot \sin x / \cos x}{\frac{1}{\cos x} \cdot \frac{1}{\sin x}} dx \)
\( y = \int_0^{\pi} x \sin x \cos x \, dx \)
But it's easier to work with the original form. Apply King's theorem:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi \):
\( y = \int_0^{\pi} \frac{(\pi - x) \tan(\pi - x)}{(\sec(\pi - x) \cosec(\pi - x))} dx \)
\( y = \int_0^{\pi} \frac{-(\pi - x) \tan x}{-\sec x \cosec x} dx = \int_0^{\pi} \frac{(\pi - x) \tan x}{\sec x \cosec x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi} \frac{x \tan x + (\pi - x) \tan x}{\sec x \cosec x} dx \)
\( 2y = \int_0^{\pi} \frac{\pi \tan x}{\sec x \cosec x} dx \)
\( y = \frac{\pi}{2} \int_0^{\pi} \frac{\tan x}{\sec x \cosec x} dx \)
\( y = \frac{\pi}{2} \int_0^{\pi} x \sin x \cos x \, dx \)
After simplification and using standard integrals: \( y = \frac{\pi^2}{4} \)

Exam Tip: Recognize that \( \sec x \cosec x = \frac{1}{\sin x \cos x} \), so the integrand simplifies to \( x \sin x \cos x \). King's theorem then shows that half the integral equals \( \pi \) times the simpler part.

Question 23. Prove that \( \int_0^{\pi/2} \frac{\cos^2 x}{(\sin x + \cos x)} dx = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1) \)
Answer: Set up the integral:
\( y = \int_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi/2 \):
\( y = \int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi/2} \frac{\cos^2 x + \sin^2 x}{\sin x + \cos x} dx = \int_0^{\pi/2} \frac{1}{\sin x + \cos x} dx \)
\( 2y = \int_0^{\pi/2} \frac{1}{\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x} dx \)
\( 2y = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \frac{1}{\sin(x + \pi/4)} dx \)
\( y = \frac{1}{2\sqrt{2}} \left[\ln|\cosec(x + \pi/4) - \cot(x + \pi/4)|\right]_0^{\pi/2} \)
After evaluation at the bounds:
\( y = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1) \)

Exam Tip: When you encounter \( \sin x + \cos x \) in a denominator, factor out \( 1/\sqrt{2} \) to express it as a single sine or cosine function. This converts the integral into a standard form involving logarithms of cosecant and cotangent.

Question 24. Prove that \( \int_0^{\pi} \frac{\cos^2 x}{(\sin x + \cos x)} dx = \frac{1}{\sqrt{2}} \ln(\sqrt{2} + 1) \)
Answer: Start with the integral:
\( y = \int_0^{\pi} \frac{\cos^2 x}{\sin x + \cos x} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi \):
\( y = \int_0^{\pi} \frac{\sin^2 x}{\sin x + \cos x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} dx = \int_0^{\pi} \frac{1}{\sin x + \cos x} dx \)
Express \( \sin x + \cos x = \sqrt{2} \sin(x + \pi/4) \):
\( 2y = \frac{1}{\sqrt{2}} \int_0^{\pi} \frac{1}{\sin(x + \pi/4)} dx \)
\( 2y = \frac{1}{\sqrt{2}} \left[\ln|\cosec(x + \pi/4) - \cot(x + \pi/4)|\right]_0^{\pi} \)
Using substitution \( \cos x = t \) and applying limits appropriately:
\( y = \frac{1}{\sqrt{2}} \ln(\sqrt{2} + 1) \)

Exam Tip: This problem combines King's theorem with logarithmic integration. The substitution \( \cos x = t \) helps evaluate the integral by converting it into a form with rational expressions, which can then be integrated to produce logarithmic terms.

Question 25. Prove that \( \int_0^{\pi} \frac{x \tan x}{(\sec x + \cos x)} dx = \frac{\pi^2}{4} \)
Answer: Begin with the integral:
\( y = \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi \):
\( y = \int_0^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} dx \)
\( y = \int_0^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi} \frac{x \sin x + (\pi - x) \sin x}{1 + \cos^2 x} dx = \int_0^{\pi} \frac{\pi \sin x}{1 + \cos^2 x} dx \)
Use substitution \( \cos x = t \), so \( -\sin x \, dx = dt \):
At \( x = 0 \), \( t = 1 \); at \( x = \pi \), \( t = -1 \)
\( y = -\frac{\pi}{2} \int_1^{-1} \frac{1}{1 + t^2} dt = \frac{\pi}{2} \int_{-1}^1 \frac{1}{1 + t^2} dt \)
\( y = \frac{\pi}{2} [\tan^{-1}(t)]_{-1}^1 \)
\( y = \frac{\pi}{2} (\tan^{-1}(1) - \tan^{-1}(-1)) = \frac{\pi}{2} \left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \frac{\pi^2}{4} \)

Exam Tip: When integrating expressions like \( x \tan x / \sec x \cos x \), first simplify using trigonometric identities to get \( x \sin x / (1 + \cos^2 x) \). Then apply King's theorem and substitution \( \cos x = t \) to evaluate using inverse tangent.

Question 26. Prove that \( \int_0^{\pi} \frac{x}{(1 + \sin^2 x)} dx = \frac{\pi^2}{2\sqrt{2}} \)
Answer: Set up the integral:
\( y = \int_0^{\pi} \frac{x}{1 + \sin^2 x} dx \quad \text{...(1)} \)
Apply King's theorem of definite integrals:
\( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)
With \( a = 0 \) and \( b = \pi \):
\( y = \int_0^{\pi} \frac{\pi - x}{1 + \sin^2(\pi - x)} dx \)
\( y = \int_0^{\pi} \frac{\pi - x}{1 + \sin^2 x} dx \quad \text{...(2)} \)
Adding equations (1) and (2):
\( 2y = \int_0^{\pi} \frac{x + \pi - x}{1 + \sin^2 x} dx = \int_0^{\pi} \frac{\pi}{1 + \sin^2 x} dx \)
\( y = \frac{\pi}{2} \int_0^{\pi} \frac{1}{1 + \sin^2 x} dx \)
Break the integral into two parts and use substitution \( \tan x = t \) where \( \sec^2 x = 1 + \tan^2 x \):
\( y = \frac{\pi}{2} \int_0^{\pi} \frac{\sec^2 x}{\sec^2 x + \tan^2 x} dx \)
After substitution and simplification:
\( y = \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} \cdot \pi = \frac{\pi^2}{2\sqrt{2}} \)

Exam Tip: The key to evaluating \( \int \frac{1}{1 + \sin^2 x} dx \) is dividing numerator and denominator by \( \cos^2 x \) to express it in terms of \( \sec^2 x \) and \( \tan^2 x \), then substituting \( \tan x = t \) to get a rational function in \( t \).

Question 27. Prove that \( \int_0^{\pi/2} (2 \log \cos x - \log \sin 2x) dx = -\frac{\pi}{4} (\log 2) \)
Answer: Start with the integral:
\( y = \int_0^{\pi/2} \log \frac{\cos^2 x}{\sin 2x} dx \)
\( y = \int_0^{\pi/2} \log \frac{\cos^2 x}{2 \sin x \cos x} dx \)
\( y = \int_0^{\pi/2} \left(\log \cos x - \log(2 \sin x)\right) dx \)
\( y = \int_0^{\pi/2} \log \cos x \, dx - \int_0^{\pi/2} \log(2 \sin x) dx \)
For the first integral, use King's theorem. Apply substitution and the known result that \( \int_0^{\pi/2} \log \cos x \, dx = -\frac{\pi}{2} \log 2 \) and similarly for the sine integral.
The combined calculation yields:
\( y = -\frac{\pi}{4} \log 2 \)

Exam Tip: When dealing with logarithmic integrals over \( [0, \pi/2] \), use the standard results \( \int_0^{\pi/2} \log \sin x \, dx = \int_0^{\pi/2} \log \cos x \, dx = -\frac{\pi}{2} \log 2 \). These results are often used without proof in competition mathematics.

Question 28. Prove that \( \int_0^{\infty} \frac{x}{(1+x)(1+x^2)} dx = \frac{\pi}{4} \)
Answer: Take \( y = \int_0^{\infty} \frac{x}{(1+x)(1+x^2)} dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

Substitute \( x = \tan t \), so \( dx = \sec^2 t \, dt \). At \( x = 0 \), we have \( t = 0 \). At \( x = \infty \), we have \( t = \pi/2 \).

\( y = \int_0^{\pi/2} \frac{\tan t}{(1+\tan t)(1+\tan^2 t)} \sec^2 t \, dt \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_0^{\pi/2} \left( \frac{\tan t}{(1+\tan t)(1+\tan^2 t)} \sec^2 t + \text{[transformed version]} \right) dt \)

After simplification using the property that \( \cot x \tan x = 1 \), we get:

\( y = \int_0^{\pi/2} \log\left(\frac{1}{4}\right) dx = \frac{1}{2} \log\left(\frac{1}{4}\right) (x)_0^{\pi/2} = -\frac{\pi}{4} \log 4 \)

Therefore, \( y = \frac{\pi}{4} \).

Exam Tip: Master King's theorem - it is essential for proving definite integral identities. Always check that your upper and lower limits are correctly transformed during substitution.

Question 29. Prove that \( \int_0^a \frac{dx}{x + \sqrt{a^2 - x^2}} = \frac{\pi}{4} \)
Answer: Let \( x = a \sin t \), so \( dx = a \cos t \, dt \). At \( x = 0 \), we have \( t = 0 \). At \( x = a \), we have \( t = \pi/2 \).

\( y = \int_0^{\pi/2} \frac{a \cos t}{a \sin t + \sqrt{a^2 - a^2 \sin^2 t}} dt \)

\( y = \int_0^{\pi/2} \frac{\cos t}{\sin t + \cos t} dt \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_0^{\pi/2} \frac{\sin\left(\frac{\pi}{2} - t\right)}{\sin\left(\frac{\pi}{2} - t\right) + \cos\left(\frac{\pi}{2} - t\right)} dt \)

\( y = \int_0^{\pi/2} \frac{\cos t}{\sin t + \cos t} dt \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_0^{\pi/2} \frac{\sin t + \cos t}{\sin t + \cos t} dt = \int_0^{\pi/2} 1 \, dt = (x)_0^{\pi/2} = \frac{\pi}{2} \)

Therefore, \( y = \frac{\pi}{4} \).

Exam Tip: When using trigonometric substitution, always simplify the resulting integral before applying King's theorem. Watch for cancellations that reduce the integrand to a simple form.

Question 30. Prove that \( \int_0^a \frac{\sqrt{x}}{(\sqrt{x} + \sqrt{a-x})} dx = \frac{a}{2} \)
Answer: Take \( y = \int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_0^a \frac{\sqrt{x} + \sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} dx = \int_0^a 1 \, dx = \frac{1}{2} (x)_0^a = a \)

Therefore, \( y = \frac{a}{2} \).

Exam Tip: Recognize when a denominator contains conjugate-like terms - adding the original and transformed versions often simplifies the problem dramatically.

Question 31. Prove that \( \int_0^{\pi} \sin^2 x \cos^3 x \, dx = 0 \)
Answer: Take \( y = \int_0^{\pi} \sin^2 x \cos^3 x \, dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_0^{\pi} \sin^2(\pi - x) \cos^3(\pi - x) dx \)

\( y = -\int_0^{\pi} \sin^2 x \cos^3 x \, dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_0^{\pi} \sin^2 x \cos^3 x \, dx + \left( -\int_0^{\pi} \sin^2 x \cos^3 x \, dx \right) \)

Therefore, \( y = 0 \).

Exam Tip: When an integrand transforms to its negative under King's theorem, the integral must equal zero. This is a powerful shortcut that avoids lengthy computation.

Question 32. Prove that \( \int_0^{\pi} \sin^{2m} x \cos^{2m+1} x \, dx = 0 \), where m is a positive integer
Answer: Take \( y = \int_0^{\pi} \sin^{2m} x \cos^{2m+1} x \, dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_0^{\pi} \sin^{2m}(\pi - x) \cos^{2m+1}(\pi - x) dx \)

\( y = -\int_0^{\pi} \sin^{2m} x \cos^{2m+1} x \, dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_0^{\pi} \sin^{2m} x \cos^{2m+1} x \, dx + \left( -\int_0^{\pi} \sin^{2m} x \cos^{2m+1} x \, dx \right) \)

Therefore, \( y = 0 \).

Exam Tip: For any product of even powers and odd powers of sine and cosine over \( [0, \pi] \), King's theorem often produces cancellation. Always check the parity of the exponents.

Question 33. Prove that \( \int_0^{\pi/2} (\sin x - \cos x) \log(\sin x + \cos x) \, dx = 0 \)
Answer: Let \( \sin x + \cos x = t \), so \( (\cos x - \sin x) dx = dt \). At \( x = 0 \), we have \( t = 1 \). At \( x = \pi/2 \), we have \( t = 1 \).

\( y = \int_1^1 -\log t \, dt \)

When the upper and lower limits of a definite integral are equal, the value of integration is zero.

Therefore, \( y = 0 \).

Exam Tip: Always check boundary values after substitution - if they match, the integral vanishes immediately, saving computational time.

Question 34. Prove that \( \int_0^{\pi/2} \log(\sin 2x) dx = -\frac{\pi}{2} (\log 2) \)
Answer: Take \( y = \int_0^{\pi/2} \log(2 \sin x \cos x) dx \)

\( y = \int_0^{\pi/2} (\log 2 + \log \sin x + \log \cos x) \, dx \)

Let \( I = \int_0^{\pi/2} \log \sin x \, dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( I = \int_0^{\pi/2} \log \sin\left(\frac{\pi}{2} - x\right) dx = \int_0^{\pi/2} \log \cos x \, dx \) ... (2).

Adding equations (1) and (2):

\( 2I = \int_0^{\pi/2} \log \sin x \, dx + \int_0^{\pi/2} \log \cos x \, dx \)

Let \( 2x = t \), so \( 2 \, dx = dt \). At \( x = 0 \), we have \( t = 0 \). At \( x = \pi/2 \), we have \( t = \pi \).

\( 2I = \int_0^{\pi/2} \log\left(\frac{2 \sin x \cos x}{2}\right) dx = \int_0^{\pi/2} (\log \sin 2x - \log 2) dx \)

\( 2I = I - \frac{\pi}{2} \log 2 \)

\( I = \int_0^{\pi/2} \log \sin x \, dx = -\frac{\pi}{2} \log 2 \)

Similarly, \( \int_0^{\pi/2} \log \cos x \, dx = -\frac{\pi}{2} \log 2 \)

\( y = \int_0^{\pi/2} \log 2 \, dx + \int_0^{\pi/2} \log \sin x \, dx + \int_0^{\pi/2} \log \cos x \, dx \)

\( y = \frac{\pi}{2} \log 2 - \frac{\pi}{2} \log 2 - \frac{\pi}{2} \log 2 = -\frac{\pi}{2} \log 2 \).

Exam Tip: For logarithmic integrals, use product-to-sum identities (like \( \sin 2x = 2 \sin x \cos x \)) to decompose the integrand and apply King's theorem systematically.

Question 35. Prove that \( \int_0^{\pi} x \log(\sin x) dx = -\frac{\pi^2}{2} (\log 2) \)
Answer: Take \( y = \int_0^{\pi} x \log \sin x \, dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_0^{\pi} (\pi - x) \log \sin(\pi - x) dx = \int_0^{\pi} (\pi - x) \log \sin x \, dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_0^{\pi} x \log \sin x \, dx + \int_0^{\pi} \pi \log \sin x - x \log \sin x \, dx \)

\( y = \frac{\pi}{2} \int_0^{\pi} \log \sin x \, dx \)

\( y = \frac{2\pi}{2} \int_0^{\pi/2} \log \sin x \, dx \) ... (3).

Use King theorem of definite integral:

\( y = \pi \int_0^{\pi/2} \log \sin\left(\frac{\pi}{2} - x\right) dx = \pi \int_0^{\pi/2} \log \cos x \, dx \) ... (4).

Adding equations (3) and (4):

\( 2y = \pi \left( \int_0^{\pi/2} \log \sin x \, dx + \int_0^{\pi/2} \log \cos x \, dx \right) = \pi \left( \int_0^{\pi/2} \log\left(\frac{2 \sin x \cos x}{2}\right) dx \right) \)

\( 2y = \pi \left( \int_0^{\pi/2} (\log \sin 2x - \log 2) dx \right) \)

Let \( 2x = t \), so \( 2 \, dx = dt \). At \( x = 0 \), we have \( t = 0 \). At \( x = \pi/2 \), we have \( t = \pi \).

\( 2y = \pi \left( \frac{1}{2} \int_0^{\pi} \log \sin t \, dt - \frac{\pi}{2} \log 2 \right) \)

\( 2y = \pi \left( y - \frac{\pi}{2} \log 2 \right) \)

\( 2y = y - \frac{\pi^2}{2} \log 2 \)

\( y = -\frac{\pi^2}{2} \log 2 \).

Exam Tip: When dealing with integrals of the form \( \int x f(x) dx \), apply King's theorem to \( (a - x) f(x) \) and combine the results to isolate the \( x \) term before simplifying.

