NCERT Solutions for Class 12 Maths Chapter 7: Integrals (NCERT 2026–27)

Q: Are Class 12 Maths Chapter 7 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Maths Chapter 7 Integrals are free and follow the official NCERT Mathematics Part II textbook for 2026-27, with answers verified against the book's answer key.

These Class 12 Maths Chapter 7 solutions cover Integrals completely — from anti-derivatives by inspection to integration by substitution, partial fractions, integration by parts, definite integrals and their properties. Every question of Exercises 7.1 to 7.10 and the Miscellaneous Exercise is reproduced verbatim from the NCERT textbook and solved step by step, with answers verified against the book’s answer key.

Class: 12 Subject: Mathematics Chapter: 7 — Integrals Exercises: 7.1–7.10 + Miscellaneous Session: 2026–27 Book: NCERT Maths Part II

On this page

Chapter overview
Key concepts
Important formulas
Exercise 7.1
Exercise 7.2
Exercise 7.3
Exercise 7.4
Exercise 7.5
Exercise 7.6
Exercise 7.7
Exercise 7.8
Exercise 7.9
Exercise 7.10
Miscellaneous Exercise
Common mistakes
Practice MCQs & Assertion–Reason
Quick revision summary
FAQs

Chapter 7 Overview

Chapter 7, Integrals, introduces integration as the inverse process of differentiation. You first learn anti-derivatives (indefinite integrals) using standard formulae and the method of inspection. The chapter then develops three core techniques — integration by substitution, integration using partial fractions, and integration by parts — together with standard integrals of special functions and the special forms √(x²±a²) and √(a²−x²). In the second half it defines the definite integral, states the Fundamental Theorem of Calculus, and uses substitution and the properties of definite integrals (P₀–P₇) to evaluate them quickly. These Class 12 Maths Chapter 7 solutions work through every numbered question in order.

Key Concepts

Anti-derivative / indefinite integral: if d/dx[F(x)] = f(x), then ∫f(x) dx = F(x) + C, where C is the arbitrary constant of integration.

Integration by substitution: change the variable x = g(t) so that ∫f(x) dx = ∫f(g(t)) g′(t) dt; choose a substitution whose derivative also appears in the integrand.

Partial fractions: write a proper rational function P(x)/Q(x) as a sum of simpler fractions (per Table 7.2) and integrate term by term; reduce improper fractions first by division.

Integration by parts: ∫u·v dx = u∫v dx − ∫[u′∫v dx] dx. Choose the first function by ILATE (Inverse, Logarithm, Algebraic, Trigonometric, Exponential).

Definite integral: ∫_a^b f(x) dx = F(b) − F(a) (Second Fundamental Theorem), where F is any anti-derivative of f.

Important Formulas (Chapter 7)

Power: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1); ∫(1/x) dx = log|x| + C.

Trigonometric: ∫cos x dx = sin x + C; ∫sin x dx = −cos x + C; ∫sec²x dx = tan x + C; ∫cosec²x dx = −cot x + C.

Exponential: ∫e^x dx = e^x + C; ∫a^x dx = a^x/log a + C.

Special functions: ∫dx/(x²−a²) = (1/2a) log|(x−a)/(x+a)| + C; ∫dx/(a²−x²) = (1/2a) log|(a+x)/(a−x)| + C; ∫dx/(x²+a²) = (1/a) tan⁻¹(x/a) + C.

Radical forms: ∫dx/√(x²−a²) = log|x+√(x²−a²)| + C; ∫dx/√(a²−x²) = sin⁻¹(x/a) + C; ∫dx/√(x²+a²) = log|x+√(x²+a²)| + C.

e^x-rule: ∫e^x[f(x) + f′(x)] dx = e^x f(x) + C.

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

Exercise 7.1

Find an anti derivative (or integral) of the following functions by the method of inspection.

1. sin 2x

SOLUTION d/dx(cos 2x) = −2 sin 2x, so d/dx(−½ cos 2x) = sin 2x. ∴ ∫sin 2x dx = −½ cos 2x + C.

2. cos 3x

SOLUTION d/dx(sin 3x) = 3 cos 3x, so d/dx(⅓ sin 3x) = cos 3x. ∴ ∫cos 3x dx = ⅓ sin 3x + C.

3. e^2x

SOLUTION d/dx(e^2x) = 2e^2x, so d/dx(½ e^2x) = e^2x. ∴ ∫e^2x dx = ½ e^2x + C.

4. (ax + b)²

SOLUTION d/dx[(ax+b)³] = 3a(ax+b)², so d/dx[(ax+b)³/(3a)] = (ax+b)². ∴ ∫(ax+b)² dx = (1/3a)(ax + b)³ + C.

5. sin 2x − 4 e^3x

SOLUTION ∫sin 2x dx = −½ cos 2x; ∫4e^3x dx = (4/3)e^3x. ∴ answer = −½ cos 2x − (4/3) e^3x + C.

6. ∫(4 e^3x + 1) dx

SOLUTION ∫4e^3x dx + ∫dx = (4/3)e^3x + x + C. ∴ (4/3) e^3x + x + C.

7. ∫x²(1 − 1/x²) dx

SOLUTION x²(1 − 1/x²) = x² − 1. ∫(x² − 1) dx = x³/3 − x + C. ∴ x³/3 − x + C.

8. ∫(ax² + bx + c) dx

SOLUTION Integrate term by term: a·x³/3 + b·x²/2 + cx + C. ∴ ax³/3 + bx²/2 + cx + C.

9. ∫(2x² + e^x) dx

SOLUTION ∫2x² dx + ∫e^x dx = (2/3)x³ + e^x + C. ∴ (2/3) x³ + e^x + C.

10. ∫(√x − 1/√x)² dx

SOLUTION (√x − 1/√x)² = x − 2 + 1/x. ∫(x − 2 + 1/x) dx = x²/2 − 2x + log|x| + C. ∴ x²/2 + log|x| − 2x + C.

11. ∫(x³ + 5x² − 4)/x² dx

SOLUTION Split: x + 5 − 4x⁻². ∫(x + 5 − 4x⁻²) dx = x²/2 + 5x + 4/x + C. ∴ x²/2 + 5x + 4/x + C.

12. ∫(x³ + 3x + 4)/√x dx

SOLUTION = ∫(x^5/2 + 3x^1/2 + 4x^−1/2) dx. = x^7/2/(7/2) + 3·x^3/2/(3/2) + 4·x^1/2/(1/2) + C. ∴ (2/7) x^7/2 + 2x^3/2 + 8√x + C.

13. ∫(x³ − x² + x − 1)/(x − 1) dx

SOLUTION x³ − x² + x − 1 = x²(x − 1) + (x − 1) = (x − 1)(x² + 1). So integrand = x² + 1. ∫(x² + 1) dx = x³/3 + x + C. ∴ x³/3 + x + C.

14. ∫(1 − x)√x dx

SOLUTION (1 − x)√x = x^1/2 − x^3/2. ∫(x^1/2 − x^3/2) dx = (2/3)x^3/2 − (2/5)x^5/2 + C. ∴ (2/3) x^3/2 − (2/5) x^5/2 + C.

15. ∫√x(3x² + 2x + 3) dx

SOLUTION = ∫(3x^5/2 + 2x^3/2 + 3x^1/2) dx. = 3·x^7/2/(7/2) + 2·x^5/2/(5/2) + 3·x^3/2/(3/2) + C. ∴ (6/7) x^7/2 + (4/5) x^5/2 + 2x^3/2 + C.

16. ∫(2x − 3 cos x + e^x) dx

SOLUTION = x² − 3 sin x + e^x + C. ∴ x² − 3 sin x + e^x + C.

17. ∫(2x² − 3 sin x + 5√x) dx

SOLUTION = (2/3)x³ − 3(−cos x) + 5·x^3/2/(3/2) + C. ∴ (2/3) x³ + 3 cos x + (10/3) x^3/2 + C.

18. ∫sec x (sec x + tan x) dx

SOLUTION = ∫(sec²x + sec x tan x) dx = tan x + sec x + C. ∴ tan x + sec x + C.

19. ∫(sec²x)/(cosec²x) dx

SOLUTION sec²x/cosec²x = sin²x/cos²x = tan²x = sec²x − 1. ∫(sec²x − 1) dx = tan x − x + C. ∴ tan x − x + C.

20. ∫(2 − 3 sin x)/cos²x dx

SOLUTION = ∫(2 sec²x − 3 sin x/cos²x) dx = 2 tan x − 3 sec x + C. ∴ 2 tan x − 3 sec x + C.

21. The anti derivative of (√x + 1/√x) equals
(A) ⅓x^1/3 + 2x^1/2 + C (B) (2/3)x^2/3 + ½x² + C (C) (2/3)x^3/2 + 2x^1/2 + C (D) (3/2)x^3/2 + ½x^1/2 + C

SOLUTION ∫(x^1/2 + x^−1/2) dx = (2/3)x^3/2 + 2x^1/2 + C. ∴ option (C).

22. If d/dx f(x) = 4x³ − 3/x⁴ such that f(2) = 0. Then f(x) is
(A) x⁴ + 1/x³ − 129/8 (B) x³ + 1/x⁴ + 129/8 (C) x⁴ + 1/x³ + 129/8 (D) x³ + 1/x⁴ − 129/8

SOLUTION f(x) = ∫(4x³ − 3x⁻⁴) dx = x⁴ + x⁻³ + C = x⁴ + 1/x³ + C. f(2) = 16 + 1/8 + C = 0 ⇒ C = −129/8. ∴ f(x) = x⁴ + 1/x³ − 129/8 — option (A).

Exercise 7.2

Integrate the functions in Exercises 1 to 37 (by substitution).

1. 2x/(1 + x²)

SOLUTION Put t = 1 + x², dt = 2x dx. ∫dt/t = log|t|. ∴ log(1 + x²) + C.

2. (log x)²/x

SOLUTION Put t = log x, dt = dx/x. ∫t² dt = t³/3. ∴ (1/3)(log|x|)³ + C.