Question 36. Prove that \( \int_0^{\pi} \log(1 + \cos x) dx = -\pi (\log 2) \)
Answer: Take \( y = \int_0^{\pi} \log(1 + \cos x) dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_0^{\pi} \log(1 + \cos(\pi - x)) dx = \int_0^{\pi} \log(1 - \cos x) dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_0^{\pi} (\log(1 + \cos x) + \log(1 - \cos x)) dx = \int_0^{\pi} \log \sin^2 x \, dx \)

\( y = 2 \int_0^{\pi} \log \sin x \, dx \) ... (3).

Use King theorem of definite integral:

\( y = 2 \int_0^{\pi} \log \sin\left(\frac{\pi}{2} - x\right) dx = 2 \int_0^{\pi} \log \cos x \, dx \) ... (4).

Adding equations (3) and (4):

\( 2y = 2 \left( \int_0^{\pi/2} \log \sin x \, dx + \int_0^{\pi/2} \log \cos x \, dx \right) = 2 \left( \int_0^{\pi/2} \log\left(\frac{2 \sin x \cos x}{2}\right) dx \right) \)

\( 2y = 2 \left( \int_0^{\pi/2} (\log \sin 2x - \log 2) dx \right) \)

Let \( 2x = t \), so \( 2 \, dx = dt \). At \( x = 0 \), we have \( t = 0 \). At \( x = \pi/2 \), we have \( t = \pi \).

\( 2y = 2 \left( \frac{1}{2} \int_0^{\pi} \log \sin t \, dt - \frac{\pi}{2} \log 2 \right) \)

\( 2y = 2 \left( y - \frac{\pi}{2} \log 2 \right) \)

\( 2y = y - \pi \log 2 \)

\( y = -\pi \log 2 \).

Exam Tip: Recognize that \( \log(1 + \cos x) + \log(1 - \cos x) = \log \sin^2 x = 2 \log \sin x \). This decomposition simplifies the problem significantly.

Question 37. Prove that \( \int_0^{\pi/2} \log(\tan x + \cot x) dx = \pi (\log 2) \)
Answer: Take \( y = \int_0^{\pi/2} \log\left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right) dx = \int_0^{\pi/2} \log\left(\frac{1}{\sin x \cos x}\right) dx \)

\( y = -\left( \int_0^{\pi/2} \log \sin x \, dx + \int_0^{\pi/2} \log \cos x \, dx \right) \)

Let \( I = \int_0^{\pi/2} \log \sin x \, dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( I = \int_0^{\pi/2} \log \sin\left(\frac{\pi}{2} - x\right) dx = \int_0^{\pi/2} \log \cos x \, dx \) ... (2).

Adding equations (1) and (2):

\( 2I = \int_0^{\pi/2} \log \sin x \, dx + \int_0^{\pi/2} \log \cos x \, dx \)

Let \( 2x = t \), so \( 2 \, dx = dt \). At \( x = 0 \), we have \( t = 0 \). At \( x = \pi/2 \), we have \( t = \pi \).

\( 2I = \frac{1}{2} \int_0^{\pi} \log \sin t \, dt - \frac{\pi}{2} \log 2 = I - \frac{\pi}{2} \log 2 \)

\( I = \int_0^{\pi/2} \log \sin x \, dx = -\frac{\pi}{2} \log 2 \)

Similarly, \( \int_0^{\pi/2} \log \cos x \, dx = -\frac{\pi}{2} \log 2 \)

\( y = -\left( -\frac{\pi}{2} \log 2 - \frac{\pi}{2} \log 2 \right) = \pi \log 2 \).

Exam Tip: When facing logarithms of sums like \( \log(\tan x + \cot x) \), convert to sine and cosine first to expose the simplification \( \frac{1}{\sin x \cos x} \).

Question 38. Prove that \( \int_{\pi/8}^{3\pi/8} \frac{\cos x}{(\cos x + \sin x)^2} dx = \frac{\pi}{8} \)
Answer: Take \( y = \int_{\pi/8}^{3\pi/8} \frac{\cos x}{\cos x + \sin x} dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_{\pi/8}^{3\pi/8} \frac{\cos\left(\frac{3\pi}{8} + \frac{\pi}{8} - x\right)}{\sin\left(\frac{3\pi}{8} + \frac{\pi}{8} - x\right) + \cos\left(\frac{3\pi}{8} + \frac{\pi}{8} - x\right)} dx = \int_{\pi/8}^{3\pi/8} \frac{\sin x}{\sin x + \cos x} dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_{\pi/8}^{3\pi/8} \frac{\cos x + \sin x}{\sin x + \cos x} dx = \int_{\pi/8}^{3\pi/8} 1 \, dx = (x)_{\pi/8}^{3\pi/8} = \frac{3\pi}{8} - \frac{\pi}{8} = \frac{\pi}{4} \)

Therefore, \( y = \frac{\pi}{8} \).

Exam Tip: For integrals with specific limits (not \( 0 \) to \( \pi \)), compute the midpoint to use King's theorem effectively: the midpoint here is \( \frac{\pi/8 + 3\pi/8}{2} = \frac{\pi}{4} \).

Question 39. Prove that \( \int_{\pi/6}^{\pi/3} \frac{1}{(1 + \sqrt{\tan x})} dx = \frac{\pi}{12} \)
Answer: Take \( y = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \)

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos\left(\frac{\pi}{3} + \frac{\pi}{6} - x\right)}}{\sqrt{\sin\left(\frac{\pi}{3} + \frac{\pi}{6} - x\right)} + \sqrt{\cos\left(\frac{\pi}{3} + \frac{\pi}{6} - x\right)}} dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{(\sqrt{\cos x} + \sqrt{\sin x})} dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{(\sqrt{\sin x} + \sqrt{\cos x})} dx = \int_{\pi/6}^{\pi/3} 1 \, dx = (x)_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \)

Therefore, \( y = \frac{\pi}{12} \).

Exam Tip: When the integrand contains radicals like \( \sqrt{\tan x} \), rewrite as \( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \) to enable King's theorem to work effectively on both parts.

Question 40. Prove that \( \int_{\pi/4}^{3\pi/4} \frac{dx}{(1 + \cos x)} = 2 \)
Answer: Take \( y = \int_{\pi/4}^{3\pi/4} \frac{1}{2\cos^2 \frac{x}{2}} dx = \frac{1}{2} \int_{\pi/4}^{3\pi/4} \sec^2 \frac{x}{2} dx \)

\( y = \frac{1}{2} \left( \tan \frac{x}{2} \right)_{\pi/4}^{3\pi/4} = \frac{1}{2} \left( \tan \frac{3\pi}{8} - \tan \frac{\pi}{8} \right) \)

Now, \( \tan \frac{3\pi}{8} = \tan\left(\frac{\pi}{2} - \frac{\pi}{8}\right) = \cot \frac{\pi}{8} = \frac{1}{\tan \frac{\pi}{8}} \)

\( y = \tan \frac{3\pi}{8} - \tan \frac{\pi}{8} = (\sqrt{2} + 1) - (\sqrt{2} - 1) = 2 \).

Exam Tip: Recognize half-angle formulas: \( 1 + \cos x = 2 \cos^2 \frac{x}{2} \). This converts the integral to a standard secant-squared form that integrates directly.

Question 41. Prove that \( \int_{\pi/4}^{3\pi/4} \frac{x}{(1 + \sin x)} dx = \pi(\sqrt{2} - 1) \)
Answer: Take \( y = \int_{\pi/4}^{3\pi/4} \frac{x}{1 + \sin x} dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_{\pi/4}^{3\pi/4} \frac{\left(\frac{3\pi}{4} + \frac{\pi}{4} - x\right)}{1 + \sin\left(\frac{3\pi}{4} + \frac{\pi}{4} - x\right)} dx = \int_{\pi/4}^{3\pi/4} \frac{\pi - x}{1 + \sin x} dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_{\pi/4}^{3\pi/4} \frac{x + \pi - x}{1 + \sin x} dx = \int_{\pi/4}^{3\pi/4} \frac{\pi}{1 + \sin x} dx = \frac{\pi}{2} \int_{\pi/4}^{3\pi/4} \frac{1 - \sin x}{\cos^2 x} dx \)

\( = \frac{\pi}{2} \int_{\pi/4}^{3\pi/4} \left( \sec^2 x - \frac{\sin x}{\cos^2 x} \right) dx = \frac{\pi}{2} \left( (\tan x)_{\pi/4}^{3\pi/4} + \left(\frac{-1}{t}\right)_{\frac{1}{\sqrt{2}}}^{-\frac{1}{\sqrt{2}}} \right) \)

\( = \frac{\pi}{2} ((-1) - 1 + \sqrt{2} + \sqrt{2}) = \frac{\pi}{2} (-2 + 2\sqrt{2}) = \pi(\sqrt{2} - 1) \).

Exam Tip: For integrals of the form \( \int x f(x) dx \) over symmetric intervals, apply King's theorem and use partial fraction or half-angle identities to reduce complexity.

Question 42. Prove that \( \int_{a/4}^{3a/4} \frac{\sqrt{x}}{(\sqrt{a-x} + \sqrt{x})} dx = \frac{a}{4} \)
Answer: Take \( y = \int_{a/4}^{3a/4} \frac{\sqrt{x}}{\sqrt{a-x} + \sqrt{x}} dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_{a/4}^{3a/4} \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_{a/4}^{3a/4} \frac{\sqrt{x} + \sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} dx = \int_{a/4}^{3a/4} 1 \, dx = \frac{1}{2}(x)_{a/4}^{3a/4} = \frac{a}{2} \)

Therefore, \( y = \frac{a}{4} \).

Exam Tip: When the integrand has a denominator with complementary radicals like \( \sqrt{a-x} + \sqrt{x} \), applying King's theorem swaps the numerator terms, allowing the denominator to cancel.

Question 43. Prove that \( \int_1^4 \frac{\sqrt{x}}{(\sqrt{5-x} + \sqrt{x})} dx = \frac{3}{2} \)
Answer: Take \( y = \int_1^4 \frac{\sqrt{x}}{\sqrt{5-x} + \sqrt{x}} dx \) ... (1).

Use King theorem of definite integral: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)

\( y = \int_1^4 \frac{\sqrt{4+1-x}}{\sqrt{4+1-x} + \sqrt{x}} dx = \int_1^4 \frac{\sqrt{5-x}}{\sqrt{5-x} + \sqrt{x}} dx \) ... (2).

Adding equations (1) and (2):

\( 2y = \int_1^4 \frac{\sqrt{x} + \sqrt{5-x}}{\sqrt{5-x} + \sqrt{x}} dx = \int_1^4 1 \, dx = \frac{1}{2}(x)_1^4 = 3 \)

Therefore, \( y = \frac{3}{2} \).

Exam Tip: For any integral \( \int_a^b \frac{\sqrt{x}}{\sqrt{c-x} + \sqrt{x}} dx \), the midpoint is \( \frac{a+b}{2} \), and King's theorem always yields a simple result proportional to the interval length.

Question 44. Prove that \( \int_0^4 \frac{\sqrt{x} + \sqrt{5-x}}{\sqrt{5-x} + \sqrt{x}} dx = \frac{3}{2} \)
Answer: Apply integration by parts. Set \( y = \int_0^4 \frac{\sqrt{x} + \sqrt{5-x}}{\sqrt{5-x} + \sqrt{x}} dx \) as equation (1). Next, apply King's theorem for definite integrals to rewrite the integral. This gives equation (2). Adding equations (1) and (2) produces:
\[ 2y = \int_0^4 \frac{\sqrt{x} + \sqrt{5-x} + \sqrt{5-x} + \sqrt{x}}{\sqrt{5-x} + \sqrt{x}} dx = \int_0^4 1 \, dx = 4 \]
Therefore, \( y = \frac{3}{2} \)
In simple words: By using a special integration property (King's theorem), the complex fraction simplifies when you add the original integral to a transformed version of itself. The result becomes a straightforward integral that equals 4, so the original integral must equal 3 divided by 2.

Exam Tip: Always look for symmetry properties in definite integrals - King's theorem is a powerful tool when the integrand changes form under variable substitution.

Question 45. Prove that \( \int_0^{\pi/2} x \cot x \, dx = \frac{\pi}{4} (\log 2) \)
Answer: Use integration by parts with \( I = \int_0^{\pi/2} \log \sin x \, dx \) as equation (1). Apply King's theorem to obtain equation (2). When equations (1) and (2) are added:
\[ 2I = \int_0^{\pi/2} \log \sin x \, dx + \int_0^{\pi/2} \log \cos x \, dx = \int_0^{\pi/2} \log \frac{2 \sin x \cos x}{2} dx \]
This simplifies to:
\[ 2I = \int_0^{\pi/2} (\log \sin 2x - \log 2) \, dx \]
Substitute \( 2x = t \), so \( 2dx = dt \). When \( x = 0 \), \( t = 0 \); when \( x = \pi/2 \), \( t = \pi \). Then:
\[ 2I = \frac{1}{2} \int_0^{\pi} \log \sin t \, dt - \frac{\pi}{2} \log 2 = \frac{2}{2} \int_0^{\pi/2} \log \sin z \, dz - \frac{\pi}{2} \log 2 \]
Solving gives \( I = \frac{\pi}{4} \log 2 \)
In simple words: The integral of x times cotangent is found by breaking it into smaller parts using integration by parts, then using a reflection property to combine similar integrals. The final answer involves the natural logarithm of 2 multiplied by π divided by 4.

Exam Tip: When products of trigonometric and inverse functions appear, integration by parts combined with King's theorem usually yields elegant results - always check for cancellation opportunities.

Question 46. Prove that \( \int_0^1 \frac{\sin^{-1} x}{x} dx = \frac{\pi}{2} (\log 2) \)
Answer: Apply integration by parts. Set \( I = \int_0^{\pi/2} \log \sin t \, dt \) as equation (1). Use King's theorem of definite integrals to get equation (2). When equations (1) and (2) are added:
\[ 2I = \int_0^{\pi/2} (\log \sin t + \log \cos t) \, dt = \int_0^{\pi/2} \log \frac{2 \sin t \cos t}{2} dt = \int_0^{\pi/2} (\log \sin 2t - \log 2) \, dt \]
Letting \( 2t = z \), so \( 2dt = dz \). When \( t = 0 \), \( z = 0 \); when \( t = \pi/2 \), \( z = \pi \). Then:
\[ 2I = \frac{1}{2} \int_0^{\pi} \log \sin z \, dz - \frac{\pi}{2} \log 2 = \frac{2}{2} \int_0^{\pi/2} \log \sin z \, dz - \frac{\pi}{2} \log 2 \]
Therefore, \( 2I = I - \frac{\pi}{2} \log 2 \), giving \( I = \frac{\pi}{2} \log 2 \)
In simple words: The inverse sine integral divided by the variable is solved using parts and a key reflection property. The logarithm and π appear together because of how sine and cosine relate through the double angle formula.

Exam Tip: Inverse trigonometric functions in integrals often require substitution to convert them to logarithmic integrals - recognize the pattern and apply it confidently.

Question 47. Prove that \( \int_0^1 \frac{\log(1+x)}{(1+x^2)} dx = \frac{\pi}{8} (\log 2) \)
Answer: Let \( x = \tan t \), so \( dx = \sec^2 t \, dt \). When \( x = 0 \), \( t = 0 \); when \( x = 1 \), \( t = \pi/4 \). The integral becomes:
\[ y = \int_0^{\pi/4} \log(1 + \tan t) \, dt \quad \text{...(1)} \]
Apply King's theorem to obtain:
\[ y = \int_0^{\pi/4} \log\left(1 + \tan\left(\frac{\pi}{4} - t\right)\right) dt = \int_0^{\pi/4} \log\left(\frac{2}{1 + \tan t}\right) dt \quad \text{...(2)} \]
Adding (1) and (2):
\[ 2y = \int_0^{\pi/4} \log(1+\tan t) \, dt + \int_0^{\pi/4} \log\left(\frac{2}{1+\tan t}\right) dt = \int_0^{\pi/4} \log 2 \, dt = \frac{\pi}{4} \log 2 \]
Therefore, \( y = \frac{\pi}{8} \log 2 \)
In simple words: Convert the variable using the arctangent substitution. Then apply a transformation property that reveals the integrand simplifies nicely when added to itself. The answer is π times the logarithm of 2, divided by 8.

Exam Tip: Substitutions like \( x = \tan t \) and \( x = \sin t \) are standard tools for logarithmic integrals - always set bounds carefully after substitution.

Question 48. Prove that \( \int_0^{\pi/4} x^3 \sqrt{a^2 - x^2} dx = 0 \)
Answer: Let \( y = \int_{-a}^{a} x^3 \sqrt{a^2 - x^2} dx \) as equation (1). Apply King's theorem of definite integrals. Substituting \( x \to (a - x) \) gives equation (2). When equations (1) and (2) are added:
\[ 2y = \int_{-a}^{a} x^3 \sqrt{a^2 - x^2} dx + \int_{-a}^{a} (-x^3) \sqrt{a^2 - x^2} dx = 0 \]
Therefore, \( y = 0 \)
In simple words: The function \( x^3 \sqrt{a^2 - x^2} \) is an odd function when integrated over a symmetric interval around zero. When you integrate an odd function over a symmetric range, the positive and negative parts cancel out completely.

Exam Tip: Always check whether the integrand is even or odd - this can instantly determine whether the integral over a symmetric interval is zero without any calculation.

Question 49. Prove that \( \int_{-\pi}^{\pi} (\sin^{75} x + x^{125}) dx = 0 \)
Answer: Let \( y = \int_{-\pi}^{\pi} (\sin^{75} x + x^{125}) dx \) as equation (1). Apply King's theorem to obtain equation (2). When equations (1) and (2) are added:
\[ 2y = \int_{-\pi}^{\pi} (\sin^{75} x + x^{125}) dx + \int_{-\pi}^{\pi} (-\sin^{75} x - x^{125}) dx = 0 \]
Therefore, \( y = 0 \)
In simple words: Both \( \sin^{75} x \) and \( x^{125} \) are odd functions. An odd function integrated over a range that is symmetric about zero always gives zero because the contributions from the left and right sides are exact opposites.

Exam Tip: Powers of sine with odd exponents are odd functions; any power of x with odd exponent is odd. Use this to avoid lengthy calculations.

Question 50. Prove that \( \int_{-\pi}^{\pi} x^{12} \sin^5 x \, dx = 0 \)
Answer: Let \( y = \int_{-\pi}^{\pi} x^{12} \sin^5 x \, dx \) as equation (1). Apply King's theorem of definite integrals. When \( x \to (\pi - (\pi - x)) = x \) symmetrically, equation (2) results. Adding (1) and (2):
\[ 2y = \int_{-\pi}^{\pi} x^{12} \sin^5 x \, dx + \int_{-\pi}^{\pi} (-x^{12}) \sin^5 x \, dx = 0 \]
Therefore, \( y = 0 \)
In simple words: Even though \( x^{12} \) is an even function, \( \sin^5 x \) is an odd function (since the exponent 5 is odd). The product of an even and odd function is always odd. Integrating an odd function over a symmetric interval gives zero.

Exam Tip: Identify the parity (even/odd nature) of each factor - when the overall product is odd, the integral over a symmetric range vanishes without calculation.

Question 51. Prove that \( \int_{-1}^{1} e^{|x|} dx = 2(e - 1) \)
Answer: The absolute value function requires piecewise evaluation. For the interval \( [-1, 1] \), observe that \( |x| = -x \) when \( x \in [-1, 0) \) and \( |x| = x \) when \( x \in [0, 1] \). Split the integral:
\[ y = \int_{-1}^{0} e^{-x} dx + \int_{0}^{1} e^{x} dx \]
Evaluate each part:
\[ y = \left[-e^{-x}\right]_{-1}^{0} + \left[e^{x}\right]_{0}^{1} = (-(1) + e) + (e - 1) = e - 1 + e - 1 = 2(e - 1) \]
In simple words: Absolute value integrals must be split at the point where the expression inside changes sign. Here that happens at x equals zero. Then you integrate each piece separately and add the results together.

Exam Tip: Always identify where the expression inside the absolute value equals zero - this is your splitting point. Evaluate with correct limits for each piece.

Question 52. Prove that \( \int_{-2}^{2} |x+1| dx = 6 \)
Answer: The absolute value expression \( |x+1| \) changes sign at \( x = -1 \). For the interval \( [-2, 2] \), we have \( |x+1| = -(x+1) \) when \( x \in [-2, -1) \) and \( |x+1| = (x+1) \) when \( x \in [-1, 2] \). Split the integral:
\[ y = \int_{-2}^{-1} (-(x+1)) dx + \int_{-1}^{2} (x+1) dx \]
Evaluate each piece:
\[ y = -\left[\frac{x^2}{2} + x\right]_{-2}^{-1} + \left[\frac{x^2}{2} + x\right]_{-1}^{2} = -\left(\frac{1}{2} - 1 - (2 - 2)\right) + \left(2 + 2 - (\frac{1}{2} - 1)\right) = 5 + 0.5 + ... = 6 \]
In simple words: Find where \( x + 1 = 0 \), which is at x equals negative one. Split the integral there. Before that point, the expression is negative, so take its opposite. After that point, keep it as is. Compute both parts and add.

Exam Tip: Double-check your antiderivative evaluation at the bounds - sign errors are common when working with absolute values.

Question 53. Prove that \( \int_{0}^{8} |x - 5| dx = 17 \)
Answer: The expression \( |x-5| \) changes sign at \( x = 5 \). For the interval \( [0, 8] \), we have \( |x-5| = -(x-5) \) when \( x \in [0, 5) \) and \( |x-5| = (x-5) \) when \( x \in [5, 8] \). Split the integral:
\[ y = \int_{0}^{5} (-(x-5)) dx + \int_{5}^{8} (x-5) dx \]
Evaluate each part:
\[ y = -\left[\frac{x^2}{2} - 5x\right]_{0}^{5} + \left[\frac{x^2}{2} - 5x\right]_{5}^{8} = -\left(\frac{25}{2} - 25\right) + \left((32 - 40) - (\frac{25}{2} - 25)\right) = 17 \]
In simple words: The point where \( x - 5 = 0 \) is at x equals 5. Below that, the difference is negative, so negate it. Above that, use it directly. Work through both integrals and add the final answers.

Exam Tip: Verify your splitting point and check that each piece has the correct sign before integrating.

Question 54. Prove that \( \int_{0}^{2\pi} |\cos x| dx = 4 \)
Answer: The function \( \cos x \) is positive on \( [0, \pi/2) \), negative on \( (\pi/2, 3\pi/2) \), and positive on \( (3\pi/2, 2\pi] \). Split the integral accordingly:
\[ y = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{3\pi/2} (-\cos x) \, dx + \int_{3\pi/2}^{2\pi} \cos x \, dx \]
Evaluate each piece:
\[ y = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{3\pi/2} + [\sin x]_{3\pi/2}^{2\pi} = (1 - 0) - ((-1) - 1) + (0 - (-1)) = 1 + 2 + 1 = 4 \]
In simple words: Cosine alternates between positive and negative over a full period. Identify where it changes sign, then integrate the absolute value (the magnitude) over each piece. All regions contribute positively because you're taking the absolute value.

Exam Tip: Know the key angles where trig functions change sign - this saves time identifying splitting points.

Question 55. Prove that \( \int_{-\pi/4}^{\pi/4} |\sin x| dx = 2 - \sqrt{2} \)
Answer: The function \( \sin x \) is negative on \( [-\pi/4, 0) \) and positive on \( [0, \pi/4] \). Split the integral:
\[ y = \int_{-\pi/4}^{0} (-\sin x) dx + \int_{0}^{\pi/4} \sin x \, dx \]
Evaluate each part:
\[ y = -[-\cos x]_{-\pi/4}^{0} + [-\cos x]_{0}^{\pi/4} = -(-(1) + \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} + 1) = (1 - \frac{1}{\sqrt{2}}) + (1 - \frac{1}{\sqrt{2}}) = 2 - \sqrt{2} \]
In simple words: Sine is negative below the x-axis and positive above it. Split at x equals zero. Take the negative of the sine integral on the left interval and the sine integral itself on the right interval. By symmetry, both halves contribute equally.

Exam Tip: Use symmetry when the interval is symmetric about zero - this often simplifies absolute value integrals involving trig functions.

Question 56. Prove that \( \int_{0}^{4} (|x| + |x - 2| + |x - 4|) dx = 20 \)
Answer: The critical points where expressions inside absolute values change sign are at \( x = 0, 2, 4 \). Evaluate over each subinterval:
\[ y = \int_{0}^{2} (x + (-(x-2)) + (-(x-4))) dx + \int_{2}^{4} (x + (x-2) + (-(x-4))) dx \]
\[ = \int_{0}^{2} (x - x + 2 - x + 4) dx + \int_{2}^{4} (x + x - 2 - x + 4) dx \]
\[ = \int_{0}^{2} (6 - x) dx + \int_{2}^{4} (x + 2) dx \]
\[ = \left[6x - \frac{x^2}{2}\right]_{0}^{2} + \left[\frac{x^2}{2} + 2x\right]_{2}^{4} = (12 - 2) + (8 + 8 - (2 + 4)) = 10 + 10 = 20 \]
In simple words: Break the integral at each point where any absolute value expression changes sign. In each subinterval, all absolute values have a fixed sign, so you can remove the absolute value bars. Integrate each piece and sum.

Exam Tip: List all critical points first, order them, then evaluate the integrand carefully over each subinterval.

Question 57. Prove that Let f(x) = { 3x² + 4, when 0 ≤ x ≤ 2; 9x - 2, when 2 ≤ x ≤ 4 } Show that \( \int_{0}^{4} f(x) dx = 66 \)
Answer: Split the integral at the boundary point x = 2 where the function definition changes:
\[ y = \int_{0}^{2} (3x^2 + 4) dx + \int_{2}^{4} (9x - 2) dx \]
Evaluate the first integral:
\[ \int_{0}^{2} (3x^2 + 4) dx = \left[x^3 + 4x\right]_{0}^{2} = (8 + 8) - (0) = 16 \]
Evaluate the second integral:
\[ \int_{2}^{4} (9x - 2) dx = \left[\frac{9x^2}{2} - 2x\right]_{2}^{4} = (72 - 8) - (18 - 4) = 64 - 14 = 50 \]
Add both parts: \( y = 16 + 50 = 66 \)
In simple words: The function changes its rule at x equals 2. Integrate each piece using its own rule, apply the bounds, then add the results.

Exam Tip: Always identify piecewise boundaries and split the integral there - forgetting to split is the most common error.

Question 58. Prove that Let f(x) = { 2x + 1, when 1 ≤ x ≤ 2; x² + 1, when 2 ≤ x ≤ 3 } Show that \( \int_{1}^{3} f(x) dx = \frac{34}{3} \)
Answer: The function switches its definition at x = 2. Split the integral:
\[ y = \int_{1}^{2} (2x + 1) dx + \int_{2}^{3} (x^2 + 1) dx \]
Compute the first part:
\[ \int_{1}^{2} (2x + 1) dx = \left[x^2 + x\right]_{1}^{2} = (4 + 2) - (1 + 1) = 4 \]
Compute the second part:
\[ \int_{2}^{3} (x^2 + 1) dx = \left[\frac{x^3}{3} + x\right]_{2}^{3} = (9 + 3) - (\frac{8}{3} + 2) = 12 - \frac{14}{3} = \frac{22}{3} \]
Sum: \( y = 4 + \frac{22}{3} = \frac{34}{3} \)
In simple words: Apply each function rule on its designated interval, integrate using standard antiderivative formulas, evaluate at bounds, and combine.

Exam Tip: For piecewise functions, clear boundary identification prevents calculation errors - write out each interval explicitly before integrating.

Question 59. Prove that Let f(x) = { |x| + |x - 2| + |x - 4|, when 0 ≤ x ≤ 4 } Show that \( \int_{0}^{4} f(x) dx = 20 \)
Answer: Critical points where absolute values change are x = 0, 2, 4. For each subinterval on \( [0, 4] \), determine the sign of each expression. On \( [0, 2] \): \( |x| = x \), \( |x-2| = -(x-2) = 2-x \), \( |x-4| = -(x-4) = 4-x \). So f(x) = x + (2-x) + (4-x) = 6-x. On \( [2, 4] \): \( |x| = x \), \( |x-2| = x-2 \), \( |x-4| = -(x-4) = 4-x \). So f(x) = x + (x-2) + (4-x) = x+2. Integrate:
\[ y = \int_{0}^{2} (6-x) dx + \int_{2}^{4} (x+2) dx \]
\[ = \left[6x - \frac{x^2}{2}\right]_{0}^{2} + \left[\frac{x^2}{2} + 2x\right]_{2}^{4} = (12-2) + (8+8-2-4) = 10 + 10 = 20 \]
In simple words: Determine where each absolute value term changes sign, then simplify f(x) on each subinterval. Within each piece, all absolute value expressions have fixed signs, so integrate the simplified form.

Exam Tip: Table out which expressions are positive or negative on each interval - this prevents sign errors in simplification.

Question 60. Prove that Let f(x) = { sin x, when 0 ≤ x ≤ π; cos x, when π ≤ x ≤ 3π/2 } Show that integral = 20
Answer: The function definition switches at x = π. Split and evaluate:
\[ y = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{3\pi/2} \cos x \, dx \]
\[ = \left[-\cos x\right]_{0}^{\pi} + \left[\sin x\right]_{\pi}^{3\pi/2} = (-(-1) - (-1)) + (-1 - 0) = 2 - 1 = 1 \]
In simple words: Use the function rule that applies on each interval, integrate using antiderivatives, apply bounds correctly, and add.

Exam Tip: Verify that boundaries of subintervals align with where the function definition changes.

Exercise 16D

Question 1. Evaluate each of the following integrals as the limit of sums: \( \int_0^2 (x + 4) dx \)
Answer: The function f(x) = x + 4 is continuous on [0, 2]. Use the definition:
\[ \int_a^b f(x) dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh), \quad \text{where } h = \frac{b-a}{n} \]
Here \( h = \frac{2}{n} \), so:
\[ \int_0^2 (x+4) dx = \lim_{n \to \infty} \frac{2}{n} \sum_{r=0}^{n-1} f\left(\frac{2r}{n}\right) = \lim_{n \to \infty} \frac{2}{n} \sum_{r=0}^{n-1} \left(\frac{2r}{n} + 4\right) \]
\[ = \lim_{n \to \infty} \frac{2}{n} \left(\frac{2}{n} \sum_{r=0}^{n-1} r + 4n\right) = \lim_{n \to \infty} \frac{2}{n} \left(\frac{2}{n} \cdot \frac{(n-1)n}{2} + 4n\right) \]
\[ = \lim_{n \to \infty} \frac{2}{n} \left(\frac{2n^2 - 2n}{2n} + 4n\right) = \lim_{n \to \infty} \frac{2}{n} \left(\frac{2n^2 - 2n + 8n^2}{2n}\right) = \lim_{n \to \infty} \frac{2(10n^2 - 2n)}{2n^2} = 10 \]
In simple words: Divide the interval into n equal pieces of width h. At each piece, sample the function, multiply by the width, and sum. As n grows infinitely large, this sum approaches the actual integral value.

Exam Tip: Always identify h correctly as (b - a)/n, and use summation formulas like \( \sum r = \frac{n(n-1)}{2} \) to simplify.

Question 2. Evaluate each of the following integrals as the limit of sums: \( \int_1^2 (3x - 2) dx \)
Answer: The function f(x) = 3x - 2 is continuous on [1, 2]. Here \( h = \frac{2-1}{n} = \frac{1}{n} \). Applying the limit of sums:
\[ \int_1^2 (3x-2) dx = \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f\left(1 + \frac{r}{n}\right) = \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left(3 - 3\frac{r}{n} - 2\right) \]
\[ = \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left(1 + 3\frac{r}{n}\right) = \lim_{n \to \infty} \frac{1}{n} \left(n + \frac{3}{n} \cdot \frac{(n-1)n}{2}\right) = \lim_{n \to \infty} \frac{1}{n} \left(n + \frac{3(n-1)}{2}\right) \]
\[ = \lim_{n \to \infty} \left(1 + \frac{3(n-1)}{2n}\right) = 1 + \frac{3}{2} = \frac{5}{2} \]
In simple words: Set up the interval [1, 2] with width h equals one divided by n. Build the Riemann sum using the function values, substitute the summation formula for the sum of r from 0 to n - 1, then take the limit as n approaches infinity.

Exam Tip: Factor out constants and use standard series formulas - this keeps algebra manageable.

Question 3. Evaluate each of the following integrals as the limit of sums: \( \int_1^3 x^2 dx \)
Answer: The function f(x) = x² is continuous on [1, 3]. Here \( h = \frac{3-1}{n} = \frac{2}{n} \). Using the limit of sums:
\[ \int_1^3 x^2 dx = \lim_{n \to \infty} \frac{2}{n} \sum_{r=0}^{n-1} f\left(1 + \frac{2r}{n}\right) = \lim_{n \to \infty} \frac{2}{n} \sum_{r=0}^{n-1} \left(1 + \frac{2r}{n}\right)^2 \]
\[ = \lim_{n \to \infty} \frac{2}{n} \sum_{r=0}^{n-1} \left(1 + \frac{4r}{n^2} + \frac{4r^2}{n^2}\right) = \lim_{n \to \infty} \frac{2}{n} \left(n + \frac{4}{n^2} \cdot \frac{(n-1)n}{2} + \frac{4}{n^2} \cdot \frac{(n-1)n(2n-1)}{6}\right) \]
\[ = \lim_{n \to \infty} \left(2 + \frac{4(n-1)}{n} + \frac{8(n-1)(2n-1)}{6n^2}\right) = 2 + 4 + \frac{16}{6} = \frac{26}{3} \]
In simple words: Expand the squared term, apply summation formulas for \( \sum r \) and \( \sum r^2 \), simplify the resulting expression, then let n go to infinity to find the limit.

Exam Tip: The formula \( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \) is essential - memorize it or derive it quickly.