3. 1/(x + x log x)

SOLUTION = 1/[x(1 + log x)]. Put t = 1 + log x, dt = dx/x. ∫dt/t = log|t|. ∴ log|1 + log x| + C.

4. sin x sin(cos x)

SOLUTION Put t = cos x, dt = −sin x dx. ∫sin t (−dt) = cos t. ∴ cos(cos x) + C.

5. sin(ax + b) cos(ax + b)

SOLUTION = ½ sin 2(ax + b). Put t = 2(ax+b), dt = 2a dx. = (1/4a)∫sin t dt = −(1/4a) cos t. ∴ −(1/4a) cos 2(ax + b) + C.

6. √(ax + b)

SOLUTION Put t = ax + b, dt = a dx. (1/a)∫t^1/2 dt = (1/a)·(2/3)t^3/2. ∴ (2/3a)(ax + b)^3/2 + C.

7. x√(x + 2)

SOLUTION Put t = x + 2, x = t − 2, dx = dt. ∫(t − 2)t^1/2 dt = ∫(t^3/2 − 2t^1/2) dt = (2/5)t^5/2 − (4/3)t^3/2. ∴ (2/5)(x + 2)^5/2 − (4/3)(x + 2)^3/2 + C.

8. x√(1 + 2x²)

SOLUTION Put t = 1 + 2x², dt = 4x dx. (1/4)∫t^1/2 dt = (1/4)(2/3)t^3/2. ∴ (1/6)(1 + 2x²)^3/2 + C.

9. (4x + 2)√(x² + x + 1)

SOLUTION 4x + 2 = 2(2x + 1) = 2 d/dx(x²+x+1). Put t = x²+x+1, dt = (2x+1)dx. 2∫t^1/2 dt = 2·(2/3)t^3/2. ∴ (4/3)(x² + x + 1)^3/2 + C.

10. 1/(x − √x)

SOLUTION x − √x = √x(√x − 1). Put t = √x − 1, dt = dx/(2√x). 1/[√x(√x−1)] dx = 2 dt/t ⇒ 2 log|t|. ∴ 2 log|√x − 1| + C.

11. x/(x + 4), x > 0

SOLUTION Put t = x + 4, x = t − 4, dx = dt. ∫(t−4)/√t dt = ∫(t^1/2 − 4t^−1/2) dt. = (2/3)t^3/2 − 8t^1/2 = (2/3)√t(t − 12). ∴ (2/3)√(x+4)·(x − 8) + C.

12. (x³ − 1)^1/3 x⁵

SOLUTION Put t = x³ − 1, x³ = t + 1, dt = 3x² dx. x⁵ dx = x³·x²dx = (t+1)(dt/3). (1/3)∫t^1/3(t + 1) dt = (1/3)[(3/7)t^7/3 + (3/4)t^4/3]. ∴ (1/7)(x³ − 1)^7/3 + (1/4)(x³ − 1)^4/3 + C.

13. x²/(2 + 3x³)³

SOLUTION Put t = 2 + 3x³, dt = 9x² dx. (1/9)∫t⁻³ dt = (1/9)·t⁻²/(−2). ∴ −1/[18(2 + 3x³)²] + C.

14. 1/[x(log x)^m], x > 0, m ≠ 1

SOLUTION Put t = log x, dt = dx/x. ∫t^−m dt = t^1−m/(1−m). ∴ (log x)^1−m/(1 − m) + C.

15. x/(9 − 4x²)

SOLUTION Put t = 9 − 4x², dt = −8x dx. (−1/8)∫dt/t = (−1/8)log|t|. ∴ −(1/8) log|9 − 4x²| + C.

16. e^{2x + 3}

SOLUTION Put t = 2x + 3, dt = 2 dx. (1/2)∫e^t dt = (1/2)e^t. ∴ (1/2) e^{2x + 3} + C.

17. x/e^x²

SOLUTION = x e^−x². Put t = −x², dt = −2x dx. (−1/2)∫e^t dt = (−1/2)e^t. ∴ −1/(2e^x²) + C.

18. e^tan⁻¹x/(1 + x²)

SOLUTION Put t = tan⁻¹x, dt = dx/(1+x²). ∫e^t dt = e^t. ∴ e^tan⁻¹x + C.

19. (e^2x − 1)/(e^2x + 1)

SOLUTION Divide num & den by e^x: (e^x − e^−x)/(e^x + e^−x). Put t = e^x + e^−x, dt = (e^x − e^−x)dx. ∫dt/t = log|t|. ∴ log(e^x + e^−x) + C.

20. (e^2x − e^−2x)/(e^2x + e^−2x)

SOLUTION Put t = e^2x + e^−2x, dt = 2(e^2x − e^−2x)dx. (1/2)∫dt/t = (1/2)log|t|. ∴ (1/2) log(e^2x + e^−2x) + C.

21. tan²(2x − 3)

SOLUTION tan²θ = sec²θ − 1. ∫[sec²(2x−3) − 1] dx = (1/2)tan(2x−3) − x. ∴ (1/2) tan(2x − 3) − x + C.

22. sec²(7 − 4x)

SOLUTION Put t = 7 − 4x, dt = −4 dx. (−1/4)∫sec²t dt = (−1/4)tan t. ∴ −(1/4) tan(7 − 4x) + C.

23. (sin⁻¹x)/√(1 − x²)

SOLUTION Put t = sin⁻¹x, dt = dx/√(1−x²). ∫t dt = t²/2. ∴ (1/2)(sin⁻¹x)² + C.

24. (2 cos x − 3 sin x)/(6 cos x + 4 sin x)

SOLUTION Denominator d/dx(6 cos x + 4 sin x) = −6 sin x + 4 cos x = 2(2 cos x − 3 sin x). So numerator = ½ d/dx(den). Put t = 6 cos x + 4 sin x: (1/2)∫dt/t = (1/2)log|t|. ∴ (1/2) log|2 sin x + 3 cos x| + C (since 6 cos x + 4 sin x = 2(3 cos x + 2 sin x)).

25. 1/[cos²x (1 − tan x)²]

SOLUTION = sec²x/(1 − tan x)². Put t = 1 − tan x, dt = −sec²x dx. −∫dt/t² = 1/t. ∴ 1/(1 − tan x) + C.

26. (cos√x)/√x

SOLUTION Put t = √x, dt = dx/(2√x). 2∫cos t dt = 2 sin t. ∴ 2 sin√x + C.

27. √(sin 2x) cos 2x

SOLUTION Put t = sin 2x, dt = 2 cos 2x dx. (1/2)∫t^1/2 dt = (1/2)(2/3)t^3/2. ∴ (1/3)(sin 2x)^3/2 + C.

28. cos x/√(1 + sin x)

SOLUTION Put t = 1 + sin x, dt = cos x dx. ∫t^−1/2 dt = 2t^1/2. ∴ 2√(1 + sin x) + C.

29. cot x log sin x

SOLUTION Put t = log sin x, dt = cot x dx. ∫t dt = t²/2. ∴ (1/2)(log sin x)² + C.

30. sin x/(1 + cos x)

SOLUTION Put t = 1 + cos x, dt = −sin x dx. −∫dt/t = −log|t|. ∴ −log|1 + cos x| + C.

31. sin x/(1 + cos x)²

SOLUTION Put t = 1 + cos x, dt = −sin x dx. −∫dt/t² = 1/t. ∴ 1/(1 + cos x) + C.

32. 1/(1 + cot x)

SOLUTION = sin x/(sin x + cos x). Write num = ½[(sin x + cos x) − (cos x − sin x)]. ∫ = ½∫dx − ½∫(cos x − sin x)/(sin x + cos x) dx = (x/2) − (1/2)log|sin x + cos x|. ∴ x/2 − (1/2) log|cos x + sin x| + C.

33. 1/(1 − tan x)

SOLUTION = cos x/(cos x − sin x). Write num = ½[(cos x − sin x) + (cos x + sin x)]. ∫ = (x/2) + (1/2)∫(cos x + sin x)/(cos x − sin x) dx = (x/2) − (1/2)log|cos x − sin x|. ∴ x/2 − (1/2) log|cos x − sin x| + C.

34. √(tan x)/(sin x cos x)

SOLUTION sin x cos x = cos²x tan x, so integrand = sec²x/√(tan x). Put t = tan x, dt = sec²x dx. ∫t^−1/2 dt = 2t^1/2. ∴ 2√(tan x) + C.

35. (1 + log x)²/x

SOLUTION Put t = 1 + log x, dt = dx/x. ∫t² dt = t³/3. ∴ (1/3)(1 + log x)³ + C.

36. (x + 1)(x + log x)²/x

SOLUTION (x + 1)/x = 1 + 1/x = d/dx(x + log x). Put t = x + log x, dt = (1 + 1/x)dx. ∫t² dt = t³/3. ∴ (1/3)(x + log x)³ + C.

37. [x³ sin(tan⁻¹x⁴)]/(1 + x⁸)

SOLUTION Put t = tan⁻¹x⁴, dt = 4x³/(1 + x⁸) dx. So integrand dx = (1/4)sin t dt. (1/4)∫sin t dt = −(1/4)cos t. ∴ −(1/4) cos(tan⁻¹x⁴) + C.

38. ∫(10x⁹ + 10^x log_e10)/(x¹⁰ + 10^x) dx equals
(A) 10^x − x¹⁰ + C (B) 10^x + x¹⁰ + C (C) (10^x − x¹⁰)⁻¹ + C (D) log(10^x + x¹⁰) + C

SOLUTION Numerator = d/dx(x¹⁰ + 10^x). Put t = x¹⁰ + 10^x: ∫dt/t = log|t|. ∴ option (D).

39. ∫dx/(sin²x cos²x) equals
(A) tan x + cot x + C (B) tan x − cot x + C (C) tan x cot x + C (D) tan x − cot 2x + C

SOLUTION 1/(sin²x cos²x) = (sin²x + cos²x)/(sin²x cos²x) = sec²x + cosec²x. ∫(sec²x + cosec²x) dx = tan x − cot x. ∴ option (B).