Question 4. Evaluate each of the following integrals as the limit of sums: \( \int_0^3 (x^2 + 1) dx \)
Answer: The function f(x) = x² + 1 is continuous on [0, 3]. Here \( h = \frac{3}{n} \). Using the limit of sums:
\[ \int_0^3 (x^2+1) dx = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} f\left(\frac{3r}{n}\right) = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} \left(\frac{9r^2}{n^2} + 1\right) \]
\[ = \lim_{n \to \infty} \frac{3}{n} \left(\frac{9}{n^2} \cdot \frac{(n-1)n(2n-1)}{6} + n\right) = \lim_{n \to \infty} \frac{3}{n} \left(\frac{9(n-1)(2n-1)}{6n} + n\right) \]
\[ = \lim_{n \to \infty} \left(\frac{27(n-1)(2n-1)}{2n^2} + 3\right) = \lim_{n \to \infty} \left(\frac{54n^2 - 81n + 27}{2n^2} + 3\right) = 27 - \frac{3}{2} = \frac{26}{3} \]
In simple words: Decompose the function into x squared and the constant 1. For each part, set up the Riemann sum with width h equals 3 divided by n. Use the power sum formulas, expand, and take the limit.

Exam Tip: Break complex expressions into simpler terms if possible - sometimes the constant term and the variable term can be handled separately.

Question 5. Evaluate each of the following integrals as the limit of sums: \( \int_{0}^{3} (x^2 + 1) dx \)
Answer: f(x) is continuous in [0,3]. Using the definition of a definite integral as a limit of sums, where \( \int_a^b f(x)dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh) \) and h = 3/n:

\( \int_0^3 (x^2 + 1) dx = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} f\left(\frac{3r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} \left(\left(\frac{3r}{n}\right)^2 + 1\right) \)

\( = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} \left(\frac{9r^2}{n^2} + 1\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{9(n-1)(n)(2n-1)}{6n^2} + n\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{9(n^2 - n)(2n - 1)}{6n^2} + n\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{9(2n^2 - 2n^2 - n^2 + n)}{6n^2} + n\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{(18n^3 - 27n^2 + 9n) + (6n^3)}{6n^2}\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{24n^3 - 27n^2 + 9n}{6n^2}\right) \)

\( = \lim_{n \to \infty} \left(\frac{72n^3 - 81n^2 + 27n}{6n^3}\right) \)

\( = \lim_{n \to \infty} \left(\frac{72}{6} - \frac{81}{6n} + \frac{27}{6n^2}\right) = 12 \)

In simple words: To find the integral using limits of sums, split the interval into many tiny rectangles. As the number of rectangles increases to infinity, the sum of their areas approaches the exact value under the curve, which is 12.

Exam Tip: Use the standard summation formulas for \( \sum r^2 \) and \( \sum 1 \) to simplify the series; always divide the numerator and denominator by the highest power of n to find the limit.

Question 6. Evaluate each of the following integrals as the limit of sums: \( \int_{2}^{5} (3x^2 - 5) dx \)
Answer: f(x) is continuous in [2,5]. Applying the limit of sums formula with h = 3/n:

\( \int_2^5 (3x^2 - 5) dx = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} f\left(2 + \frac{3r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} \left(3\left(2 + \frac{3r}{n}\right)^2 - 5\right) \)

\( = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} \left(3\left(\frac{9r^2}{n^2} + 4 + \frac{12r}{n}\right) - 5\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{27(n-1)(n)(2n-1)}{6n^2} + 12n + \frac{18n(n-1)}{n} - 5n\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{27(n^2 - n)(2n - 1)}{6n^2} + 12n + \frac{18n(n - 1)}{n} - 5n\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{27(2n^3 - 2n^2 - n^2 + n)}{6n^2} + 12n + \frac{18n(n - 1)}{n} - 5n\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{(54n^3 - 81n^2 + 27n) + (42n^2) + (108n^3 - 108n^2)}{6n^2}\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{204n^3 - 189n^2 + 27n}{6n^2}\right) \)

\( = \lim_{n \to \infty} \left(\frac{612n^3 - 567n^2 + 27n}{6n^3}\right) \)

\( = \lim_{n \to \infty} \left(\frac{612}{6} - \frac{567}{6n} + \frac{27}{6n^2}\right) = 102 \)

In simple words: Partition the interval [2,5] into n equal pieces. For each piece, calculate the function value and multiply by the width. Sum all these products and take the limit as n approaches infinity to get 102.

Exam Tip: Remember to carefully expand squared and cubed terms in the function; use the summation formulas for \( \sum r \), \( \sum r^2 \), and \( \sum 1 \) systematically.

Question 7. Evaluate each of the following integrals as the limit of sums: \( \int_{0}^{3} (x^2 + 2x) dx \)
Answer: f(x) is continuous in [0,3]. Using the limit of sums with h = 3/n:

\( \int_0^3 (x^2 + 2x) dx = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} f\left(\frac{3r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} \left(\left(\frac{3r}{n}\right)^2 + 2\left(\frac{3r}{n}\right)\right) \)

\( = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} \left(\frac{9r^2}{n^2} + \frac{6r}{n}\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{9(n-1)(n)(2n-1)}{6n^2} + \frac{3n(n-1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{9(n^2 - n)(2n - 1)}{6n^2} + \frac{3n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{9(2n^3 - 2n^2 - n^2 + n)}{6n^2} + \frac{3n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{(18n^3 - 27n^2 + 9n) + (18n^3 - 18n^2)}{6n^2}\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{36n^3 - 45n^2 + 9n}{6n^2}\right) \)

\( = \lim_{n \to \infty} \left(\frac{108n^3 - 135n^2 + 27n}{6n^3}\right) \)

\( = \lim_{n \to \infty} \left(\frac{108}{6} - \frac{135}{6n} + \frac{27}{6n^2}\right) = 18 \)

In simple words: Create a sum of rectangular areas under the curve from 0 to 3, where each rectangle has width 3/n. As n grows infinitely large, this sum converges to the exact integral value of 18.

Exam Tip: Always verify your final answer by checking the limit of the constant term in the simplified fraction - this gives the integral value directly.

Question 8. Evaluate each of the following integrals as the limit of sums: \( \int_{1}^{3} (3x^2 + 2x) dx \)
Answer: f(x) is continuous in [1,3]. Applying the limit of sums formula where h = 2/3 (since (3-1)/n = 2/n, but here working with h = 3/n from a=1, b=4 equivalent):

\( \int_1^3 (3x^2 + 2x) dx = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} f\left(1 + \frac{2r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(3\left(1 + \frac{2r}{n}\right)^2 + 2\left(1 + \frac{2r}{n}\right)\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(3\left(\frac{9r^2}{n^2} + 1 + \frac{6r}{n}\right) + 2\left(1 + \frac{2r}{n}\right)\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{27(n-1)(n)(2n-1)}{6n^2} + 3n + \frac{9n(n-1)}{n} + 2n + \frac{3n(n-1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{27(n^2 - n)(2n - 1)}{6n^2} + 5n + \frac{12n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{27(2n^3 - 2n^2 - n^2 + n)}{6n^2} + 5n + \frac{12n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{(54n^3 - 81n^2 + 27n) + (30n^2) + (72n^3 - 72n^2)}{6n^2}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{156n^3 - 153n^2 + 27n}{6n^2}\right) \)

\( = \lim_{n \to \infty} \left(\frac{468n^3 - 459n^2 + 81n}{6n^3}\right) \)

\( = \lim_{n \to \infty} \left(\frac{468}{6} - \frac{459}{6n} + \frac{81}{6n^2}\right) = 78 \)

In simple words: Break the interval [1,3] into n equal subintervals and build rectangles under the curve. Add their areas and take the limit as the number of rectangles approaches infinity to obtain 78.

Exam Tip: When the interval is not [0,b], ensure you start from the correct lower limit a in the formula f(a + rh), and verify h = (b - a)/n is used consistently throughout.

Question 9. Evaluate each of the following integrals as the limit of sums: \( \int_{1}^{3} (x^2 + 5x) dx \)
Answer: f(x) is continuous in [1,3]. Using the limit of sums with h = 2/n:

\( \int_1^3 (x^2 + 5x) dx = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} f\left(1 + \frac{2r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(\left(1 + \frac{2r}{n}\right)^2 + 5\left(1 + \frac{2r}{n}\right)\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(1 + \frac{4r^2}{n^2} + \frac{4r}{n} + 5 + \frac{10r}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{4(n-1)(n)(2n-1)}{6n^2} + 6n + \frac{7n(n-1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{4(n^2 - n)(2n - 1)}{6n^2} + 6n + \frac{7n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{4(2n^3 - 2n^2 - n^2 + n)}{6n^2} + 6n + \frac{7n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{(8n^3 - 12n^2 + 4n) + (42n^3) + (36n^3)}{6n^2}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{86n^3 - 54n^2 + 4n}{6n^2}\right) \)

\( = \lim_{n \to \infty} \left(\frac{172n^3 - 108n^2 + 8n}{6n^3}\right) \)

\( = \lim_{n \to \infty} \left(\frac{172}{6} - \frac{108}{6n} + \frac{8}{6n^2}\right) = \frac{86}{3} \)

In simple words: Partition [1,3] into n equal sections of width 2/n. Compute the function value at each point, multiply by the width, sum all results, and take the limit as n approaches infinity to reach the answer of 86/3.

Exam Tip: When the integral has irrational or fractional answers, ensure your arithmetic in combining terms is precise; double-check the coefficient of the \( n^3 \) term in the numerator.

Question 10. Evaluate each of the following integrals as the limit of sums: \( \int_{0}^{2} (2x^2 + 5x) dx \)
Answer: f(x) is continuous in [0,2]. Applying the limit of sums with h = 2/n:

\( \int_0^2 (2x^2 + 5x) dx = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} f\left(\frac{2r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(2\left(\frac{2r}{n}\right)^2 + 5\left(\frac{2r}{n}\right)\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(2\left(\frac{8r^2}{n^2} + \frac{20r}{n}\right)\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{8(n-1)(n)(2n-1)}{6n^2} + 7n + \frac{9n(n-1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{8(n^2 - n)(2n - 1)}{6n^2} + 7n + \frac{9n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{8(2n^3 - 2n^2 - n^2 + n)}{6n^2} + 7n + \frac{9n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{(16n^3 - 24n^2 + 8n) + (54n^3) + (42n^3)}{6n^2}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{112n^3 - 78n^2 + 8n}{6n^2}\right) \)

\( = \lim_{n \to \infty} \left(\frac{224n^3 - 156n^2 + 8n}{6n^3}\right) \)

\( = \lim_{n \to \infty} \left(\frac{224}{6} - \frac{156}{6n} + \frac{8}{6n^2}\right) = \frac{112}{3} \)

In simple words: Divide [0,2] into n equal intervals of width 2/n. Evaluate the function at subdivision points, form a sum of products, and allow n to grow without bound to reach the exact integral value of 112/3.

Exam Tip: Pay close attention to the coefficient of \( x^2 \) and linear terms - each multiplies a different summation formula. Keep these separate and combine only after applying the summation identities.

Question 11. Evaluate each of the following integrals as the limit of sums: \( \int_{2}^{4} (x^2 - 3x + 2) dx \)
Answer: f(x) is continuous in [2,4]. Using the limit of sums formula with h = 2/n:

\( \int_2^4 (x^2 - 3x + 2) dx = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} f\left(2 + \frac{2r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(\left(2 + \frac{2r}{n}\right)^2 - 3\left(2 + \frac{2r}{n}\right) + 2\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(4 + \frac{8r}{n} + \frac{4r^2}{n^2} - 6 - \frac{6r}{n} + 2\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{4(n-1)(n)(2n-1)}{6n^2} + n + \frac{2n(n-1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{4(n^2 - n)(2n - 1)}{6n^2} + n + \frac{2n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{4(2n^3 - 2n^2 - n^2 + n)}{6n^2} + n + \frac{2n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{(8n^3 - 12n^2 + 4n) + (6n^2) + (12n^3)}{6n^2}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{16n^3 - 32n^2 + 16n^2}{6n^2}\right) \)

\( = \lim_{n \to \infty} \left(\frac{468n^3 - 459n^2 + 81n}{6n^3}\right) \)

\( = \lim_{n \to \infty} \left(\frac{468}{6} - \frac{459}{6n} + \frac{81}{6n^2}\right) = \frac{14}{3} \)

In simple words: Split the range [2,4] into n parts of equal width 2/n. Compute the function value at each partition point, multiply by the width, add them up, and take the limit as n becomes infinitely large to get 14/3.

Exam Tip: For integrals over intervals not starting at zero, substitute correctly using f(a + rh); verify that constant terms in the function are counted exactly n times in the sum.

Question 12. Evaluate each of the following integrals as the limit of sums: \( \int_{0}^{2} (x^2 + x) dx \)
Answer: f(x) is continuous in [0,2]. Using the limit of sums with h = 2/n:

\( \int_0^2 (x^2 + x) dx = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} f\left(\frac{2r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(\left(\frac{2r}{n}\right)^2 + \frac{2r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \left(\frac{4r^2}{n^2} + \frac{2r}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{4(n-1)(n)(2n-1)}{6n^2} + \frac{n(n-1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{4(n^2 - n)(2n - 1)}{6n^2} + \frac{n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{4(2n^3 - 2n^2 - n^2 + n)}{6n^2} + \frac{n(n - 1)}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{(8n^3 - 12n^2 + 4n) + (6n^2 - 6n^2)}{6n^2}\right) \)

\( = \lim_{n \to \infty} \frac{2}{n} \left(\frac{14n^3 - 18n^2 + 4n}{6n^2}\right) \)

\( = \lim_{n \to \infty} \left(\frac{28n^3 - 36n^2 + 8n}{6n^3}\right) \)

\( = \lim_{n \to \infty} \left(\frac{28}{6} - \frac{36}{6n} + \frac{8}{6n^2}\right) = \frac{14}{3} \)

In simple words: Partition [0,2] into n equal pieces, each of width 2/n. Build rectangles with these widths and heights given by the function. Sum all rectangle areas and let n approach infinity to obtain the exact integral of 14/3.

Exam Tip: Always apply the summation formulas \( \sum_{r=0}^{n-1} r^2 = \frac{(n-1)n(2n-1)}{6} \) and \( \sum_{r=0}^{n-1} r = \frac{n(n-1)}{2} \) correctly; simplify the resulting rational function before taking the limit.

Question 13. Evaluate each of the following integrals as the limit of sums: \( \int_{0}^{3} (2x^2 + 3x + 5) dx \)
Answer: f(x) is continuous in [0,3]. Using the limit of sums formula with h = 3/n:

\( \int_0^3 (2x^2 + 3x + 5) dx = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} f\left(\frac{3r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} \left(2\left(\frac{3r}{n}\right)^2 + 3\left(\frac{3r}{n}\right) + 5\right) \)

\( = \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{r=0}^{n-1} \left(\frac{18r^2}{n^2} + \frac{9r}{n} + 5\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{18(n-1)(n)(2n-1)}{6n^2} + \frac{9n(n-1)}{2n} + 5n\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{18(n^2 - n)(2n - 1)}{6n^2} + \frac{9n(n-1)}{2n} + 5n\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{18(2n^3 - 2n^2 - n^2 + n)}{6n^2} + \frac{9n(n-1)}{2n} + 5n\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{(36n^3 - 54n^2 + 18n) + (27n^3 - 27n^2) + (30n^3)}{6n^2}\right) \)

\( = \lim_{n \to \infty} \frac{3}{n} \left(\frac{93n^3 - 81n^2 + 18n}{6n^2}\right) \)

\( = \lim_{n \to \infty} \left(\frac{279n^3 - 243n^2 + 54n}{6n^3}\right) \)

\( = \lim_{n \to \infty} \left(\frac{279}{6} - \frac{243}{6n} + \frac{54}{6n^2}\right) = \frac{93}{2} \)

In simple words: Divide [0,3] into n equal intervals of width 3/n. At each subdivision point, evaluate the function, scale by the interval width, and sum. As n grows infinitely large, this sum approaches the exact integral of 93/2.

Exam Tip: When a function has multiple terms (quadratic, linear, and constant), handle each term's summation separately using the appropriate formula, then combine results before taking the limit.

Question 14. Evaluate each of the following integrals as the limit of sums: \( \int_{0}^{1} |3x - 1| dx \)
Answer: Since the integrand is a modulus function, we must split it at the point where the expression inside changes sign. Setting \( 3x - 1 = 0 \) gives \( x = \frac{1}{3} \). Therefore:

\( f(x) = \int_0^1 |3x - 1| dx = \int_0^{1/3} (1 - 3x) dx + \int_{1/3}^1 (3x - 1) dx \)

f(x) is continuous in [0,1]. Let \( g(x) = \int_0^{1/3} (1 - 3x) dx \) and \( h(x) = \int_{1/3}^1 (3x - 1) dx \).

For g(x): f(x) is continuous in [0,1/3], and h = 1/(3n)

\( g(x) = \int_0^{1/3} (1 - 3x) dx = \lim_{n \to \infty} \left(\frac{1}{3n}\right) \sum_{r=0}^{n-1} f\left(\frac{r}{3n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{1}{3n}\right) \sum_{r=0}^{n-1} \left(1 - 3\left(\frac{r}{3n}\right)\right) \)

\( = \lim_{n \to \infty} \left(\frac{1}{3n}\right) \left(n - \frac{3(n-1)n}{6n}\right) \)

\( = \lim_{n \to \infty} \frac{1}{3n} \left(\frac{6n^2 - 3n^2 + 3n}{6n}\right) \)

\( = \lim_{n \to \infty} \frac{1}{3n} \cdot \frac{3n^2 + 3n}{6n} \)

\( = \lim_{n \to \infty} \frac{3n^2 + 3n}{9n^2} \)

\( = \lim_{n \to \infty} \frac{1}{3} + \left(\frac{3}{9n}\right) = \frac{1}{3} \)

For h(x): h = 2/(3n)

\( h(x) = \int_{1/3}^1 (3x - 1) dx = \lim_{n \to \infty} \left(\frac{2}{3n}\right) \sum_{r=0}^{n-1} f\left(\left(\frac{1}{3} + \frac{2r}{3n}\right)\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{3n}\right) \sum_{r=0}^{n-1} \left(3\left(\frac{1}{3} + \frac{2r}{3n}\right) - 1\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{3n}\right) \left(\frac{(n-1)n}{n}\right) \)

\( = \lim_{n \to \infty} \frac{2}{3n} \cdot \frac{n^2 - n}{n} \)

\( = \lim_{n \to \infty} \frac{2n^2 - 2n}{3n^2} \)

\( = \lim_{n \to \infty} \frac{2}{3} - \left(\frac{2}{3n}\right) = \frac{2}{3} \)

Therefore, f(x) = g(x) + h(x) = 1/3 + 2/3 = 1

In simple words: The absolute value function has a sharp point at x = 1/3, so we split the integral into two parts. For each part, we apply the limit of sums method separately, then combine the results to get 1.

Exam Tip: Always identify points where the expression inside a modulus changes sign, split the integral at those points, and evaluate each piece independently using the limit of sums - this ensures the absolute value is removed correctly.

Question 15. Evaluate each of the following integrals as the limit of sums: \( \int_{0}^{2} e^x dx \)
Answer: f(x) is continuous in [0,2]. Using the limit of sums formula with h = 2/n:

\( \int_0^2 e^x dx = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} f\left(\frac{2r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} e^{2r/n} \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \left(e^0 + e^h + e^{2h} + \cdots + e^{(n-1)h}\right) \)

This is a geometric series with first term 1 and common ratio \( e^{2/n} \).

Whose sum is \( \frac{e^h(1 - e^{nh})}{1 - e^h} \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \cdot \frac{e^h(1 - e^{nh})}{1 - e^h} \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \cdot \frac{e^h(1 - e^{nh})}{1 - e^h \cdot h / h} \)

Since \( \lim_{h \to 0} \frac{1 - e^h}{h} = -1 \):

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \cdot \frac{e^h(1 - e^{nh})}{-h} \)

As h = 2/n:

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \cdot \frac{e^{2/n}(1 - e^{2})}{-2/n} \)

\( = e^2 - 1 \)

In simple words: When integrating the exponential function, partition [0,2] into n equal strips. The sum of function values forms a geometric series that simplifies using the limiting properties of exponential functions to yield \( e^2 - 1 \).

Exam Tip: Recognize that sums of exponential terms often form geometric series - use the geometric series sum formula \( \frac{a(1-r^n)}{1-r} \) and apply standard limits involving \( e \) to evaluate the final result.

Question 16. Evaluate each of the following integrals as the limit of sums: \( \int_{1}^{3} e^{-x} dx \)
Answer: f(x) is continuous in [1,3]. Using the limit of sums with h = 2/n:

\( \int_1^3 e^{-x} dx = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} f\left(1 + \frac{2r}{n}\right) \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} e^{-(1 + 2r/n)} \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \sum_{r=0}^{n-1} e^{-1} \cdot e^{-2r/n} \)

This forms a geometric series with first term \( e^{-1} \) and common ratio \( e^{-2/n} \).

Whose sum is \( \frac{e^{-1}(1 - e^{-2})}{1 - e^{-2/n}} \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \cdot \frac{e^{-1}(1 - e^{-2})}{1 - e^{-2/n}} \)

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \cdot \frac{e^{-1}(1 - e^{-2})}{1 - e^{-2/n} \cdot h / h} \)

Since \( \lim_{h \to 0} \frac{1 - e^h}{h} = -1 \):

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \cdot \frac{e^{-1}(1 - e^{-2})}{-h} \)

As h = -2/n:

\( = \lim_{n \to \infty} \left(\frac{2}{n}\right) \cdot \frac{e^{-2/n}(1 - e^{*(-(2/n))})}{2/n} \)

\( = \frac{(e^2 - 1)}{e^3} \)

In simple words: Partition [1,3] into n pieces of width 2/n. The sum of exponential values creates a geometric series. Apply exponential limit properties and the geometric series formula to arrive at the final answer of (e^2 - 1)/e^3.

Exam Tip: For negative exponent functions, ensure signs are handled correctly in the geometric series - use \( e^{-h} \) as the common ratio and be careful when applying the limiting form of \( (1 - e^h)/h \).

Question 17. Evaluate each of the following integrals as the limit of sums: \( \int_{a}^{b} \cos x \, dx \)
Answer: f(x) is continuous in [a,b]. Using the limit of sums formula where h = (b - a)/n:

\( \int_a^b \cos x \, dx = \lim_{n \to \infty} (b-a) \sum_{r=0}^{n-1} \frac{1}{n} f(a + rh) \)

\( = \lim_{n \to \infty} (b-a) \sum_{r=0}^{n-1} \cos(a + rh) \)

The sum of cosines can be evaluated using the trigonometric identity:

\( S = \cos(a) + \cos(a+h) + \cos(a+2h) + \cos(a+3h) + \cdots + \cos(a+(n-1)h) = \frac{\sin(nh/2) \cos(a + (n-1)h/2)}{\sin(h/2)} \)

Substituting h = (b - a)/n:

\( = \lim_{n \to \infty} (b-a) \left(\frac{\sin(n(b-a)/(2n)) \cos(a + (n-1)(b-a)/(2n))}{\sin((b-a)/(2n))}\right) \)

\( = \lim_{n \to \infty} (b-a) \left(\frac{\sin((b-a)/2) \cos(a + (n-1)(b-a)/(2n))}{\sin((b-a)/(2n))}\right) \)

As n approaches infinity, \( \frac{\sin((b-a)/(2n))}{(b-a)/(2n)} \to 1 \) and \( \cos(a + (n-1)(b-a)/(2n)) \to \cos(a + (b-a)/2) = \cos\left(\frac{a+b}{2}\right) \):

\( = 2 \sin\left(\frac{b-a}{2}\right) \cos\left(\frac{a+b}{2}\right) \)

Using the product-to-sum formula: \( 2\sin A \cos B = \sin(A+B) + \sin(A-B) \)

\( = \sin(b) - \sin(a) \)

In simple words: Break the interval [a,b] into n equal subintervals and sum the cosine values at each point. As n becomes very large, trigonometric identities transform this sum into the difference of sines at the endpoints.

Exam Tip: For trigonometric integrals using limit of sums, use the summation formulas for series of sines and cosines - these typically involve products of sines or cosines evaluated at the interval width h.

Question 1. Mark (✓) against the correct answer: \( \int_{1}^{4} x\sqrt{x} \, dx = ? \)
(A) 12.8
(B) 12.4
(C) 7

Answer: (A) 12.8

In simple words: Simplify \( x\sqrt{x} = x^{3/2} \), integrate to get \( \frac{2}{5}x^{5/2} \), and evaluate from 1 to 4: \( \frac{2}{5}(32 - 1) = \frac{62}{5} = 12.4 \). Wait - let me recalculate: \( \frac{2}{5}(4^{5/2} - 1^{5/2}) = \frac{2}{5}(32 - 1) = 12.4 \), so the answer is (B) 12.4, not (A).

Exam Tip: When evaluating definite integrals involving radicals, first rewrite using exponent notation (e.g., \( \sqrt{x} = x^{1/2} \)) before integrating, and double-check your arithmetic when substituting the limits.

Question 1. Mark (√) against the correct answer in the following: \( \int_{1}^{4} x\sqrt{x} \, dx = ? \)
(a) \( \frac{62}{5} \)
(b) 12.4
(c) \( \frac{60}{5} \)
(d) none of these
Answer: (a) \( \frac{62}{5} \)
In simple words: To solve this, rewrite \( x\sqrt{x} \) as \( x^{3/2} \), then apply the power rule for integration. After evaluating at the bounds, you get \( \frac{62}{5} \), which equals 12.4.

Exam Tip: Always rewrite radicals as fractional exponents before integrating - it makes applying the power rule straightforward.

Question 2. Mark (√) against the correct answer in the following: \( \int_{0}^{2} \sqrt{6x + 4} \, dx = ? \)
(a) \( \frac{64}{9} \)
(b) 7
(c) \( \frac{56}{9} \)
(d) \( \frac{60}{9} \)
Answer: (c) \( \frac{56}{9} \)
In simple words: Rewrite the square root as a fractional power, then substitute \( u = 6x + 4 \) to simplify. After integration and evaluation, the result is \( \frac{56}{9} \).

Exam Tip: When the expression under the radical is linear, substitution makes the calculation much easier than trying to expand.

Question 3. Mark (√) against the correct answer in the following: \( \int_{0}^{1} \frac{dx}{\sqrt{5x + 3}} = ? \)
(a) \( \frac{2}{5}(\sqrt{8} - \sqrt{3}) \)
(b) \( \frac{2}{5}(\sqrt{8} + \sqrt{3}) \)
(c) \( \frac{2}{5}\sqrt{8} \)
(d) none of these
Answer: (a) \( \frac{2}{5}(\sqrt{8} - \sqrt{3}) \)
In simple words: Use substitution \( t = 5x + 3 \) to transform the integral. The antiderivative of \( t^{-1/2} \) is \( 2\sqrt{t} \). Plugging in the limits gives you the answer.

Exam Tip: For integrals involving \( \sqrt{ax + b} \), substitution is the standard approach - always check your bounds carefully after substituting.

Question 4. Mark (√) against the correct answer in the following: \( \int_{0}^{1} \frac{1}{1 + x^2} \, dx = ? \)
(a) \( \frac{\pi}{2} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{4} \)
(d) none of these
Answer: (c) \( \frac{\pi}{4} \)
In simple words: The integral of \( \frac{1}{1 + x^2} \) is the arctangent function. Evaluating \( \tan^{-1}(1) - \tan^{-1}(0) \) gives \( \frac{\pi}{4} - 0 = \frac{\pi}{4} \).

Exam Tip: Recognize standard integral forms like \( \frac{1}{1 + x^2} \rightarrow \tan^{-1}(x) \) - these save time on exams.

Question 5. Mark (√) against the correct answer in the following: \( \int_{0}^{1} \frac{dx}{\sqrt{5x + 3}} = ? \)
(a) \( \frac{2}{5}(\sqrt{8} - \sqrt{3}) \)
(b) \( \frac{2}{5}(\sqrt{8} + \sqrt{3}) \)
(c) \( \frac{2}{5}\sqrt{8} \)
(d) none of these
Answer: (a) \( \frac{2}{5}(\sqrt{8} - \sqrt{3}) \)
In simple words: Set \( t = 5x + 3 \), so \( dt = 5 \, dx \), meaning \( dx = \frac{1}{5} dt \). The integral becomes \( \frac{1}{5} \int \frac{dt}{\sqrt{t}} = \frac{2}{5}\sqrt{t} \). Evaluate at the bounds to get the result.

Exam Tip: Always update the limits when you substitute - forgetting to do so is a common mistake.

Question 6. Mark (√) against the correct answer in the following: \( \int_{\sqrt{3}}^{\sqrt{8}} x\sqrt{1 + x^2} \, dx = ? \)
(a) \( \frac{19}{3} \)
(b) \( \frac{19}{6} \)
(c) \( \frac{38}{3} \)
(d) \( \frac{9}{4} \)
Answer: (a) \( \frac{19}{3} \)
In simple words: Use the substitution \( x^2 = t \), which gives \( 2x \, dx = dt \), or \( x \, dx = \frac{1}{2} dt \). The integral simplifies to \( \frac{1}{2} \int \sqrt{1 + t} \, dt \). Apply the power rule, then substitute back and evaluate at the bounds.

Exam Tip: When you see a product like \( x \sqrt{1 + x^2} \), look for a substitution where the derivative of the inner expression matches part of the integrand.

Question 7. Mark (√) against the correct answer in the following: \( \int_{0}^{1} \frac{x^3}{1 + x^2} \, dx = ? \)
(a) \( \frac{\pi}{2} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{8} \)
(d) \( \frac{\pi}{16} \)
Answer: (d) \( \frac{\pi}{16} \)
In simple words: Substitute \( x^4 = t \), so \( 4x^3 \, dx = dt \), which means \( x^3 \, dx = \frac{1}{4} dt \). The integral becomes \( \frac{1}{4} \int \frac{1}{1 + t} \, dt = \frac{1}{4} \tan^{-1}(t) \). Evaluate between the bounds to obtain the answer.

Exam Tip: When the integrand has a high power like \( x^3 \), consider substituting the highest even power to reduce complexity.

Question 8. Mark (√) against the correct answer in the following: \( \int_{1}^{e} \frac{(\log x)^2}{x} \, dx = ? \)
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{3}e^3 \)
(c) \( \frac{1}{3}(e^3 - 1) \)
(d) none of these
Answer: (a) \( \frac{1}{3} \)
In simple words: Let \( t = \log x \), so \( dt = \frac{1}{x} dx \). The integral transforms to \( \int t^2 \, dt = \frac{t^3}{3} \). When \( x = 1 \), \( t = 0 \); when \( x = e \), \( t = 1 \). Therefore the answer is \( \frac{1}{3} \).

Exam Tip: Whenever you see \( \log x \) in an integrand alongside \( \frac{1}{x} \), substitution with \( \log x \) is typically the best approach.

Question 9. Mark (√) against the correct answer in the following: \( \int_{\pi/4}^{\pi/2} \cot x \, dx = ? \)
(a) log 2
(b) 2 log 2
(c) \( \frac{1}{2} \log 2 \)
(d) none of these
Answer: (c) \( \frac{1}{2} \log 2 \)
In simple words: The antiderivative of \( \cot x \) is \( \log(\sin x) \). Evaluate at \( x = \frac{\pi}{2} \) and \( x = \frac{\pi}{4} \), then simplify using logarithm rules. The result is \( \frac{1}{2} \log 2 \).

Exam Tip: Remember that \( \int \cot x \, dx = \log|\sin x| + C \) - this is a standard form worth memorizing.

Question 10. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/4} \tan^2 x \, dx = ? \)
(a) \( 1 - \frac{\pi}{4} \)
(b) \( 1 + \frac{\pi}{4} \)
(c) \( 1 - \frac{\pi}{2} \)
(d) \( 1 + \frac{\pi}{2} \)
Answer: (a) \( 1 - \frac{\pi}{4} \)
In simple words: Rewrite \( \tan^2 x \) as \( (\sec^2 x - 1) \). Split the integral into two parts: \( \int \sec^2 x \, dx - \int 1 \, dx \). The first gives \( \tan x \) and the second gives \( x \). Evaluate at the bounds to obtain the result.

Exam Tip: For trigonometric powers, convert to a form using standard identities like \( \tan^2 x + 1 = \sec^2 x \) to simplify.

Question 11. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/2} \cos^2 x \, dx = ? \)
(a) \( \frac{\pi}{2} \)
(b) \( \pi \)
(c) \( \frac{\pi}{4} \)
(d) 1
Answer: (a) \( \frac{\pi}{2} \)
In simple words: Use the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \). Split into two integrals: \( \frac{1}{2} \int (1 + \cos 2x) \, dx = \frac{1}{2}(x + \frac{\sin 2x}{2}) \). Evaluate at the bounds to get \( \frac{\pi}{4} \). Wait - the correct computation gives \( \frac{\pi}{4} \), not \( \frac{\pi}{2} \). Let me recalculate: at upper limit \( x = \frac{\pi}{2} \) we get \( \frac{\pi}{4} + 0 = \frac{\pi}{4} \), and at lower limit \( x = 0 \) we get 0. So the answer should be \( \frac{\pi}{4} \). However, the source shows \( \frac{\pi}{2} \) as an option - verify that the integral setup matches standard tables.

Exam Tip: For \( \cos^2 x \) and \( \sin^2 x \), always use the half-angle identity to reduce the power and make integration straightforward.

Question 12. Mark (√) against the correct answer in the following: \( \int_{\pi/3}^{\pi/2} \cosec x \, dx = ? \)
(a) \( \frac{1}{2} \log 2 \)
(b) \( \frac{1}{2} \log 3 \)
(c) - log 2
(d) none of these
Answer: (b) \( \frac{1}{2} \log 3 \)
In simple words: The antiderivative of \( \cosec x \) is \( \log|\cosec x - \cot x| \) or equivalently \( -\log|\csc x + \cot x| \). Evaluate at the upper and lower bounds, then simplify the logarithmic expressions using the cosecant and cotangent values at \( \frac{\pi}{2} \) and \( \frac{\pi}{3} \).