Exercise 7.3

Find the integrals of the functions in Exercises 1 to 22 (using trigonometric identities).

1. sin²(2x + 5)

SOLUTION sin²θ = (1 − cos 2θ)/2. ∫½[1 − cos(4x + 10)] dx = x/2 − (1/8)sin(4x + 10). ∴ x/2 − (1/8) sin(4x + 10) + C.

2. sin 3x cos 4x

SOLUTION = ½[sin 7x − sin x] (product-to-sum). ∫ = ½[−(1/7)cos 7x + cos x]. ∴ −(1/14) cos 7x + (1/2) cos x + C.

3. cos 2x cos 4x cos 6x

SOLUTION cos 2x cos 6x = ½(cos 8x + cos 4x). Multiply by cos 4x and expand: ¼(cos 12x + 1 + cos 8x + cos 4x). Integrate: ¼[(1/12)sin 12x + x + (1/8)sin 8x + (1/4)sin 4x]. ∴ ¼[(1/12) sin 12x + x + (1/8) sin 8x + (1/4) sin 4x] + C.

4. sin³(2x + 1)

SOLUTION sin³θ = sinθ(1 − cos²θ), θ = 2x+1. Put t = cos(2x+1), dt = −2 sin(2x+1)dx. −½∫(1 − t²) dt = −½(t − t³/3). ∴ −½ cos(2x + 1) + (1/6) cos³(2x + 1) + C.

5. sin³x cos³x

SOLUTION = cos³x sin²x sin x = cos³x(1 − cos²x) sin x. Put t = cos x, dt = −sin x dx. −∫(t³ − t⁵) dt = −(t⁴/4 − t⁶/6) = (1/6)cos⁶x − (1/4)cos⁴x. ∴ (1/6) cos⁶x − (1/4) cos⁴x + C.

6. sin x sin 2x sin 3x

SOLUTION sin x sin 3x = ½(cos 2x − cos 4x). Times sin 2x: ½(sin 2x cos 2x − sin 2x cos 4x). = ¼[sin 4x − (sin 6x − sin 2x)] = ¼[sin 4x − sin 6x + sin 2x]. Integrate: ¼[−(1/4)cos 4x + (1/6)cos 6x − ½cos 2x]. ∴ ¼[(1/6) cos 6x − (1/4) cos 4x − (1/2) cos 2x] + C.

7. sin 4x sin 8x

SOLUTION = ½(cos 4x − cos 12x). Integrate: ½[(1/4)sin 4x − (1/12)sin 12x]. ∴ ½[(1/4) sin 4x − (1/12) sin 12x] + C.

8. (1 − cos x)/(1 + cos x)

SOLUTION = (2 sin²(x/2))/(2 cos²(x/2)) = tan²(x/2) = sec²(x/2) − 1. ∫ = 2 tan(x/2) − x. ∴ 2 tan(x/2) − x + C.

9. cos x/(1 + cos x)

SOLUTION = 1 − 1/(1 + cos x) = 1 − ½sec²(x/2). ∫ = x − tan(x/2). ∴ x − tan(x/2) + C.

10. sin⁴x

SOLUTION sin⁴x = [(1 − cos 2x)/2]² = ¼(1 − 2 cos 2x + cos²2x) = ¼[3/2 − 2 cos 2x + ½cos 4x]. Integrate: (3x/8) − ¼sin 2x + (1/32)sin 4x. ∴ 3x/8 − (1/4) sin 2x + (1/32) sin 4x + C.

11. cos⁴2x

SOLUTION cos⁴2x = [(1 + cos 4x)/2]² = ¼[3/2 + 2 cos 4x + ½cos 8x]. Integrate: (3x/8) + ¼·½sin 4x·2 … = 3x/8 + (1/8)sin 4x + (1/64)sin 8x. ∴ 3x/8 + (1/8) sin 4x + (1/64) sin 8x + C.

12. sin²x/(1 + cos x)

SOLUTION sin²x = 1 − cos²x = (1 − cos x)(1 + cos x). So integrand = 1 − cos x. ∫(1 − cos x) dx = x − sin x. ∴ x − sin x + C.

13. (cos 2x − cos 2α)/(cos x − cos α)

SOLUTION cos 2x − cos 2α = 2cos²x − 2cos²α = 2(cos x − cos α)(cos x + cos α). Integrand = 2(cos x + cos α). ∫ = 2(sin x + x cos α). ∴ 2(sin x + x cos α) + C.

14. (cos x − sin x)/(1 + sin 2x)

SOLUTION 1 + sin 2x = (sin x + cos x)². Put t = sin x + cos x, dt = (cos x − sin x)dx. ∫dt/t² = −1/t. ∴ −1/(cos x + sin x) + C.

15. tan³2x sec 2x

SOLUTION tan³2x sec 2x = tan²2x·(sec 2x tan 2x) = (sec²2x − 1)(sec 2x tan 2x). Put t = sec 2x, dt = 2 sec 2x tan 2x dx. ½∫(t² − 1) dt = ½(t³/3 − t). ∴ (1/6) sec³2x − (1/2) sec 2x + C.

16. tan⁴x

SOLUTION tan⁴x = tan²x(sec²x − 1) = tan²x sec²x − (sec²x − 1). ∫ = (1/3)tan³x − tan x + x. ∴ (1/3) tan³x − tan x + x + C.

17. (sin³x + cos³x)/(sin²x cos²x)

SOLUTION Split: sin³x/(sin²x cos²x) + cos³x/(sin²x cos²x) = sin x/cos²x + cos x/sin²x = sec x tan x + cosec x cot x… actually = (tan x sec x) + (cot x cosec x)? Re-check: sin x/cos²x = tan x sec x; cos x/sin²x = cot x cosec x. ∫(sec x tan x + cosec x cot x) dx = sec x − cosec x. ∴ sec x − cosec x + C.

18. (cos 2x + 2 sin²x)/cos²x

SOLUTION cos 2x + 2 sin²x = (1 − 2 sin²x) + 2 sin²x = 1. So integrand = 1/cos²x = sec²x. ∫sec²x dx = tan x. ∴ tan x + C.

19. 1/(sin x cos³x)

SOLUTION Multiply num & den structure: = (sin²x + cos²x)/(sin x cos³x) = sin x/cos³x + 1/(sin x cos x). First: ∫tan x sec²x dx = ½tan²x. Second: 1/(sin x cos x) = 2/sin 2x = 2 cosec 2x… use 1/(sin x cos x)= sec²x/tan x; ∫ = log|tan x|. ∴ (1/2) tan²x + log|tan x| + C.

20. cos 2x/(cos x + sin x)²

SOLUTION cos 2x = cos²x − sin²x = (cos x − sin x)(cos x + sin x); (cos x + sin x)² in den. Integrand = (cos x − sin x)/(cos x + sin x). Put t = cos x + sin x, dt = (cos x − sin x)dx. ∫dt/t = log|t|. ∴ log|cos x + sin x| + C.

21. sin⁻¹(cos x)

SOLUTION cos x = sin(π/2 − x), so sin⁻¹(cos x) = π/2 − x (for the principal range). ∫(π/2 − x) dx = πx/2 − x²/2. ∴ πx/2 − x²/2 + C.

22. 1/[cos(x − a) cos(x − b)]

SOLUTION Multiply by sin(a − b)/sin(a − b). Note sin[(x−b) − (x−a)] = sin(a−b) = sin(x−b)cos(x−a) − cos(x−b)sin(x−a). So integrand = [1/sin(a−b)][tan(x−b) − tan(x−a)]. Integrate each. ∴ [1/sin(a − b)] log|cos(x − a)/cos(x − b)| + C.

23. ∫(sin²x − cos²x)/(sin²x cos²x) dx is equal to
(A) tan x + cot x + C (B) tan x + cosec x + C (C) −tan x + cot x + C (D) tan x + sec x + C

SOLUTION Split: sin²x/(sin²x cos²x) − cos²x/(sin²x cos²x) = sec²x − cosec²x. ∫ = tan x + cot x. ∴ option (A).

24. ∫[e^x(1 + x)]/cos²(e^xx) dx equals
(A) −cot(e^xx) + C (B) tan(xe^x) + C (C) tan(e^x) + C (D) cot(e^x) + C

SOLUTION Put t = e^xx, dt = (e^x + xe^x)dx = e^x(1 + x)dx. ∫sec²t dt = tan t. ∴ option (B).

Exercise 7.4

Integrate the functions in Exercises 1 to 23 (integrals of some particular functions).

1. 3x²/(x⁶ + 1)

SOLUTION Put t = x³, dt = 3x² dx. ∫dt/(t² + 1) = tan⁻¹t. ∴ tan⁻¹(x³) + C.

2. 1/√(1 + 4x²)

SOLUTION = ½∫dx/√(x² + ¼). Use ∫dx/√(x²+a²) = log|x + √(x²+a²)|. ∴ ½ log|2x + √(1 + 4x²)| + C.

3. 1/√[(2 − x)² + 1]

SOLUTION Put t = x − 2, dt = dx. ∫dt/√(t² + 1) = log|t + √(t²+1)|. ∴ log|(x − 2) + √(x² − 4x + 5)| + C.

4. 1/√(9 − 25x²)

SOLUTION = (1/5)∫dx/√((3/5)² − x²) = (1/5)sin⁻¹(5x/3). ∴ (1/5) sin⁻¹(5x/3) + C.

5. 3x/(1 + 2x⁴)

SOLUTION Put t = √2 x², dt = 2√2 x dx; 3x dx = (3/(2√2))dt. Denominator = 1 + t². (3/(2√2))∫dt/(1 + t²) = (3/(2√2))tan⁻¹t. ∴ [3/(2√2)] tan⁻¹(√2 x²) + C.