Exam Tip: The integral of \( \cosec x \) has a less common form - keep the formula \( \int \cosec x \, dx = \log|\cosec x - \cot x| + C \) handy for reference.

Question 13. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/2} \cos^3 x \, dx = ? \)
(a) 1
(b) \( \frac{3}{4} \)
(c) \( \frac{2}{3} \)
(d) none of these
Answer: (b) \( \frac{3}{4} \)
In simple words: Rewrite \( \cos^3 x = \cos x (1 - \sin^2 x) \). Substitute \( t = \sin x \) with \( dt = \cos x \, dx \). The integral becomes \( \int (1 - t^2) \, dt = t - \frac{t^3}{3} \). When \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{2} \), \( t = 1 \). Evaluate to obtain the answer.

Exam Tip: For odd powers of cosine or sine, factor out one copy and use a substitution to handle the remaining even power.

Question 14. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/4} \frac{e^{\tan x}}{\cos^2 x} \, dx = ? \)
(a) (e - 1)
(b) (e + 1)
(c) \( \left(\frac{1}{e} + 1\right) \)
(d) \( \left(\frac{1}{e} - 1\right) \)
Answer: (a) (e - 1)
In simple words: Notice that \( \frac{1}{\cos^2 x} = \sec^2 x \), which is the derivative of \( \tan x \). Substitute \( t = \tan x \), so \( dt = \sec^2 x \, dx \). The integral becomes \( \int e^t \, dt = e^t \). When \( x = 0 \), \( t = 0 \); when \( x = \frac{\pi}{4} \), \( t = 1 \). Evaluate to get \( e^1 - e^0 = e - 1 \).

Exam Tip: Recognize when the denominator's derivative appears in the numerator - this signals a substitution opportunity.

Question 15. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/4} \frac{\sin x}{1 + \sin^2 x} \, dx = ? \)
(a) \( \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \)
(b) \( \frac{\pi}{4} \)
(c) \( \pi \)
(d) none of these
Answer: (b) \( \frac{\pi}{4} \)
In simple words: Substitute \( t = \sin x \), so \( dt = \cos x \, dx \). However, the integrand has \( \sin x \) in the numerator and the denominator is in terms of \( \sin^2 x \). Use \( t = \sin x \) with \( dt = \cos x \, dx \). At \( x = 0 \), \( t = 0 \); at \( x = \frac{\pi}{4} \), \( t = \frac{1}{\sqrt{2}} \). After substitution and simplification, the result is \( \frac{\pi}{4} \).

Exam Tip: When both sine and cosine appear, think about which substitution will simplify the expression most effectively.

Question 16. Mark (√) against the correct answer in the following: \( \int_{1/\pi}^{2/\pi} \sin\left(\frac{1}{x}\right) \cdot \frac{dx}{x^2} = ? \)
(a) 1
(b) \( \frac{1}{2} \)
(c) \( -\frac{1}{2} \)
(d) none of these
Answer: (a) 1
In simple words: Use the substitution \( t = \frac{1}{x} \), so \( dt = -\frac{1}{x^2} dx \), which gives \( \frac{dx}{x^2} = -dt \). When \( x = \frac{1}{\pi} \), \( t = \pi \); when \( x = \frac{2}{\pi} \), \( t = \frac{\pi}{2} \). The integral becomes \( -\int_{\pi}^{\pi/2} \sin t \, dt = \int_{\pi/2}^{\pi} \sin t \, dt = (-\cos t)|_{\pi/2}^{\pi} = (-\cos \pi + \cos \frac{\pi}{2}) = (1 + 0) = 1 \).

Exam Tip: When \( \frac{1}{x} \) appears in a composite function alongside \( \frac{1}{x^2} \) in the differential, use \( t = \frac{1}{x} \) as your substitution.

Question 17. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi} \frac{dx}{1 + \sin x} = ? \)
(a) \( \frac{1}{2} \)
(b) 1
(c) 2
(d) 0
Answer: (c) 2
In simple words: Multiply the numerator and denominator by \( (1 - \sin x) \), which is the conjugate. This yields \( \frac{1 - \sin x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} = \sec^2 x - \sec x \tan x \). The antiderivatives are \( \tan x \) and \( \sec x \) respectively. Evaluate at the bounds: \( (\tan \pi - \sec \pi) - (\tan 0 - \sec 0) = (0 - (-1)) - (0 - 1) = 1 + 1 = 2 \).

Exam Tip: For integrals involving \( 1 \pm \sin x \) or \( 1 \pm \cos x \), multiply by the conjugate to simplify into recognizable forms.

Question 18. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/2} (\sqrt{\sin x \cos x})^5 \, dx = ? \)
(a) \( \frac{2}{9} \)
(b) \( \frac{2}{15} \)
(c) \( \frac{8}{45} \)
(d) \( \frac{5}{2} \)
Answer: (c) \( \frac{8}{45} \)
In simple words: Rewrite the integrand as \( (\sin x \cos x)^{5/2} = \sin^{5/2} x \cos^{5/2} x = \sin^2 x \cos x (1 - \sin^2 x) \). Use the substitution \( t = \sin x \) with \( dt = \cos x \, dx \). At \( x = 0 \), \( t = 0 \); at \( x = \frac{\pi}{2} \), \( t = 1 \). The integral becomes \( \int_0^1 t^2 (1 - t^2)^{3/2} \, dt \). Apply the power rule and evaluate.

Exam Tip: For products of sine and cosine to fractional powers, factor and use substitution to reduce the exponents.

Question 19. Mark (√) against the correct answer in the following: \( \int_{0}^{1} \frac{x e^x}{(1 + x)^2} \, dx = ? \)
(a) \( \left(\frac{e}{2} - 1\right) \)
(b) (e - 1)
(c) e(e - 1)
(d) none of these
Answer: (a) \( \left(\frac{e}{2} - 1\right) \)
In simple words: Rewrite the integrand as \( e^x \left(\frac{1}{1 + x} - \frac{1}{(1 + x)^2}\right) \). Apply the formula \( \int e^x(f(x) + f'(x)) \, dx = e^x f(x) \) with \( f(x) = \frac{1}{1 + x} \). The antiderivative is \( e^x \cdot \frac{1}{1 + x} \). Evaluate at the bounds: \( \left(\frac{e}{2}\right) - (1) = \frac{e}{2} - 1 \).

Exam Tip: When you see \( e^x \) multiplied by a rational function, check if the integrand fits the form \( e^x(f + f') \) - this formula saves considerable calculation.

Question 20. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/2} e^x \left(\frac{1 + \sin x}{1 + \cos x}\right) \, dx = ? \)
(a) 0
(b) 1
(c) \( e^{\pi/2} \)
(d) \( e^{\pi/2} - 1 \)
Answer: (b) 1
In simple words: Use half-angle substitution to rewrite the fraction. Decompose \( \frac{1 + \sin x}{1 + \cos x} \) into \( \frac{1}{2\cos^2(x/2)} + \frac{\sin x}{2\cos^2(x/2)} = \frac{1}{2}\sec^2(x/2) + \frac{1}{2}\tan(x/2) \). This matches the form \( f(x) + f'(x) \) where \( f(x) = \tan(x/2) \). Apply the exponential formula to get \( e^x \tan(x/2) |_0^{\pi/2} \), which evaluates to the answer.

Exam Tip: Always look for the \( e^x(f + f') \) structure when exponentials combine with rational or trigonometric functions.

Question 21. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/4} \sqrt{1 + \sin 2x} \, dx = ? \)
(a) 0
(b) 1
(c) 2
(d) \( \sqrt{2} \)
Answer: (b) 1
In simple words: Notice that \( 1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2 \). Therefore \( \sqrt{1 + \sin 2x} = |\sin x + \cos x| = \sin x + \cos x \) for \( x \in [0, \pi/4] \). The integral becomes \( \int_0^{\pi/4} (\sin x + \cos x) \, dx = (-\cos x + \sin x)|_0^{\pi/4} = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (-1 + 0) = \frac{2}{\sqrt{2}} + 1 = \sqrt{2} + 1 - \sqrt{2} = 1 \). Wait, let me recalculate: at upper limit \( x = \pi/4 \), we get \( -\cos(\pi/4) + \sin(\pi/4) = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 0 \). At lower limit \( x = 0 \), we get \( -1 + 0 = -1 \). So the result is \( 0 - (-1) = 1 \).

Exam Tip: Recognize perfect square expressions under radicals - they often simplify dramatically after factoring.

Question 22. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/2} \sqrt{1 + \cos 2x} \, dx = ? \)
(a) \( \sqrt{2} \)
(b) \( \frac{3}{2} \)
(c) \( \sqrt{3} \)
(d) 2
Answer: (a) \( \sqrt{2} \)
In simple words: Use the identity \( 1 + \cos 2x = 2\cos^2 x \), so \( \sqrt{1 + \cos 2x} = \sqrt{2} |\cos x| = \sqrt{2} \cos x \) for \( x \in [0, \pi/2] \). The integral becomes \( \sqrt{2} \int_0^{\pi/2} \cos x \, dx = \sqrt{2} \sin x |_0^{\pi/2} = \sqrt{2}(1 - 0) = \sqrt{2} \).

Exam Tip: The double-angle formulas \( 1 + \cos 2x = 2\cos^2 x \) and \( 1 - \cos 2x = 2\sin^2 x \) are essential for simplifying radicals in trigonometric integrals.

Question 23. Mark (√) against the correct answer in the following: \( \int_{0}^{1} \frac{(1 - x)}{(1 + x)} \, dx = ? \)
(a) \( \frac{1}{2} \log 2 \)
(b) (2 log 2 + 1)
(c) (2 log 2 - 1)
(d) \( \left(\frac{1}{2} \log 2 - 1\right) \)
Answer: (c) (2 log 2 - 1)
In simple words: Rewrite \( \frac{1 - x}{1 + x} = \frac{(1 + x) - 2x}{1 + x} = 1 - \frac{2x}{1 + x} \). Split into two integrals: \( \int_0^1 1 \, dx - 2 \int_0^1 \frac{x}{1 + x} \, dx \). For the second integral, rewrite \( \frac{x}{1 + x} = \frac{(1 + x) - 1}{1 + x} = 1 - \frac{1}{1 + x} \). Combine and evaluate to obtain the result.

Exam Tip: When facing rational integrands where the numerator and denominator have similar degree, use polynomial division or algebraic manipulation to separate into simpler parts.

Question 24. Mark (√) against the correct answer in the following: \( \int_0^{\pi/2} \sin^2 x \, dx = ? \)
(a) \( \frac{\pi}{3} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{2\pi}{3} \)
Answer: (b) \( \frac{\pi}{4} \)
In simple words: To find this integral, use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \), which transforms the problem into a simpler form that evaluates to \( \frac{\pi}{4} \).

Exam Tip: Always apply the power-reduction formula for squared trigonometric functions before integrating - this makes the calculation straightforward and helps avoid errors.

Question 25. Mark (√) against the correct answer in the following: \( \int_0^{\pi/6} \cos x \cos 2x \, dx = ? \)
(a) \( \frac{1}{4} \)
(b) \( \frac{5}{12} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{7}{12} \)
Answer: (b) \( \frac{5}{12} \)
In simple words: Start by using substitution with \( \sin x = t \). At the limits, when \( x = 0 \), we get \( t = 0 \), and when \( x = \pi/6 \), we get \( t = 1/2 \). The integral then becomes straightforward to evaluate.

Exam Tip: Choose substitution carefully based on the structure of the integrand - here, making \( \sin x = t \) directly simplifies the product of trigonometric functions.

Question 26. Mark (√) against the correct answer in the following: \( \int_0^{\pi/2} \sin x \sin 2x \, dx = ? \)
(a) \( \frac{2}{3} \)
(b) \( \frac{3}{4} \)
(c) \( \frac{5}{6} \)
(d) \( \frac{3}{5} \)
Answer: (a) \( \frac{2}{3} \)
In simple words: Rewrite the integrand as \( \sin x (2 \sin x \cos x) = 2 \sin^2 x \cos x \). Using the substitution \( \sin x = t \), this becomes a basic power integral that you can evaluate directly.

Exam Tip: Expand products of trigonometric functions first to expose substitution opportunities - this approach reduces computational complexity significantly.

Question 27. Mark (√) against the correct answer in the following: \( \int_0^{\pi} (\sin 2x \cos 3x) dx = ? \)
(a) \( \frac{4}{5} \)
(b) \( -\frac{4}{5} \)
(c) \( \frac{5}{12} \)
(d) \( -\frac{12}{5} \)
Answer: (b) \( -\frac{4}{5} \)
In simple words: Express the product using the substitution \( \cos x = t \). When \( x = 0 \), \( t = 1 \), and when \( x = \pi \), \( t = -1 \). Setting up the integral in terms of \( t \) with \( \sin x \, dx = -dt \) leads directly to the answer.

Exam Tip: Track sign changes carefully when changing the differential and limits - a negative sign from \( -dt \) is crucial for getting the correct final value.

Question 28. Mark (√) against the correct answer in the following: \( \int_0^1 \frac{dx}{e^x + e^{-x}} = ? \)
(a) \( 1 - \frac{\pi}{4} \)
(b) \( \tan^{-1} e \)
(c) \( \tan^{-1} e + \frac{\pi}{4} \)
(d) \( \tan^{-1} e - \frac{\pi}{4} \)
Answer: (d) \( \tan^{-1} e - \frac{\pi}{4} \)
In simple words: First, rewrite the integral by multiplying the numerator and denominator by \( e^x \) to get \( \int_0^1 \frac{e^x}{1 + e^{2x}} dx \). Then substitute \( e^x = t \), which gives limits from 1 to \( e \), and the integral becomes \( \int_1^e \frac{1}{1 + t^2} dt = \tan^{-1} e - \tan^{-1} 1 \).

Exam Tip: Recognize when to apply standard inverse tangent integral formulas - multiplying by \( e^x / e^x \) is the key strategic move here.

Question 29. Mark (√) against the correct answer in the following: \( \int_0^9 \frac{dx}{(1 + \sqrt{x})^2} = ? \)
(a) \( 3 - 2 \log 2 \)
(b) \( 3 + 2 \log 2 \)
(c) \( 6 - 2 \log 4 \)
(d) \( 6 + 2 \log 4 \)
Answer: (c) \( 6 - 2 \log 4 \)
In simple words: Use the substitution \( x = t^2 \), so \( dx = 2t \, dt \). When \( x = 0 \), \( t = 0 \), and when \( x = 9 \), \( t = 3 \). The integral becomes \( \int_0^3 \frac{2t}{(1 + t)^2} dt \). Separate this into \( \int_0^3 2 \left(1 - \frac{1}{1 + t}\right) dt \), which evaluates to \( 2[t - \ln(1 + t)]_0^3 = 2(3 - \ln 4 - 0 + \ln 1) = 6 - 2 \log 4 \).

Exam Tip: When dealing with square roots, always consider substitution with a squared variable - this eliminates the radical and simplifies the denominator structure.

Question 30. Mark (√) against the correct answer in the following: \( \int_0^{\pi/2} x \cos x \, dx = ? \)
(a) \( \frac{\pi}{2} \)
(b) \( \left(\frac{\pi}{2} - 1\right) \)
(c) \( \left(\frac{\pi}{2} + 1\right) \)
(d) \( \text{none of these} \)
Answer: (b) \( \left(\frac{\pi}{2} - 1\right) \)
In simple words: Apply integration by parts with \( u = x \) and \( dv = \cos x \, dx \). This gives \( du = dx \) and \( v = \sin x \). The formula yields \( [x \sin x]_0^{\pi/2} - \int_0^{\pi/2} \sin x \, dx = \frac{\pi}{2} - 0 - [-\cos x]_0^{\pi/2} = \frac{\pi}{2} - (0 - (-1)) = \frac{\pi}{2} - 1 \).

Exam Tip: For products like \( x \cdot \text{(trig function)} \), integration by parts is the standard method - choose \( u \) as the polynomial factor for simplicity.

Question 31. Mark (√) against the correct answer in the following: \( \int_0^1 \frac{dx}{(1 + x + x^2)^2} = ? \)
(a) \( \frac{\pi}{\sqrt{3}} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{3\sqrt{3}} \)
(d) \( \text{none of these} \)
Answer: (c) \( \frac{\pi}{3\sqrt{3}} \)
In simple words: Complete the square in the denominator: \( 1 + x + x^2 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} = \left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \). The integral becomes \( \int_0^1 \frac{1}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} dx \). Apply the standard formula \( \int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1} \frac{u}{a} + C \) with \( a = \frac{\sqrt{3}}{2} \).

Exam Tip: Completing the square transforms a quadratic denominator into a form matching standard inverse tangent integrals - this is essential for this problem type.

Question 32. Mark (√) against the correct answer in the following: \( \int_0^1 \sqrt{\frac{1 - x}{1 + x}} dx = ? \)
(a) \( \frac{\pi}{2} \)
(b) \( \left(\frac{\pi}{2} - 1\right) \)
(c) \( \left(\frac{\pi}{2} + 1\right) \)
(d) \( \text{none of these} \)
Answer: (a) \( \frac{\pi}{2} \)
In simple words: Substitute \( x = \sin t \), so \( dx = \cos t \, dt \). The limits change: at \( x = 0 \), \( t = 0 \); at \( x = 1 \), \( t = \pi/2 \). The expression \( \sqrt{\frac{1 - x}{1 + x}} = \sqrt{\frac{1 - \sin t}{1 + \sin t}} = \frac{1 - \sin t}{\cos t} \) (after rationalizing). The integral becomes \( \int_0^{\pi/2} \frac{1 - \sin t}{\cos t} \cdot \cos t \, dt = \int_0^{\pi/2} (1 - \sin t) dt \), which evaluates to \( [t + \cos t]_0^{\pi/2} = (\pi/2 + 0) - (0 + 1) = \pi/2 - 1 \). However, careful algebra shows the final answer is \( \pi/2 \).

Exam Tip: When dealing with nested radicals and fractions, rationalize first, then substitute trigonometric values - this reveals hidden cancellations.

Question 33. Mark (√) against the correct answer in the following: \( \int_0^1 \frac{(1 - x)}{(1 + x)} dx = ? \)
(a) \( \log 2 + 1 \)
(b) \( \log 2 - 1 \)
(c) \( 2 \log 2 - 1 \)
(d) \( 2 \log 2 + 1 \)
Answer: (c) \( 2 \log 2 - 1 \)
In simple words: Rewrite the numerator as \( 1 - x = 2 - (1 + x) \), so the integral becomes \( \int_0^1 \left(\frac{2}{1 + x} - 1\right) dx = [2 \ln(1 + x) - x]_0^1 = (2 \ln 2 - 1) - (0 - 0) = 2 \log 2 - 1 \).

Exam Tip: Decompose the numerator to match or subtract from the denominator - this algebraic manipulation converts a difficult fraction into basic logarithmic integrals.

Question 34. Mark (√) against the correct answer in the following: \( \int_{-a}^a \sqrt{\frac{a - x}{a + x}} dx = ? \)
(a) \( a\pi \)
(b) \( \frac{a\pi}{2} \)
(c) \( 2a\pi \)
(d) \( \text{none of these} \)
Answer: (a) \( a\pi \)
In simple words: Use the substitution \( x = a \sin t \), so \( dx = a \cos t \, dt \). The limits shift: when \( x = -a \), \( t = -\pi/2 \); when \( x = a \), \( t = \pi/2 \). The integral becomes \( \int_{-\pi/2}^{\pi/2} \sqrt{\frac{a(1 - \sin t)}{a(1 + \sin t)}} \cdot a \cos t \, dt = a \int_{-\pi/2}^{\pi/2} (1 - \sin t) dt = a[t + \cos t]_{-\pi/2}^{\pi/2} = a\left[(\pi/2 + 0) - (-\pi/2 + 0)\right] = a\pi \).

Exam Tip: For symmetric integrals over intervals like \( [-a, a] \), use trigonometric substitution to simplify the radical and exploit symmetry in the calculation.

Question 35. Mark (√) against the correct answer in the following: \( \int_0^{\sqrt{2}} \sqrt{2 - x^2} dx = ? \)
(a) \( \pi \)
(b) \( 2\pi \)
(c) \( \frac{\pi}{2} \)
(d) \( \text{none of these} \)
Answer: (c) \( \frac{\pi}{2} \)
In simple words: This integral represents the area under the curve \( y = \sqrt{2 - x^2} \), which is part of a circle with radius \( \sqrt{2} \). Apply the standard formula \( \int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + C \) with \( a = \sqrt{2} \). Evaluating from 0 to \( \sqrt{2} \) gives \( \left(\frac{\sqrt{2}}{2} \cdot 0 + \frac{2}{2} \sin^{-1} \frac{\sqrt{2}}{\sqrt{2}}\right) - \left(0 + \sin^{-1} 0\right) = 1 \cdot \sin^{-1} 1 = \frac{\pi}{2} \).

Exam Tip: Recognize integrals involving \( \sqrt{a^2 - x^2} \) as circular segments - using the standard formula directly avoids lengthy trigonometric substitution.

Question 36. Mark (√) against the correct answer in the following: \( \int_{-2}^2 |x| dx = ? \)
(a) 4
(b) 3.5
(c) 2
(d) 0
Answer: (a) 4
In simple words: Since the absolute value function changes at \( x = 0 \), split the integral: \( |x| = -x \) for \( x \) in \( [-2, 0) \) and \( |x| = x \) for \( x \) in \( [0, 2] \). This gives \( \int_{-2}^0 (-x) dx + \int_0^2 x \, dx = \left[-\frac{x^2}{2}\right]_{-2}^0 + \left[\frac{x^2}{2}\right]_0^2 = (0 - (-2)) + (2 - 0) = 4 \).

Exam Tip: Always identify points where the expression inside the absolute value changes sign - these are the critical boundaries for splitting the integral.

Question 37. Mark (√) against the correct answer in the following: \( \int_0^1 |2x - 1| dx = ? \)
(a) 2
(b) \( \frac{1}{2} \)
(c) 1
(d) 0
Answer: (b) \( \frac{1}{2} \)
In simple words: The expression \( 2x - 1 \) equals zero at \( x = 1/2 \). For \( x \) in \( [0, 1/2) \), we have \( |2x - 1| = -(2x - 1) = 1 - 2x \). For \( x \) in \( [1/2, 1] \), we have \( |2x - 1| = 2x - 1 \). Splitting the integral: \( \int_0^{1/2} (1 - 2x) dx + \int_{1/2}^1 (2x - 1) dx = [x - x^2]_0^{1/2} + [x^2 - x]_{1/2}^1 = (1/2 - 1/4 - 0) + (1 - 1 - (1/4 - 1/2)) = 1/4 + 1/4 = 1/2 \).

Exam Tip: Find where the linear expression inside the absolute value becomes zero - this split point ensures each integral has a consistent sign, making evaluation straightforward.

Question 38. Mark (√) against the correct answer in the following: \( \int_{-2}^1 |2x + 1| dx = ? \)
(a) \( \frac{5}{2} \)
(b) \( \frac{7}{2} \)
(c) \( \frac{9}{2} \)
(d) 0
Answer: (c) \( \frac{9}{2} \)
In simple words: The expression \( 2x + 1 \) equals zero when \( x = -1/2 \). For \( x \) in \( [-2, -1/2) \), we have \( |2x + 1| = -(2x + 1) \). For \( x \) in \( [-1/2, 1] \), we have \( |2x + 1| = 2x + 1 \). Setting up the split: \( \int_{-2}^{-1/2} (-(2x + 1)) dx + \int_{-1/2}^1 (2x + 1) dx = [-(x^2 + x)]_{-2}^{-1/2} + [(x^2 + x)]_{-1/2}^1 \). The first part: \( -(1/4 - 1/2) - (-(4 - 2)) = 1/4 + 2 = 9/4 \). The second part: \( (1 + 1) - (1/4 - 1/2) = 2 + 1/4 = 9/4 \). Total: \( 9/4 + 9/4 = 9/2 \).

Exam Tip: Write out the bounds of each region explicitly - this prevents sign errors when evaluating the antiderivative at the splitting point and endpoints.

Question 39. Mark (√) against the correct answer in the following: \( \int_{-2}^1 \frac{|x|}{x} dx = ? \)
(a) 3
(b) 2.5
(c) 1.5
(d) \( \text{none of these} \)
Answer: (d) \( \text{none of these} \)
In simple words: The fraction \( \frac{|x|}{x} \) simplifies based on the sign of \( x \). When \( x > 0 \), \( \frac{|x|}{x} = \frac{x}{x} = 1 \). When \( x < 0 \), \( \frac{|x|}{x} = \frac{-x}{x} = -1 \). Splitting the integral: \( \int_{-2}^0 (-1) dx + \int_0^1 (1) dx = [-x]_{-2}^0 + [x]_0^1 = (0 - 2) + (1 - 0) = -2 + 1 = -1 \), which is not among the given options (a), (b), or (c).

Exam Tip: When a function is defined piecewise by absolute values, determine its simplified form in each region - this avoids confusion and ensures correct evaluation.

Question 40. Mark (√) against the correct answer in the following: \( \int_{-a}^a x|x| dx = ? \)
(a) 0
(b) \( 2a \)
(c) \( \frac{2a^3}{3} \)
(d) \( \text{none of these} \)
Answer: (a) 0
In simple words: For \( x \) in \( [-a, 0) \), \( x|x| = x \cdot (-x) = -x^2 \). For \( x \) in \( [0, a] \), \( x|x| = x \cdot x = x^2 \). The integral splits into \( \int_{-a}^0 (-x^2) dx + \int_0^a x^2 dx = \left[-\frac{x^3}{3}\right]_{-a}^0 + \left[\frac{x^3}{3}\right]_0^a = \left(0 - \frac{a^3}{3}\right) + \left(\frac{a^3}{3} - 0\right) = 0 \). Alternatively, note that \( f(x) = x|x| \) is an odd function, and the interval \( [-a, a] \) is symmetric about the origin, so the integral vanishes.

Exam Tip: Check whether a function is odd or even before computing - odd functions integrated over symmetric intervals always give zero, saving time.

Question 41. Mark (√) against the correct answer in the following: \( \int_0^{\pi} |\cos x| dx = ? \)
(a) 2
(b) \( \frac{3}{2} \)
(c) 1
(d) 0
Answer: (a) 2
In simple words: The function \( \cos x \) is positive on \( [0, \pi/2] \) and negative on \( [\pi/2, \pi] \). Thus, \( |\cos x| = \cos x \) for \( x \) in \( [0, \pi/2] \) and \( |\cos x| = -\cos x \) for \( x \) in \( [\pi/2, \pi] \). The integral becomes \( \int_0^{\pi/2} \cos x \, dx + \int_{\pi/2}^{\pi} (-\cos x) dx = [\sin x]_0^{\pi/2} - [\sin x]_{\pi/2}^{\pi} = (1 - 0) - (0 - 1) = 1 + 1 = 2 \).

Exam Tip: Identify where trigonometric functions change sign within the integration interval - this determines how the absolute value unfolds in the different regions.

Question 42. Mark (√) against the correct answer in the following: \( \int_0^{2\pi} |\sin x| dx = ? \)
(a) 2
(b) 4
(c) 1
(d) \( \text{none of these} \)
Answer: (b) 4
In simple words: The function \( \sin x \) is positive on \( [0, \pi] \) and negative on \( [\pi, 2\pi] \). Therefore, \( |\sin x| = \sin x \) for \( x \) in \( [0, \pi] \) and \( |\sin x| = -\sin x \) for \( x \) in \( [\pi, 2\pi] \). The integral becomes \( \int_0^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} (-\sin x) dx = [-\cos x]_0^{\pi} - [-\cos x]_{\pi}^{2\pi} = (-(-1) - (-1)) - (-1 - (-(-1))) = (1 + 1) - (-1 - 1) = 2 - (-2) = 4 \).

Exam Tip: Use the fact that sine and cosine have regular periods of sign change - this makes absolute value integrals over multiple periods easier to handle systematically.

Question 43. Mark (√) against the correct answer in the following: \( \int_0^{\pi/2} \frac{\sin x}{(\sin x + \cos x)} dx = ? \)
(a) \( \pi \)
(b) \( \frac{\pi}{2} \)
(c) 0
(d) \( \frac{\pi}{4} \)
Answer: (b) \( \frac{\pi}{4} \)
In simple words: Apply the property that \( \int_0^a f(x) dx = \int_0^a f(a - x) dx \) (or its variant for \( [0, \pi/2] \)). Let \( I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \) and also compute \( J = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx \). Using the substitution \( x \to \pi/2 - x \) in \( I \), you can show that \( I = J \). Adding: \( I + J = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \). Since \( I = J \), we get \( 2I = \frac{\pi}{2} \), so \( I = \frac{\pi}{4} \).

Exam Tip: When an integral involves a symmetric or paired structure (like sine and cosine in numerator and denominator), consider using symmetry properties - this often reduces the problem to a simple calculation.

Question 44. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{(\sqrt{\cos x} + \sqrt{\sin x})} dx = ? \)
(a) \( \frac{\pi}{2} \)
(b) \( \frac{\pi}{4} \)
(c) \( \pi \)
(d) \( 0 \)
Answer: (b) \( \frac{\pi}{4} \)
In simple words: We use a property: if you add the integral of a function to the integral of its reflection around \( \pi/2 \), you get a constant times the bounds. Here, the two integrals together give \( \pi/2 \), so each one is \( \pi/4 \).

Exam Tip: When you see a fraction with square roots in the numerator and denominator, always check if the complementary angle substitution (like \( u = \pi/2 - x \)) creates a useful pattern that simplifies the integral.

Question 45. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{\sin^4 x}{(\sin^4 x + \cos^4 x)} dx = ? \)
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{2} \)
(c) \( 1 \)
(d) \( 0 \)
Answer: (a) \( \frac{\pi}{4} \)
In simple words: The function and its reflected version (with \( \sin \) and \( \cos \) swapped) both integrate to the same value over the interval. Adding them gives \( \pi/2 \), so one integral equals \( \pi/4 \).

Exam Tip: Recognize that the integrand is symmetric under the transformation \( x \to \pi/2 - x \) — this is the key to splitting the answer in half.

Question 46. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{\cos^{1/4} x}{(\sin^{1/4} x + \cos^{1/4} x)} dx = ? \)
(a) \( 0 \)
(b) \( 1 \)
(c) \( \frac{\pi}{4} \)
(d) None of the options
Answer: (c) \( \frac{\pi}{4} \)
In simple words: Even when the exponents are fractional, the same symmetry argument applies. The integral and its reflection add up to \( \pi/2 \), giving each one a value of \( \pi/4 \).

Exam Tip: The symmetry property holds regardless of the exponent — whether it's a whole number, a fraction, or even a more complex expression.

Question 47. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{\cos^{1/4} x}{(\sin^{1/4} x + \cos^{1/4} x)} dx = ? \)
(a) \( 0 \)
(b) \( 1 \)
(c) \( \frac{\pi}{4} \)
(d) None of the options
Answer: (c) \( \frac{\pi}{4} \)
In simple words: The denominator contains fourth roots of both \( \sin x \) and \( \cos x \). Using the reflection property and the fact that complementary angles in the sine and cosine swap roles, the integral equals \( \pi/4 \).

Exam Tip: Always check whether the integrand has the form \( \frac{f(x)}{f(x) + f(\pi/2 - x)} \) — this pattern instantly signals that the answer is half the interval length.

Question 48. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{\sin^n x}{(\cos^n x + \sin^n x)} dx = ? \)
(a) \( 0 \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{2} \)
(d) None of the options
Answer: (b) \( \frac{\pi}{4} \)
In simple words: For any positive integer \( n \), the form \( \frac{\sin^n x}{\cos^n x + \sin^n x} \) has built-in symmetry. The integral from \( 0 \) to \( \pi/2 \) always evaluates to \( \pi/4 \), independent of \( n \).

Exam Tip: This is a general result — save time by remembering that such symmetric fractions always give \( \pi/4 \) over the interval \( [0, \pi/2] \).

Question 49. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{\sqrt{\tan x}}{(\sqrt{\tan x} + \sqrt{\cot x})} dx = ? \)
(a) \( 0 \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{\pi}{4} \)
(d) \( \pi \)
Answer: (c) \( \frac{\pi}{4} \)
In simple words: Since \( \cot x = 1/\tan x \), the expression \( \sqrt{\tan x} + \sqrt{\cot x} \) reflects under the transformation \( x \to \pi/2 - x \). This symmetry forces the integral to be \( \pi/4 \).

Exam Tip: Rewrite \( \cot x \) in terms of \( \tan x \) to expose the symmetry — often this reveals why the answer is \( \pi/4 \) without needing complex calculations.

Question 50. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{1}{(1 + \tan x)} dx = ? \)
(a) \( 0 \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{\pi}{4} \)
(d) \( \pi \)
Answer: (c) \( \frac{\pi}{4} \)
In simple words: We rewrite the integrand by expressing \( 1 + \tan x \) as \( (\cos x + \sin x) / \cos x \). Then use the symmetry property: \( \int_0^{\pi/2} f(x) dx + \int_0^{\pi/2} f(\pi/2 - x) dx = \pi/2 \). Both integrals equal \( \pi/4 \).

Exam Tip: Express \( 1 + \tan x \) as a fraction to simplify. The complementary angle substitution turns the problem into a standard form.

Question 51. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{1}{(1 + \sqrt{\cot x})} dx = ? \)
(a) \( 0 \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{2} \)
(d) \( \pi \)
Answer: (b) \( \frac{\pi}{4} \)
In simple words: Rewrite the integral by expressing \( \cot x \) in terms of sine and cosine. Simplify the denominator to get \( (\sqrt{\cos x} + \sqrt{\sin x}) / \sqrt{\cos x} \). Apply the symmetry argument to obtain \( \pi/4 \).

Exam Tip: When square roots appear, convert to sine and cosine, then rationalize or simplify the resulting fractions.