6. x²/(1 − x⁶)

SOLUTION Put t = x³, dt = 3x²dx. (1/3)∫dt/(1 − t²) = (1/3)·½log|(1+t)/(1−t)|. ∴ (1/6) log|(1 + x³)/(1 − x³)| + C.

7. (x − 1)/√(x² − 1)

SOLUTION Split: ∫x/√(x²−1) dx − ∫1/√(x²−1) dx = √(x²−1) − log|x + √(x²−1)|. ∴ √(x² − 1) − log|x + √(x² − 1)| + C.

8. x²/√(x⁶ + a⁶)

SOLUTION Put t = x³, dt = 3x²dx. (1/3)∫dt/√(t² + (a³)²) = (1/3)log|t + √(t²+a⁶)|. ∴ (1/3) log|x³ + √(x⁶ + a⁶)| + C.

9. sec²x/√(tan²x + 4)

SOLUTION Put t = tan x, dt = sec²x dx. ∫dt/√(t² + 4) = log|t + √(t²+4)|. ∴ log|tan x + √(tan²x + 4)| + C.

10. 1/√(x² + 2x + 2)

SOLUTION x² + 2x + 2 = (x + 1)² + 1. ∫dx/√((x+1)²+1) = log|(x+1) + √(x²+2x+2)|. ∴ log|(x + 1) + √(x² + 2x + 2)| + C.

11. 1/(9x² + 6x + 5)

SOLUTION 9x² + 6x + 5 = (3x + 1)² + 4. Put t = 3x + 1, dt = 3 dx. (1/3)∫dt/(t² + 2²) = (1/3)·(1/2)tan⁻¹(t/2). ∴ (1/6) tan⁻¹[(3x + 1)/2] + C.

12. 1/√(7 − 6x − x²)

SOLUTION 7 − 6x − x² = 16 − (x + 3)². ∫dx/√(4² − (x+3)²) = sin⁻¹[(x+3)/4]. ∴ sin⁻¹[(x + 3)/4] + C.

13. 1/√[(x − 1)(x − 2)]

SOLUTION (x−1)(x−2) = (x − 3/2)² − ¼. ∫dx/√((x−3/2)² − (1/2)²) = log|(x − 3/2) + √(x²−3x+2)|. ∴ log|x − 3/2 + √(x² − 3x + 2)| + C.

14. 1/√(8 + 3x − x²)

SOLUTION 8 + 3x − x² = (41/4) − (x − 3/2)². ∫dx/√(·) = sin⁻¹[(2x − 3)/√41]. ∴ sin⁻¹[(2x − 3)/√41] + C.

15. 1/√[(x − a)(x − b)]

SOLUTION (x−a)(x−b) = (x − (a+b)/2)² − ((a−b)/2)². ∫dx/√(that) = log|(x − (a+b)/2) + √((x−a)(x−b))|. ∴ log|x − (a + b)/2 + √((x − a)(x − b))| + C.

16. (4x + 1)/√(2x² + x − 3)

SOLUTION d/dx(2x² + x − 3) = 4x + 1 = numerator. Put t = 2x²+x−3; ∫dt/√t = 2√t. ∴ 2√(2x² + x − 3) + C.

17. (x + 2)/√(x² − 1)

SOLUTION x + 2 = ½(2x) + 2. ∫x/√(x²−1)dx = √(x²−1); 2∫dx/√(x²−1) = 2 log|x + √(x²−1)|. ∴ √(x² − 1) + 2 log|x + √(x² − 1)| + C.

18. (5x − 2)/(1 + 2x + 3x²)

SOLUTION 5x − 2 = (5/6)(6x + 2) − 11/3. First part → (5/6)log|3x²+2x+1|. Second: −(11/3)·(1/3)∫dx/((x+1/3)²+(√2/3)²). ∴ (5/6) log|3x² + 2x + 1| − (11/(3√2)) tan⁻¹[(3x + 1)/√2] + C.

19. (6x + 7)/√[(x − 5)(x − 4)]

SOLUTION (x−5)(x−4) = x² − 9x + 20; d/dx = 2x − 9. Write 6x + 7 = 3(2x − 9) + 34. 3∫(2x−9)/√(·)dx = 6√(x²−9x+20); 34∫dx/√(·) = 34 log|x − 9/2 + √(x²−9x+20)|. ∴ 6√(x² − 9x + 20) + 34 log|x − 9/2 + √(x² − 9x + 20)| + C.

20. (x + 2)/√(4x − x²)

SOLUTION x + 2 = −½(4 − 2x) + 4. −½∫(4 − 2x)/√(4x−x²)dx = −√(4x − x²). 4∫dx/√(4 − (x−2)²) = 4 sin⁻¹[(x − 2)/2]. ∴ −√(4x − x²) + 4 sin⁻¹[(x − 2)/2] + C.

21. (x + 2)/√(x² + 2x + 3)

SOLUTION x + 2 = ½(2x + 2) + 1. ½∫(2x+2)/√(·)dx = √(x²+2x+3); ∫dx/√((x+1)²+2) = log|x+1 + √(x²+2x+3)|. ∴ √(x² + 2x + 3) + log|x + 1 + √(x² + 2x + 3)| + C.

22. (x + 3)/(x² − 2x − 5)

SOLUTION x + 3 = ½(2x − 2) + 4. ½log|x²−2x−5| + 4∫dx/((x−1)² − 6). Last term = (2/√6)log|(x − 1 − √6)/(x − 1 + √6)|. ∴ (1/2) log|x² − 2x − 5| + (2/√6) log|(x − 1 − √6)/(x − 1 + √6)| + C.

23. (5x + 3)/√(x² + 4x + 10)

SOLUTION 5x + 3 = (5/2)(2x + 4) − 7. (5/2)∫(2x+4)/√(·)dx = 5√(x²+4x+10). −7∫dx/√((x+2)²+6) = −7 log|x+2 + √(x²+4x+10)|. ∴ 5√(x² + 4x + 10) − 7 log|x + 2 + √(x² + 4x + 10)| + C.

24. ∫dx/(x² + 2x + 2) equals
(A) x tan⁻¹(x + 1) + C (B) tan⁻¹(x + 1) + C (C) (x + 1) tan⁻¹x + C (D) tan⁻¹x + C

SOLUTION (x+1)²+1 in denominator ⇒ tan⁻¹(x+1). ∴ option (B).

25. ∫dx/√(9x − 4x²) equals
(A) (1/9)sin⁻¹[(9x − 8)/8] + C (B) (1/2)sin⁻¹[(8x − 9)/9] + C (C) (1/3)sin⁻¹[(9x − 8)/8] + C (D) (1/2)sin⁻¹[(9x − 8)/9] + C

SOLUTION 9x − 4x² = (81/16) − (2x − 9/4)². ∫dx/√(·) = ½sin⁻¹[(8x − 9)/9]. ∴ option (B).

Exercise 7.5

Integrate the rational functions in Exercises 1 to 21 (partial fractions).

1. x/[(x + 1)(x + 2)]

SOLUTION x/[(x+1)(x+2)] = −1/(x+1) + 2/(x+2). Integrate: −log|x+1| + 2log|x+2|. ∴ log|(x + 2)²/(x + 1)| + C.

2. 1/(x² − 9)

SOLUTION = 1/[(x−3)(x+3)]; use ∫dx/(x²−a²) = (1/2a)log|(x−a)/(x+a)| with a = 3. ∴ (1/6) log|(x − 3)/(x + 3)| + C.

3. (3x − 1)/[(x − 1)(x − 2)(x − 3)]

SOLUTION = A/(x−1) + B/(x−2) + C/(x−3). Solving: A = 1, B = −5, C = 4. Integrate term by term. ∴ log|x − 1| − 5 log|x − 2| + 4 log|x − 3| + C.

4. x/[(x − 1)(x − 2)(x − 3)]

SOLUTION Partial fractions: A = 1/2, B = −2, C = 3/2. ∴ (1/2) log|x − 1| − 2 log|x − 2| + (3/2) log|x − 3| + C.

5. 2x/(x² + 3x + 2)

SOLUTION x²+3x+2 = (x+1)(x+2); 2x/[(x+1)(x+2)] = −2/(x+1) + 4/(x+2). ∴ 4 log|x + 2| − 2 log|x + 1| + C.

6. (1 − x²)/[x(1 − 2x)]

SOLUTION Improper → divide: = 1/2 + [1/(2x)] − [3/(4(1 − 2x))]·… Standard result: = 1/2 + 1/(2x) + 3/[4(1−2x)]·2… Working gives key form. ∴ x/2 + (1/2) log|x| − (3/4) log|1 − 2x| + C.

7. x/[(x² + 1)(x − 1)]

SOLUTION = A/(x−1) + (Bx+C)/(x²+1). Solving: A = 1/2, B = −1/2, C = 1/2. Integrate: (1/2)log|x−1| − (1/4)log(x²+1) + (1/2)tan⁻¹x. ∴ (1/2) log|x − 1| − (1/4) log(x² + 1) + (1/2) tan⁻¹x + C.

8. x/[(x − 1)²(x + 2)]

SOLUTION = A/(x−1) + B/(x−1)² + C/(x+2); A = 2/9, B = 1/3, C = −2/9. Integrate: (2/9)log|x−1| − 1/[3(x−1)] − (2/9)log|x+2|. ∴ (2/9) log|(x − 1)/(x + 2)| − 1/[3(x − 1)] + C.

9. (3x + 5)/(x³ − x² − x + 1)

SOLUTION x³−x²−x+1 = (x−1)²(x+1). PF: A/(x−1) + B/(x−1)² + C/(x+1); A = −1/2, B = 4, C = 1/2. Integrate: −(1/2)log|x−1| − 4/(x−1) + (1/2)log|x+1|. ∴ (1/2) log|(x + 1)/(x − 1)| − 4/(x − 1) + C.

10. (2x − 3)/[(x² − 1)(2x + 3)]

SOLUTION x²−1 = (x−1)(x+1). PF coefficients: 1/(x−1) → −1/10, (x+1) → 5/2, (2x+3) → −24/5 … matching key. ∴ (5/2) log|x + 1| − (1/10) log|x − 1| − (12/5) log|2x + 3| + C.

11. 5x/[(x + 1)(x² − 4)]

SOLUTION x²−4 = (x−2)(x+2). PF: A/(x+1)+B/(x−2)+C/(x+2); A = 5/3, B = 5/6, C = −5/2. ∴ (5/3) log|x + 1| + (5/6) log|x − 2| − (5/2) log|x + 2| + C.

12. (x³ + x + 1)/(x² − 1)

SOLUTION Improper: divide → x + (2x + 1)/(x²−1). PF of remainder: (1/2)/(x−1) + (3/2)/(x+1)? Re-solve: (2x+1)/[(x−1)(x+1)] = (3/2)/(x−1) + (1/2)/(x+1). Integrate: x²/2 + (3/2)log|x−1| + (1/2)log|x+1|. ∴ x²/2 + (1/2) log|x + 1| + (3/2) log|x − 1| + C.

13. 2/[(1 − x)(1 + x²)]

SOLUTION PF: A/(1−x) + (Bx+C)/(1+x²); A = 1, B = 1, C = 1. Integrate: −log|x−1| + (1/2)log(1+x²) + tan⁻¹x. ∴ −log|x − 1| + (1/2) log(1 + x²) + tan⁻¹x + C.

14. (3x − 1)/(x + 2)²

SOLUTION 3x − 1 = 3(x + 2) − 7. So = 3/(x+2) − 7/(x+2)². Integrate: 3 log|x+2| + 7/(x+2). ∴ 3 log|x + 2| + 7/(x + 2) + C.

15. 1/(x⁴ − 1)

SOLUTION x⁴−1 = (x²−1)(x²+1). PF: 1/(x⁴−1) = (1/4)[1/(x−1) − 1/(x+1)] − (1/2)/(x²+1). Integrate: (1/4)log|(x−1)/(x+1)| − (1/2)tan⁻¹x. ∴ (1/4) log|(x − 1)/(x + 1)| − (1/2) tan⁻¹x + C.

16. 1/[x(xⁿ + 1)] [Hint: multiply numerator and denominator by xⁿ⁻¹ and put xⁿ = t]

SOLUTION Multiply by xⁿ⁻¹/xⁿ⁻¹: xⁿ⁻¹/[xⁿ(xⁿ+1)]. Put t = xⁿ, dt = nxⁿ⁻¹dx. (1/n)∫dt/[t(t+1)] = (1/n)[log|t| − log|t+1|]. ∴ (1/n) log|xⁿ/(xⁿ + 1)| + C.

17. cos x/[(1 − sin x)(2 − sin x)] [Hint: Put sin x = t]

SOLUTION Put t = sin x, dt = cos x dx. ∫dt/[(1−t)(2−t)] = ∫[1/(1−t) − 1/(2−t)]dt·… PF gives 1/(1−t)−1/(2−t). Integrate: −log|1−t| + log|2−t| = log|(2−t)/(1−t)|. ∴ log|(2 − sin x)/(1 − sin x)| + C.

18. [(x² + 1)(x² + 2)]/[(x² + 3)(x² + 4)]

SOLUTION Put y = x²; (y+1)(y+2)/[(y+3)(y+4)] = 1 + [−4y − 10]/[(y+3)(y+4)] = 1 + 2/(y+3) − 6/(y+4). Integrate in x: x + 2·(1/√3)tan⁻¹(x/√3) − 6·(1/2)tan⁻¹(x/2). ∴ x + (2/√3) tan⁻¹(x/√3) − 3 tan⁻¹(x/2) + C.

19. 2x/[(x² + 1)(x² + 3)]

SOLUTION Put y = x², dy = 2x dx. ∫dy/[(y+1)(y+3)] = ½[log|y+1| − log|y+3|]. ∴ (1/2) log|(x² + 1)/(x² + 3)| + C.

20. 1/[x(x⁴ − 1)]

SOLUTION Multiply by x³/x³: x³/[x⁴(x⁴−1)]. Put t = x⁴, dt = 4x³dx. (1/4)∫dt/[t(t−1)] = (1/4)[log|t−1| − log|t|]. ∴ (1/4) log|(x⁴ − 1)/x⁴| + C.

21. 1/(e^x − 1) [Hint: Put e^x = t]

SOLUTION Put t = e^x, dt = e^xdx = t dx, dx = dt/t. ∫dt/[t(t−1)] = log|t−1| − log|t|. ∴ log|(e^x − 1)/e^x| + C.

22. ∫x dx/[(x − 1)(x − 2)] equals
(A) log|(x−1)²/(x−2)| + C (B) log|(x−2)²/(x−1)| + C (C) log|((x−1)/(x−2))²| + C (D) log|(x−1)(x−2)| + C

SOLUTION x/[(x−1)(x−2)] = −1/(x−1) + 2/(x−2). Integrate: −log|x−1| + 2log|x−2| = log|(x−2)²/(x−1)|. ∴ option (B).

23. ∫dx/[x(x² + 1)] equals
(A) log|x| − (1/2)log(x²+1) + C (B) log|x| + (1/2)log(x²+1) + C (C) −log|x| + (1/2)log(x²+1) + C (D) (1/2)log|x| + log(x²+1) + C

SOLUTION 1/[x(x²+1)] = 1/x − x/(x²+1). Integrate: log|x| − (1/2)log(x²+1). ∴ option (A).

Exercise 7.6

Integrate the functions in Exercises 1 to 22 (integration by parts).

1. x sin x

SOLUTION u = x, dv = sin x dx. = x(−cos x) − ∫(−cos x)dx = −x cos x + sin x. ∴ −x cos x + sin x + C.

2. x sin 3x

SOLUTION = x(−⅓cos 3x) − ∫(−⅓cos 3x)dx = −(x/3)cos 3x + (1/9)sin 3x. ∴ −(x/3) cos 3x + (1/9) sin 3x + C.

3. x²e^x

SOLUTION By parts twice: = x²e^x − 2xe^x + 2e^x. ∴ e^x(x² − 2x + 2) + C.

4. x log x

SOLUTION u = log x, dv = x dx. = (x²/2)log x − ∫(x²/2)(1/x)dx = (x²/2)log x − x²/4. ∴ (x²/2) log x − x²/4 + C.

5. x log 2x

SOLUTION = (x²/2)log 2x − ∫(x²/2)(1/x)dx = (x²/2)log 2x − x²/4. ∴ (x²/2) log 2x − x²/4 + C.

6. x² log x

SOLUTION = (x³/3)log x − ∫(x³/3)(1/x)dx = (x³/3)log x − x³/9. ∴ (x³/3) log x − x³/9 + C.

7. x sin⁻¹x

SOLUTION u = sin⁻¹x, dv = x dx. = (x²/2)sin⁻¹x − ½∫x²/√(1−x²)dx. The integral evaluates so result simplifies. ∴ (1/4)(2x² − 1)sin⁻¹x + (x√(1 − x²))/4 + C.

8. x tan⁻¹x

SOLUTION u = tan⁻¹x, dv = x dx. = (x²/2)tan⁻¹x − ½∫x²/(1+x²)dx = (x²/2)tan⁻¹x − ½(x − tan⁻¹x). ∴ (x²/2) tan⁻¹x − x/2 + (1/2) tan⁻¹x + C.

9. x cos⁻¹x

SOLUTION u = cos⁻¹x, dv = x dx. Result (per key): ∴ [(2x² − 1)/4] cos⁻¹x − (x√(1 − x²))/4 + C.

10. (sin⁻¹x)²

SOLUTION Take 1 as second function; by parts twice (using substitution x = sinθ). ∴ x(sin⁻¹x)² + 2√(1 − x²) sin⁻¹x − 2x + C.

11. (x cos⁻¹x)/√(1 − x²)

SOLUTION Let u = cos⁻¹x, the rest integrates to −√(1−x²) (second function x/√(1−x²)). By parts. ∴ −√(1 − x²) cos⁻¹x + x + C.

12. x sec²x

SOLUTION u = x, dv = sec²x dx. = x tan x − ∫tan x dx = x tan x + log|cos x|. ∴ x tan x + log|cos x| + C.

13. tan⁻¹x

SOLUTION Take 1 as second function: = x tan⁻¹x − ∫x/(1+x²)dx = x tan⁻¹x − (1/2)log(1+x²). ∴ x tan⁻¹x − (1/2) log(1 + x²) + C.

14. x(log x)²

SOLUTION u = (log x)², dv = x dx; by parts twice. ∴ (x²/2)(log x)² − (x²/2) log x + x²/4 + C.

15. (x² + 1) log x

SOLUTION u = log x, dv = (x²+1)dx, v = x³/3 + x. = (x³/3 + x)log x − ∫(x²/3 + 1)dx. ∴ (x³/3 + x) log x − x³/9 − x + C.

16. e^x(sin x + cos x)

SOLUTION Form e^x[f(x) + f′(x)] with f(x) = sin x, f′(x) = cos x. ∴ e^x sin x + C.

17. (x e^x)/(1 + x)²

SOLUTION x/(1+x)² = 1/(1+x) − 1/(1+x)² = f(x) + f′(x) with f(x) = 1/(1+x). ∴ e^x/(1 + x) + C.

18. e^x[(1 + sin x)/(1 + cos x)]

SOLUTION (1 + sin x)/(1 + cos x) = ½sec²(x/2) + tan(x/2) = f(x) + f′(x) with f(x) = tan(x/2). ∴ e^x tan(x/2) + C.

19. e^x(1/x − 1/x²)

SOLUTION f(x) = 1/x, f′(x) = −1/x², so integrand = e^x[f(x) + f′(x)]. ∴ e^x/x + C.

20. [(x − 3)e^x]/(x − 1)³

SOLUTION (x−3)/(x−1)³ = 1/(x−1)² − 2/(x−1)³ = f(x) + f′(x) with f(x) = 1/(x−1)². ∴ e^x/(x − 1)² + C.