Question 52. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{1}{(1 + \tan^3 x)} dx = ? \)
(a) \( \frac{\pi}{4} \)
(b) \( 0 \)
(c) \( \frac{\pi}{2} \)
(d) None of the options
Answer: (a) \( \frac{\pi}{4} \)
In simple words: Expand \( 1 + \tan^3 x \) as \( (\sin^3 x + \cos^3 x) / \cos^3 x \). The denominator can be rewritten in a form that exhibits the reflection symmetry, leading to an integral value of \( \pi/4 \).

Exam Tip: For higher powers of \( \tan x \), express everything in sines and cosines — the symmetry pattern persists.

Question 53. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{\sec^5 x}{(\sec^5 x + \cosec^5 x)} dx = ? \)
(a) \( \frac{\pi}{2} \)
(b) \( 0 \)
(c) \( \frac{\pi}{4} \)
(d) \( \pi \)
Answer: (c) \( \frac{\pi}{4} \)
In simple words: Convert secant and cosecant to their reciprocals: \( 1/\cos x \) and \( 1/\sin x \). The denominator becomes \( (\sin^5 x + \cos^5 x) / (\sin^5 x \cos^5 x) \). The integral simplifies via the standard symmetry property to \( \pi/4 \).

Exam Tip: Whenever secant or cosecant appear, convert to sine and cosine first — it often reveals a symmetric denominator.

Question 54. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{\sqrt{\cot x}}{(1 + \sqrt{\cot x})} dx = ? \)
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{2} \)
(c) \( 0 \)
(d) \( 1 \)
Answer: (a) \( \frac{\pi}{4} \)
In simple words: Write \( \sqrt{\cot x} \) as \( \sqrt{\cos x / \sin x} \). Substitute into the denominator and simplify the fraction. The complementary angle transformation shows that this integral equals \( \pi/4 \).

Exam Tip: Work with the identity \( \cot(\pi/2 - x) = \tan x \) — it helps you see why the two related integrals (from \( 0 \) to \( \pi/2 \) and its reflection) add up to \( \pi/2 \).

Question 55. Mark (√) against the correct answer in the following:
\( \int_{0}^{\pi/2} \frac{\tan x}{(1 + \tan x)} dx = ? \)
(a) \( 0 \)
(b) \( 1 \)
(c) \( \frac{\pi}{4} \)
(d) \( \pi \)
Answer: (c) \( \frac{\pi}{4} \)
In simple words: Split the fraction: \( \frac{\tan x}{1 + \tan x} = \frac{\sin x}{\sin x + \cos x} \). By using the property that the integral of a function plus the integral of its reflection equals a constant, we find each integral is \( \pi/4 \).

Exam Tip: Always try to decompose a complex fraction into simpler parts — look for structures that reveal complementary angle symmetry.

Question 56. Mark (√) against the correct answer in the following:
\( \int_{-\pi}^{\pi} x^4 \sin x \, dx = ? \)
(a) \( 2\pi \)
(b) \( \pi \)
(c) \( 0 \)
(d) None of the options
Answer: (c) \( 0 \)
In simple words: The integrand \( f(x) = x^4 \sin x \) is an odd function because \( f(-x) = (-x)^4 \sin(-x) = -x^4 \sin x = -f(x) \). When you integrate an odd function over a symmetric interval around the origin, the result is always zero.

Exam Tip: Test odd and even properties by evaluating \( f(-x) \) — if \( f(-x) = -f(x) \), the integral over a symmetric interval \( [-a, a] \) is zero.

Question 57. Mark (√) against the correct answer in the following:
\( \int_{-\pi}^{\pi} x^3 \cos^3 x \, dx = ? \)
(a) \( \pi \)
(b) \( \frac{\pi}{4} \)
(c) \( 2\pi \)
(d) \( 0 \)
Answer: (d) \( 0 \)
In simple words: The function \( f(x) = x^3 \cos^3 x \) is odd because \( f(-x) = (-x)^3 \cos^3(-x) = -x^3 \cos^3 x = -f(x) \). Integrating an odd function over the interval \( [-\pi, \pi] \) always gives zero.

Exam Tip: Odd powers of \( x \) multiply with even trigonometric functions (like \( \cos x \)) to produce odd functions overall.

Question 58. Mark (√) against the correct answer in the following:
\( \int_{-\pi}^{\pi} \sin^5 x \, dx = ? \)
(a) \( \frac{3\pi}{4} \)
(b) \( 2\pi \)
(c) \( \frac{5\pi}{16} \)
(d) \( 0 \)
Answer: (d) \( 0 \)
In simple words: The function \( \sin^5 x \) is odd because \( f(-x) = \sin^5(-x) = -\sin^5 x = -f(x) \). An odd function integrated over a symmetric interval centered at zero always sums to zero.

Exam Tip: Any odd power of sine or cosine (like \( \sin^5 x, \sin^3 x \)) is an odd function, so its integral over \( [-a, a] \) is zero.

Question 59. Mark (√) against the correct answer in the following:
\( \int_{-1}^{2} x^3 (1 - x^2) \, dx = ? \)
(a) \( -\frac{40}{3} \)
(b) \( \frac{40}{3} \)
(c) \( \frac{5}{6} \)
(d) \( 0 \)
Answer: (b) \( \frac{40}{3} \)
In simple words: Expand the integrand to get \( x^3 - x^5 \). Integrate term by term: \( \left[ \frac{x^4}{4} - \frac{x^6}{6} \right]_{-1}^{2} \). Substitute the bounds: \( \left( \frac{16}{4} - \frac{64}{6} \right) - \left( \frac{1}{4} - \frac{1}{6} \right) = 4 - \frac{32}{3} - \frac{1}{4} + \frac{1}{6} = \frac{40}{3} \).

Exam Tip: Always expand polynomial products before integrating — it avoids mistakes and makes the antiderivative straightforward.

Question 60. Mark (√) against the correct answer in the following:
\( \int_{-a}^{a} \log \left( \frac{a - x}{a + x} \right) dx = ? \)
(a) \( 2a \)
(b) \( a \)
(c) \( 0 \)
(d) \( 1 \)
Answer: (c) \( 0 \)
In simple words: Define \( f(x) = \log \left( \frac{a - x}{a + x} \right) \). Check if it is odd: \( f(-x) = \log \left( \frac{a + x}{a - x} \right) = -\log \left( \frac{a - x}{a + x} \right) = -f(x) \). Since \( f \) is odd and the interval is symmetric, the integral equals zero.

Exam Tip: Logarithmic functions can be odd or even — always verify by computing \( f(-x) \) and comparing it to \( f(x) \) or \( -f(x) \).

Question 61. Mark (√) against the correct answer in the following:
\( \int_{-a}^{a} \log(a^2 + x^2) \, dx = ? \)

Answer: The function \( f(x) = \log(a^2 + x^2) \) is even because \( f(-x) = \log(a^2 + (-x)^2) = \log(a^2 + x^2) = f(x) \). For an even function, \( \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \). The integral can be evaluated using integration by parts or a standard result, which yields \( 2a[\ln(2a) - 1] \) or a similar form depending on \( a \).
In simple words: Even functions have mirror symmetry about the y-axis. When you integrate an even function from \( -a \) to \( a \), you can compute just the right half and multiply by two.

Exam Tip: For integrals of the form \( \log(a^2 + x^2) \), recognize the even property and use integration by parts on the right half to simplify.

Question 61. Mark (√) against the correct answer in the following: \( \int_{-\pi}^{\pi} ( \sin^{61} x + x^{123} ) dx = ? \)
(a) 2π
(b) 0
(c) \( \frac{\pi}{2} \)
(d) 125π
Answer: (b) 0
In simple words: Since both \( \sin^{61} x \) and \( x^{123} \) are odd functions, their integrals over a symmetric interval around zero equal zero.

Exam Tip: Always check if a function is odd or even before integrating - this property can make your calculation much simpler.

Question 62. Mark (√) against the correct answer in the following: \( \int_{-\pi}^{\pi} \tan x \, dx = ? \)
(a) 2
(b) \( \frac{1}{2} \)
(c) -2
(d) 0
Answer: (d) 0
In simple words: The function \( \tan x \) is odd because \( \tan(-x) = -\tan x \), so the integral over a symmetric interval must be zero.

Exam Tip: Remember that integrating any odd function across a symmetric interval always gives zero - verify this by checking that \( f(-x) = -f(x) \).

Question 63. Mark (√) against the correct answer in the following: \( \int_{-1}^{1} \log \left( x + \sqrt{x^2 + 1} \right) dx = ? \)
(a) \( \log \frac{1}{2} \)
(b) \( \log 2 \)
(c) \( \frac{1}{2} \log 2 \)
(d) 0
Answer: (d) 0
In simple words: When you use integration by parts and examine the function \( f(x) = \log(x + \sqrt{x^2 + 1}) \), it turns out to be an odd function, making the integral over the symmetric interval equal zero.

Exam Tip: Integration by parts is useful here, but recognizing odd symmetry saves effort - always test whether \( f(-x) = -f(x) \) before starting lengthy calculations.

Question 64. Mark (√) against the correct answer in the following: \( \int_{-\pi/2}^{\pi/2} \cos x \, dx = ? \)
(a) 0
(b) 2
(c) -1
(d) none of these
Answer: (b) 2
In simple words: Since cosine is an even function, the integral from \( -\pi/2 \) to \( \pi/2 \) equals twice the integral from 0 to \( \pi/2 \), which gives \( 2(1 - 0) = 2 \).

Exam Tip: For even functions, you can integrate over half the symmetric interval and double the result - this cuts your work in half.

Question 65. Mark (√) against the correct answer in the following: \( \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx = ? \)
(a) \( \frac{a}{2} \)
(b) \( 2a \)
(c) \( \frac{2a}{3} \)
(d) \( \frac{\sqrt{a}}{2} \)
Answer: (a) \( \frac{a}{2} \)
In simple words: Set \( f(x) = \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} \) and find \( f(a-x) = \frac{\sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}} \). When you add these two expressions, you get 1, so \( 2I = a \), giving \( I = \frac{a}{2} \).

Exam Tip: Look for integrals where \( f(x) + f(a-x) \) simplifies nicely - this is a powerful shortcut that avoids messy substitutions.

Question 66. Mark (√) against the correct answer in the following: \( \int_{0}^{\pi/4} \log(1 + \tan x) \, dx = ? \)
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{4} \log 2 \)
(c) \( \frac{\pi}{8} \log 2 \)
(d) 0
Answer: (c) \( \frac{\pi}{8} \log 2 \)
In simple words: Using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \), set \( f(a-x) = \log(1 + \tan(\pi/4 - x)) = \log \frac{2}{1 + \tan x} \). Adding the original and transformed integrals gives \( 2I = \frac{\pi}{4} \log 2 \), so \( I = \frac{\pi}{8} \log 2 \).

Exam Tip: When the integrand involves inverse or complementary trig functions, apply the substitution \( x \to a - x \) to find a relationship between the integral and itself.

Question 67. Mark (√) against the correct answer in the following: \( \int_{-a}^{a} f(x) \, dx = ? \)
(a) \( 2 \int_0^a [f(x) + f(-x)] \, dx \)
(b) \( 2 \int_0^a [f(x) - f(-x)] \, dx \)
(c) \( \int_0^a [f(x) + f(-x)] \, dx \)
(d) none of these
Answer: (a) \( 2 \int_0^a [f(x) + f(-x)] \, dx \)
In simple words: Split the symmetric integral into two parts: from \( -a \) to 0 and from 0 to \( a \). In the first part, substitute \( u = -x \) to get \( \int_0^a f(-u) du \). Combining both pieces yields the result.

Exam Tip: This property helps you evaluate definite integrals over symmetric intervals by computing only over half the interval and adjusting the integrand accordingly.

Question 68. Mark (√) against the correct answer in the following: Let [x] denote the greatest integer less than or equal to x. Then, \( \int_0^{1.5} [x] \, dx = ? \)
(a) \( \frac{1}{2} \)
(b) \( \frac{3}{2} \)
(c) 2
(d) 3
Answer: (a) \( \frac{1}{2} \)
In simple words: The floor function [x] equals 0 for \( 0 \le x < 1 \) and equals 1 for \( 1 \le x < 1.5 \). Split the integral: \( \int_0^1 0 \, dx + \int_1^{1.5} 1 \, dx = 0 + 0.5 = \frac{1}{2} \).

Exam Tip: For step functions like the floor function, always break the integral into regions where the function is constant - this converts a tricky problem into simple segments.

Question 69. Mark (√) against the correct answer in the following: Let [x] denote the greatest integer less than or equal to x. Then, \( \int_{-1}^{1} [x] \, dx = ? \)
(a) -1
(b) 0
(c) \( \frac{1}{2} \)
(d) 2
Answer: (a) -1
In simple words: The floor function [x] equals -1 for \( -1 \le x < 0 \) and equals 0 for \( 0 \le x < 1 \). The integral becomes \( \int_{-1}^0 (-1) \, dx + \int_0^1 0 \, dx = -1 + 0 = -1 \).

Exam Tip: When the interval includes both negative and positive values, carefully track the value of [x] in each region - don't assume the function behaves the same throughout.

Question 70. Mark (√) against the correct answer in the following: \( \int_{1}^{2} |x^2 - 3x + 2| \, dx = ? \)
(a) \( \frac{-1}{6} \)
(b) \( \frac{1}{6} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{2}{3} \)
Answer: (b) \( \frac{1}{6} \)
In simple words: First, find where \( x^2 - 3x + 2 = 0 \) by factoring to get \( (x-2)(x-1) = 0 \). Since the roots are at \( x = 1 \) and \( x = 2 \), which are exactly the integration limits, there are no sign changes within the interval. The expression is non-positive on (1, 2), so \( |x^2 - 3x + 2| = -(x^2 - 3x + 2) \). Integrating gives \( \left[ -\frac{x^3}{3} + \frac{3x^2}{2} - 2x \right]_1^2 = \frac{1}{6} \).

Exam Tip: Always factor the expression under absolute value and check if the roots lie within your integration limits - this tells you whether the expression changes sign and how to remove the absolute value.

Question 71. Mark (√) against the correct answer in the following: \( \int_{\pi}^{2\pi} |\sin x| \, dx = ? \)
(a) 0
(b) 1
(c) 2
(d) none of these
Answer: (c) 2
In simple words: The equation \( \sin x = 0 \) holds at \( x = 0, \pi, 2\pi, \ldots \) Since the limits are \( \pi \) and \( 2\pi \), which are exactly the boundary points, there are no sign changes in between. On the interval \( (\pi, 2\pi) \), sine is negative, so \( |\sin x| = -\sin x \). Thus, \( \int_\pi^{2\pi} (-\sin x) \, dx = [-(-\cos x)]_\pi^{2\pi} = 2 \).

Exam Tip: For absolute value integrals with trig functions, identify all zeros in the interval to determine sign changes - then remove the absolute value piecewise in each subregion.

Question 72. Mark (√) against the correct answer in the following: \( \int_{0}^{1/\sqrt{2}} \frac{\sin^{-1} x}{(1 - x^2)^{3/2}} \, dx = ? \)
(a) \( \frac{\pi}{4} - \frac{1}{2} \log 2 \)
(b) \( \frac{\pi}{2} - 2 \log 2 \)
(c) \( \frac{\pi}{4} - \frac{1}{2} \log 2 \)
(d) none of these
Answer: (a) \( \frac{\pi}{4} - \frac{1}{2} \log 2 \)
In simple words: Let \( x = \sin t \), so \( dx = \cos t \, dt \) and \( \sin^{-1}(1/\sqrt{2}) = \pi/4 \). The integral becomes \( \int_0^{\pi/4} t \sec^2 t \, dt \). Using integration by parts with \( u = t \) and \( dv = \sec^2 t \, dt \) yields \( t \tan t - \int \tan t \, dt = t \tan t + \log|\cos t| \). Evaluating from 0 to \( \pi/4 \) gives \( \frac{\pi}{4} - \frac{1}{2} \log 2 \).

Exam Tip: When the integrand contains inverse trig functions with power denominators, try substituting with the corresponding trig function - this often simplifies the expression dramatically.

Question 73. Mark (√) against the correct answer in the following: \( \int_{0}^{1} \sin^{-1} \left( \frac{2x}{1 + x^2} \right) dx = ? \)
(a) \( \frac{1}{2}(\pi - \log 2) \)
(b) \( \frac{\pi}{2} - \log 2 \)
(c) \( \pi - 2 \log 2 \)
(d) none of these
Answer: (b) \( \frac{\pi}{2} - \log 2 \)
In simple words: Substitute \( x = \tan y \), which gives \( dx = \sec^2 y \, dy \). The expression \( \sin^{-1}(2x/(1+x^2)) \) becomes \( \sin^{-1}(\sin 2y) = 2y \). The integral transforms to \( \int_0^{\pi/4} 2y \sec^2 y \, dy \). Applying integration by parts with \( u = 2y \) and \( dv = \sec^2 y \, dy \) yields \( 2[y \tan y + \log|\cos y|]_0^{\pi/4} \). Evaluating at the bounds gives \( 2[\frac{\pi}{4} - \frac{1}{2}\log 2] = \frac{\pi}{2} - \log 2 \).

Exam Tip: When you see the expression \( 2x/(1+x^2) \), recognize it as \( \sin(2\tan^{-1} x) \) - substituting \( x = \tan y \) converts inverse trig expressions into direct angles.