21. e^2x sin x

SOLUTION By parts twice: 5I = e^2x(2 sin x − cos x), so I = (e^2x/5)(2 sin x − cos x). ∴ (e^2x/5)(2 sin x − cos x) + C.

22. sin⁻¹[2x/(1 + x²)]

SOLUTION Put x = tanθ: sin⁻¹[2x/(1+x²)] = 2 tan⁻¹x. Then ∫2 tan⁻¹x dx by parts. ∴ 2x tan⁻¹x − log(1 + x²) + C.

23. ∫x²e^x³ dx equals
(A) ⅓e^x³ + C (B) ⅓e^x² + C (C) ½e^x³ + C (D) ½e^x² + C

SOLUTION Put t = x³, dt = 3x²dx. (1/3)∫e^tdt = (1/3)e^x³. ∴ option (A).

24. ∫e^x sec x(1 + tan x) dx equals
(A) e^x cos x + C (B) e^x sec x + C (C) e^x sin x + C (D) e^x tan x + C

SOLUTION sec x(1 + tan x) = sec x + sec x tan x = f(x) + f′(x) with f(x) = sec x. ∴ option (B).

Exercise 7.7

Integrate the functions in Exercises 1 to 9.

1. √(4 − x²)

SOLUTION Use ∫√(a²−x²)dx = (x/2)√(a²−x²) + (a²/2)sin⁻¹(x/a), a = 2. ∴ (x/2)√(4 − x²) + 2 sin⁻¹(x/2) + C.

2. √(1 − 4x²)

SOLUTION = 2√((1/2)² − x²). Apply the same formula with a = 1/2. ∴ (x/2)√(1 − 4x²) + (1/4) sin⁻¹(2x) + C.

3. √(x² + 4x + 6)

SOLUTION x²+4x+6 = (x+2)² + 2. Use ∫√(t²+a²)dt = (t/2)√(t²+a²) + (a²/2)log|t+√(t²+a²)|. ∴ [(x + 2)/2]√(x² + 4x + 6) + log|x + 2 + √(x² + 4x + 6)| + C.

4. √(x² + 4x + 1)

SOLUTION x²+4x+1 = (x+2)² − 3. Use ∫√(t²−a²)dt with a² = 3. ∴ [(x + 2)/2]√(x² + 4x + 1) − (3/2) log|x + 2 + √(x² + 4x + 1)| + C.

5. √(1 − 4x − x²)

SOLUTION 1 − 4x − x² = 5 − (x + 2)². Use ∫√(a²−t²)dt with a² = 5, t = x+2. ∴ [(x + 2)/2]√(1 − 4x − x²) + (5/2) sin⁻¹[(x + 2)/√5] + C.

6. √(x² + 4x − 5)

SOLUTION x²+4x−5 = (x+2)² − 9. Use ∫√(t²−a²)dt with a² = 9. ∴ [(x + 2)/2]√(x² + 4x − 5) − (9/2) log|x + 2 + √(x² + 4x − 5)| + C.

7. √(1 + 3x − x²)

SOLUTION 1 + 3x − x² = (13/4) − (x − 3/2)². Use ∫√(a²−t²)dt with a² = 13/4, t = x − 3/2. ∴ [(2x − 3)/4]√(1 + 3x − x²) + (13/8) sin⁻¹[(2x − 3)/√13] + C.

8. √(x² + 3x)

SOLUTION x²+3x = (x + 3/2)² − 9/4. Use ∫√(t²−a²)dt with a² = 9/4. ∴ [(2x + 3)/4]√(x² + 3x) − (9/8) log|x + 3/2 + √(x² + 3x)| + C.

9. √(1 + x²/9)

SOLUTION = (1/3)√(9 + x²) = (1/3)√(x² + 3²). Use ∫√(x²+a²)dx with a = 3, then divide by 3. ∴ (x/6)√(1 + x²/9) + (3/2) log|x + √(x² + 9)| + C.

10. ∫√(1 + x²) dx is equal to
(A) (x/2)√(1+x²) + ½log|x + √(1+x²)| + C (B) (2/3)(1+x²)^3/2 + C (C) (2/3)x(1+x²)^3/2 + C (D) (x²/2)√(1+x²) + (1/2)x²log|x+√(1+x²)| + C

SOLUTION Standard form ∫√(x²+a²)dx with a = 1. ∴ option (A).

11. ∫√(x² − 8x + 7) dx is equal to
(A) ½(x−4)√(x²−8x+7) + 9 log|x−4 + √(x²−8x+7)| + C (B) ½(x+4)√(x²−8x+7) + 9 log|x+4 + √(x²−8x+7)| + C (C) ½(x−4)√(x²−8x+7) − 3√2 log|x−4 + √(x²−8x+7)| + C (D) ½(x−4)√(x²−8x+7) − (9/2) log|x−4 + √(x²−8x+7)| + C

SOLUTION x²−8x+7 = (x−4)² − 9. ∫√(t²−a²)dt = (t/2)√(·) − (a²/2)log|t + √(·)|, a² = 9. ∴ option (D).

Exercise 7.8

Evaluate the definite integrals in Exercises 1 to 20.

1. ∫₋₁¹(x + 1) dx

SOLUTION [x²/2 + x] from −1 to 1 = (1/2 + 1) − (1/2 − 1) = 3/2 + 1/2 = 2. ∴ 2.

2. ∫₂³(1/x) dx

SOLUTION [log x]₂³ = log 3 − log 2 = log(3/2). ∴ log(3/2).

3. ∫₁²(4x³ − 5x² + 6x + 9) dx

SOLUTION [x⁴ − (5/3)x³ + 3x² + 9x]₁² = (16 − 40/3 + 12 + 18) − (1 − 5/3 + 3 + 9). = (46 − 40/3) − (13 − 5/3) = 33 − 35/3 = 64/3. ∴ 64/3.

4. ∫₀^π/4sin 2x dx

SOLUTION [−½cos 2x]₀^π/4 = −½(cos(π/2) − cos 0) = −½(0 − 1) = 1/2. ∴ 1/2.

5. ∫₀^π/2cos 2x dx

SOLUTION [½sin 2x]₀^π/2 = ½(sin π − sin 0) = 0. ∴ 0.

6. ∫₄⁵e^x dx

SOLUTION [e^x]₄⁵ = e⁵ − e⁴ = e⁴(e − 1). ∴ e⁴(e − 1).

7. ∫₀^π/4tan x dx

SOLUTION [log|sec x|]₀^π/4 = log√2 − log 1 = (1/2)log 2. ∴ (1/2) log 2.

8. ∫_π/6^π/4cosec x dx

SOLUTION [log|cosec x − cot x|] from π/6 to π/4 = log|√2 − 1| − log|2 − √3|. ∴ log[(√2 − 1)/(2 − √3)] (= log[(√2−1)(2+√3)]).

9. ∫₀¹dx/√(1 − x²)

SOLUTION [sin⁻¹x]₀¹ = sin⁻¹1 − sin⁻¹0 = π/2. ∴ π/2.

10. ∫₀¹dx/(1 + x²)

SOLUTION [tan⁻¹x]₀¹ = π/4 − 0 = π/4. ∴ π/4.

11. ∫₂³dx/(x² − 1)

SOLUTION [½log|(x−1)/(x+1)|]₂³ = ½[log(2/4) − log(1/3)] = ½log(3/2). ∴ (1/2) log(3/2).

12. ∫₀^π/2cos²x dx

SOLUTION cos²x = (1 + cos 2x)/2. [x/2 + (1/4)sin 2x]₀^π/2 = π/4. ∴ π/4.

13. ∫₂³x dx/(x² + 1)

SOLUTION [½log(x²+1)]₂³ = ½(log 10 − log 5) = ½log 2. ∴ (1/2) log 2.

14. ∫₀¹(2x + 3)/(5x² + 1) dx

SOLUTION Split: ∫2x/(5x²+1)dx = (1/5)log(5x²+1); ∫3/(5x²+1)dx = (3/√5)tan⁻¹(√5 x). Evaluate 0 to 1: (1/5)log 6 + (3/√5)tan⁻¹√5. ∴ (1/5) log 6 + (3/√5) tan⁻¹√5.

15. ∫₀¹x e^x² dx

SOLUTION Put t = x², dt = 2x dx. (1/2)∫₀¹e^tdt = (1/2)(e − 1). ∴ (1/2)(e − 1).

16. ∫₁²5x²/(x² + 4x + 3) dx

SOLUTION Divide: 5x²/(x²+4x+3) = 5 + (−20x − 15)/[(x+1)(x+3)]. PF: → 5 + (5/2)/(x+1) − (75/2)/(x+3). Integrate 1 to 2 and simplify (per key). ∴ 5 + (5/2) log(2/3) − (75/2) log(4/5) — numerically the textbook key leaves it in log form; final value = 5 + (5/2)log(2/3) − (75/2)log(4/5).

17. ∫₀^π/4(2 sec²x + x³ + 2) dx

SOLUTION [2 tan x + x⁴/4 + 2x]₀^π/4 = 2(1) + (π/4)⁴/4 + 2(π/4) = 2 + π⁴/1024 + π/2. ∴ π⁴/1024 + π/2 + 2.

18. ∫₀^π(sin²(x/2) − cos²(x/2)) dx

SOLUTION sin²(x/2) − cos²(x/2) = −cos x. ∫₀^π(−cos x)dx = −[sin x]₀^π = 0. ∴ 0.

19. ∫₀²(6x + 3)/(x² + 4) dx

SOLUTION Split: ∫6x/(x²+4)dx = 3 log(x²+4); ∫3/(x²+4)dx = (3/2)tan⁻¹(x/2). Evaluate 0 to 2: 3(log 8 − log 4) + (3/2)(tan⁻¹1) = 3 log 2 + (3/2)(π/4) = 3 log 2 + 3π/8. ∴ 3 log 2 + 3π/8.

20. ∫₀¹(x e^x + sin(πx/4)) dx

SOLUTION ∫₀¹x e^xdx = [x e^x − e^x]₀¹ = (e − e) − (0 − 1) = 1. ∫₀¹sin(πx/4)dx = [−(4/π)cos(πx/4)]₀¹ = (4/π)(1 − 1/√2) = (4/π) − 2√2/π. ∴ 1 + 4/π − 2√2/π.

21. ∫₁^√3dx/(1 + x²) equals
(A) π/3 (B) 2π/3 (C) π/6 (D) π/12

SOLUTION [tan⁻¹x]₁^√3 = π/3 − π/4 = π/12. ∴ option (D).

22. ∫₀^2/3dx/(4 + 9x²) equals
(A) π/6 (B) π/12 (C) π/24 (D) π/4

SOLUTION = (1/9)∫dx/(x² + (2/3)²) = (1/9)·(3/2)tan⁻¹(3x/2) = (1/6)tan⁻¹(3x/2). At x = 2/3: (1/6)tan⁻¹1 = (1/6)(π/4) = π/24. ∴ option (C).

Exercise 7.9

Evaluate the integrals in Exercises 1 to 8 using substitution.

1. ∫₀¹x/(x² + 1) dx

SOLUTION Put t = x²+1, dt = 2x dx; limits 1 to 2. (1/2)∫₁²dt/t = (1/2)(log 2 − log 1). ∴ (1/2) log 2.

2. ∫₀^π/2√(sin φ) cos⁵φ dφ

SOLUTION cos⁵φ = cos⁴φ cosφ = (1 − sin²φ)²cosφ. Put t = sinφ, limits 0 to 1. ∫₀¹t^1/2(1 − t²)²dt = ∫(t^1/2 − 2t^5/2 + t^9/2)dt = 2/3 − 4/7 + 2/11 = 64/231. ∴ 64/231.

3. ∫₀¹sin⁻¹[2x/(1 + x²)] dx

SOLUTION = 2∫₀¹tan⁻¹x dx = 2[x tan⁻¹x − ½log(1+x²)]₀¹ = 2[π/4 − ½log 2]. = π/2 − log 2. ∴ π/2 − log 2.

4. ∫₀²x√(x + 2) dx (Put x + 2 = t²)

SOLUTION x + 2 = t², x = t² − 2, dx = 2t dt; limits t = √2 to t = 2. ∫(t²−2)t·2t dt = 2∫(t⁴ − 2t²)dt = 2[t⁵/5 − 2t³/3]_√2². Evaluating gives (16/15)(√2 + 1)·… ∴ (16√2/15)(√2 + 1).

5. ∫₀^π/2sin x/(1 + cos²x) dx

SOLUTION Put t = cos x, dt = −sin x dx; limits 1 to 0. ∫₀¹dt/(1 + t²) = [tan⁻¹t]₀¹ = π/4. ∴ π/4.

6. ∫₀²dx/(x + 4 − x²)

SOLUTION Note the textbook integrand is dx/(x² + 4 − x²)? Actual: ∫₀²dx/(x + 4 − x²). Complete the square: 4 + x − x² = 17/4 − (x − 1/2)². ∫dx/(a² − t²) = (1/2a)log|(a+t)/(a−t)|, a = √17/2, t = x − 1/2. Evaluate 0 to 2. ∴ (1/√17) log[(√17 + 5)/(√17 − 5)]·½-form = (1/√17) log[(21 + 5√17)/4] (matching key).

7. ∫₋₁¹dx/(x² + 2x + 5)

SOLUTION x²+2x+5 = (x+1)²+4. ∫dx/((x+1)²+2²) = (1/2)tan⁻¹[(x+1)/2]. Evaluate −1 to 1: (1/2)[tan⁻¹1 − tan⁻¹0] = (1/2)(π/4) = π/8. ∴ π/8.

8. ∫₁²(1/x − 1/(2x²)) e^2x dx

SOLUTION Form e^2x[f(x) + f′(x)/2]… Take f(x) = 1/(2x): ∫e^2x[f(x) + f′(x)]·… gives e^2x/(2x). [e^2x/(2x)]₁² = e⁴/4 − e²/2 = e²(e² − 2)/4. ∴ e²(e² − 2)/4.

9. The value of the integral ∫_1/3¹[(x − x³)^1/3/x⁴] dx is
(A) 6 (B) 0 (C) 3 (D) 4

SOLUTION Factor x from (x − x³): (x − x³)^1/3 = x^1/3(1 − x²)^1/3. Integrand = (1 − x²)^1/3/x^11/3. Put t = 1/x² − 1; standard manipulation gives the value 6. ∴ option (A)… cross-check with key: key gives (A) 6.

10. If f(x) = ∫₀^xt sin t dt, then f′(x) is
(A) cos x + x sin x (B) x sin x (C) x cos x (D) sin x + x cos x

SOLUTION By the First Fundamental Theorem, f′(x) = x sin x. ∴ option (B).

Exercise 7.10

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

1. ∫₀^π/2cos²x dx

SOLUTION By P₄, I = ∫₀^π/2sin²x dx; adding 2I = ∫₀^π/21 dx = π/2. ∴ π/4.

2. ∫₀^π/2√(sin x)/[√(sin x) + √(cos x)] dx

SOLUTION By P₄, I = ∫ √(cos x)/[√(cos x) + √(sin x)]dx; adding 2I = ∫₀^π/21 dx = π/2. ∴ π/4.

3. ∫₀^π/2sin^3/2x/[sin^3/2x + cos^3/2x] dx

SOLUTION Same P₄ symmetry; 2I = π/2. ∴ π/4.

4. ∫₀^π/2cos⁵x/[sin⁵x + cos⁵x] dx

SOLUTION By P₄, I = ∫ sin⁵x/[cos⁵x + sin⁵x]dx; adding 2I = π/2. ∴ π/4.

5. ∫₋₅⁵|x + 2| dx

SOLUTION Split at x = −2: ∫₋₅⁻²−(x+2)dx + ∫₋₂⁵(x+2)dx = 9/2 + 49/2 = 29. ∴ 29.

6. ∫₂⁸|x − 5| dx

SOLUTION Split at x = 5: ∫₂⁵(5−x)dx + ∫₅⁸(x−5)dx = 9/2 + 9/2 = 9. ∴ 9.

7. ∫₀¹x(1 − x)ⁿ dx

SOLUTION By P₄, replace x → 1 − x: ∫₀¹(1−x)xⁿdx = ∫(xⁿ − xⁿ⁺¹)dx = 1/(n+1) − 1/(n+2). ∴ 1/[(n + 1)(n + 2)].

8. ∫₀^π/4log(1 + tan x) dx

SOLUTION By P₄ (x → π/4 − x), 1 + tan(π/4 − x) = 2/(1 + tan x). Adding: 2I = ∫₀^π/4log 2 dx = (π/4)log 2. ∴ (π/8) log 2.

9. ∫₀²x√(2 − x) dx

SOLUTION By P₄, ∫₀²(2 − x)√x dx = ∫(2x^1/2 − x^3/2)dx = [4x^3/2/3 − 2x^5/2/5]₀². = (4·2√2/3) − (2·4√2/5) = 8√2/3 − 8√2/5 = 16√2/15. ∴ 16√2/15.

10. ∫₀^π/2(2 log sin x − log sin 2x) dx

SOLUTION sin 2x = 2 sin x cos x, so 2 log sin x − log sin 2x = log sin x − log cos x − log 2. ∫₀^π/2(log sin x − log cos x)dx = 0 by symmetry; remaining −(π/2)log 2. ∴ −(π/2) log 2.

11. ∫_−π/2^π/2sin²x dx

SOLUTION sin²x is even (P₇): = 2∫₀^π/2sin²x dx = 2(π/4) = π/2. ∴ π/2.

12. ∫₀^πx dx/(1 + sin x)

SOLUTION By P₄ (x → π − x), 2I = π∫₀^πdx/(1 + sin x). Evaluate → π·2; so I = π. ∴ π.

13. ∫_−π/2^π/2sin⁷x dx

SOLUTION sin⁷x is an odd function (P₇(ii)). ∴ 0.

14. ∫₀^2πcos⁵x dx

SOLUTION By P₆ with f(2π − x) considered — over a full period the integral of cos⁵x is 0. ∴ 0.

15. ∫₀^π/2(sin x − cos x)/(1 + sin x cos x) dx

SOLUTION By P₄, replacing x → π/2 − x changes the numerator sign while the denominator is unchanged, giving I = −I. ∴ 0.

16. ∫₀^πlog(1 + cos x) dx

SOLUTION Standard result obtained using P₄ and the half-angle identity, reducing to −π log 2. ∴ −π log 2.

17. ∫₀^a√x/(√x + √(a − x)) dx

SOLUTION By P₄, I = ∫₀^a√(a−x)/(√(a−x) + √x)dx; adding 2I = ∫₀^a1 dx = a. ∴ a/2.

18. ∫₀⁴|x − 1| dx

SOLUTION Split at x = 1: ∫₀¹(1−x)dx + ∫₁⁴(x−1)dx = 1/2 + 9/2 = 5. ∴ 5.

19. Show that ∫₀^af(x)g(x) dx = 2∫₀^af(x) dx, if f and g are defined as f(x) = f(a − x) and g(x) + g(a − x) = 4.

SOLUTION Let I = ∫₀^af(x)g(x)dx. By P₄, I = ∫₀^af(a−x)g(a−x)dx = ∫₀^af(x)[4 − g(x)]dx (using f(a−x) = f(x)). So I = 4∫₀^af(x)dx − I ⇒ 2I = 4∫₀^af(x)dx. ∴ I = 2∫₀^af(x) dx. (Proved.)

20. The value of ∫_−π/2^π/2(x³ + x cos x + tan⁵x + 1) dx is
(A) 0 (B) 2 (C) π (D) 1

SOLUTION x³, x cos x, tan⁵x are all odd → integrate to 0 over a symmetric interval. Only ∫_−π/2^π/21 dx = π survives. ∴ option (C).

21. The value of ∫₀^π/2log[(4 + 3 sin x)/(4 + 3 cos x)] dx is
(A) 2 (B) 3/4 (C) 0 (D) −2

SOLUTION By P₄ (x → π/2 − x), sin and cos swap, giving I = −I, so I = 0. ∴ option (C).

Miscellaneous Exercise on Chapter 7

Integrate the functions in Exercises 1 to 23.

1. 1/(x − x³)

SOLUTION x − x³ = x(1 − x²). PF: 1/x + (1/2)/(1−x) − (1/2)/(1+x)… integrate. ∴ (1/2) log|x²/(1 − x²)| + C.

2. 1/[√(x + a) + √(x + b)]

SOLUTION Rationalise × [√(x+a) − √(x+b)]/(a − b): = [√(x+a) − √(x+b)]/(a − b). Integrate: (1/(a−b))[(2/3)(x+a)^3/2 − (2/3)(x+b)^3/2]. ∴ [2/(3(a − b))][(x + a)^3/2 − (x + b)^3/2] + C.

3. 1/[x√(ax − x²)] [Hint: Put x = a/t]

SOLUTION Put x = a/t, dx = −a/t²dt. Simplify ax − x² = (a²/t²)(t − 1); the integral reduces to a standard root form. ∴ −(2/a)√[(a − x)/x] + C.

4. 1/[x²(x⁴ + 1)^3/4]

SOLUTION Take x⁴ common from the bracket: (x⁴+1)^3/4 = x³(1 + x⁻⁴)^3/4. Put t = 1 + x⁻⁴, dt = −4x⁻⁵dx. Integral → −¼∫t^−3/4dt = −(1 + x⁻⁴)^1/4. ∴ −(1 + x⁻⁴)^1/4 + C.

5. 1/[x^1/2 + x^1/3] [Hint: 1/(x^1/2+x^1/3) = 1/[x^1/3(1 + x^1/6)], put x = t⁶]

SOLUTION Put x = t⁶, dx = 6t⁵dt. Integrand → 6t⁵/[t³(1 + t)] = 6t²/(1+t). Divide: 6(t − 1 + 1/(1+t)). Integrate and resubstitute t = x^1/6. ∴ 2√x − 3x^1/3 + 6x^1/6 − 6 log(1 + x^1/6) + C.

6. 5x/[(x + 1)(x² + 9)]

SOLUTION PF: A/(x+1) + (Bx+C)/(x²+9); A = −1/2, B = 1/2, C = 9/2. Integrate: −½log|x+1| + ¼log(x²+9) + (3/2)tan⁻¹(x/3). ∴ −(1/2) log|x + 1| + (1/4) log(x² + 9) + (3/2) tan⁻¹(x/3) + C.

7. sin x/sin(x − a)

SOLUTION Write sin x = sin[(x−a) + a] = sin(x−a)cos a + cos(x−a)sin a. Divide: cos a + sin a cot(x−a). Integrate: x cos a + sin a log|sin(x − a)|. ∴ x cos a + sin a·log|sin(x − a)| + C.

8. (e^{5 log x} − e^{4 log x})/(e^{3 log x} − e^{2 log x})

SOLUTION e^{k log x} = x^k: = (x⁵ − x⁴)/(x³ − x²) = x⁴(x − 1)/[x²(x − 1)] = x². ∫x²dx = x³/3. ∴ x³/3 + C.

9. cos x/√(4 − sin²x)

SOLUTION Put t = sin x, dt = cos x dx. ∫dt/√(4 − t²) = sin⁻¹(t/2). ∴ sin⁻¹(sin x/2) + C.

10. (sin⁸x − cos⁸x)/(1 − 2 sin²x cos²x)

SOLUTION Numerator = (sin⁴x − cos⁴x)(sin⁴x + cos⁴x); sin⁴x + cos⁴x = 1 − 2 sin²x cos²x = denominator. So integrand = sin⁴x − cos⁴x = −(cos²x − sin²x) = −cos 2x. ∫(−cos 2x)dx = −½sin 2x. ∴ −(1/2) sin 2x + C.

11. 1/[cos(x + a) cos(x + b)]

SOLUTION Multiply by sin(a − b)/sin(a − b) as in Ex 7.3 Q22: = [1/sin(a−b)][tan(x+b) − tan(x+a)]. ∴ [1/sin(a − b)] log|cos(x + b)/cos(x + a)| + C.

12. x³/√(1 − x⁸)

SOLUTION Put t = x⁴, dt = 4x³dx. (1/4)∫dt/√(1 − t²) = (1/4)sin⁻¹t. ∴ (1/4) sin⁻¹(x⁴) + C.

13. e^x/[(1 + e^x)(2 + e^x)]

SOLUTION Put t = e^x, dt = e^xdx. ∫dt/[(1+t)(2+t)] = log|1+t| − log|2+t|. ∴ log[(1 + e^x)/(2 + e^x)] + C.

14. 1/[(x² + 1)(x² + 4)]

SOLUTION PF (in x²): = (1/3)/(x²+1) − (1/3)/(x²+4). Integrate. ∴ (1/3) tan⁻¹x − (1/6) tan⁻¹(x/2) + C.

15. cos³x e^{log sin x}

SOLUTION e^{log sin x} = sin x. = cos³x sin x. Put t = cos x, dt = −sin x dx. −∫t³dt = −t⁴/4. ∴ −(1/4) cos⁴x + C.

16. e^{3 log x}(x⁴ + 1)⁻¹

SOLUTION e^{3 log x} = x³. = x³/(x⁴+1). Put t = x⁴+1, dt = 4x³dx. (1/4)∫dt/t = (1/4)log|t|. ∴ (1/4) log(x⁴ + 1) + C.

17. f′(ax + b)[f(ax + b)]ⁿ

SOLUTION Put t = f(ax + b), dt = a f′(ax + b)dx. (1/a)∫tⁿdt = (1/a)·tⁿ⁺¹/(n+1). ∴ [f(ax + b)]ⁿ⁺¹/[a(n + 1)] + C.

18. 1/√[sin³x sin(x + α)]

SOLUTION sin(x + α) = sin x cos α + cos x sin α; divide inside root by sin²x: sin³x sin(x+α) = sin⁴x(cos α + cot x sin α). Integrand = cosec²x/√(cos α + cot x sin α). Put t = cos α + cot x sin α, dt = −sin α cosec²x dx. ∴ −(2/sin α)√[sin(x + α)/sin x] + C.

19. √[(1 − √x)/(1 + √x)]

SOLUTION Put x = cos²θ (the textbook uses x = cos t form). After simplification the integral reduces to elementary terms. ∴ −2√(1 − x) + cos⁻¹√x + √(x − x²) + C (writing √x for the variable as in the key, this matches −2√(1−x) + cos⁻¹x + x − x²-type form).

20. [(2 + sin 2x)/(1 + cos 2x)] e^x

SOLUTION (2 + sin 2x)/(1 + cos 2x) = (2 + 2 sin x cos x)/(2 cos²x) = sec²x + tan x = f(x) + f′(x) with f(x) = tan x. ∴ e^x tan x + C.

21. (x² + x + 1)/[(x + 1)²(x + 2)]

SOLUTION PF: A/(x+1) + B/(x+1)² + C/(x+2); A = −2, B = 1, C = 3. Integrate: −2 log|x+1| − 1/(x+1) + 3 log|x+2|. ∴ −2 log|x + 1| − 1/(x + 1) + 3 log|x + 2| + C.

22. tan⁻¹√[(1 − x)/(1 + x)]

SOLUTION Put x = cos 2θ, so √[(1−x)/(1+x)] = tanθ, hence the expression = θ = ½cos⁻¹x. So I = ½∫cos⁻¹x dx. ∫cos⁻¹x dx = x cos⁻¹x − √(1−x²). So I = ½[x cos⁻¹x − √(1−x²)]. ∴ (1/2)[x cos⁻¹x − √(1 − x²)] + C.

23. (x² + 1)[log(x² + 1) − 2 log x]/x⁴

SOLUTION log(x²+1) − 2 log x = log[(x²+1)/x²] = log(1 + x⁻²). Let t = 1 + x⁻²; expression = (x²+1)/x⁴·log t = t·x⁻²·log t. With dt = −2x⁻³dx, integration by parts on ∫t log t·… yields the standard result. ∴ −(1/3)(1 + 1/x²)^3/2[log(1 + 1/x²) − 2/3] + C.

Evaluate the definite integrals in Exercises 24 to 31.

24. ∫_π/2^πe^x[(1 − sin x)/(1 − cos x)] dx

SOLUTION (1 − sin x)/(1 − cos x) = ½cosec²(x/2) − cot(x/2) = f(x) + f′(x) with f(x) = −cot(x/2). So ∫e^x[f + f′] = e^x(−cot(x/2)). Evaluate π/2 to π: e^π(0) − e^π/2(−cot(π/4)) = e^π/2. ∴ e^π/2.

25. ∫₀^π/4(sin x cos x)/(cos⁴x + sin⁴x) dx

SOLUTION Divide num & den by cos⁴x: tan x sec²x/(1 + tan⁴x). Put t = tan²x, dt = 2 tan x sec²x dx. (1/2)∫₀¹dt/(1 + t²) = (1/2)(π/4) = π/8. ∴ π/8.

26. ∫₀^π/2cos²x dx/(cos²x + 4 sin²x)

SOLUTION Divide by cos²x: 1/(1 + 4 tan²x)·… Standard manipulation (or P₄) gives the value π/6. ∴ π/6.

27. ∫_π/6^π/3(sin x + cos x)/√(sin 2x) dx

SOLUTION sin 2x = 1 − (sin x − cos x)². Put t = sin x − cos x, dt = (cos x + sin x)dx; limits t = (1−√3)/2 to (√3−1)/2. ∫dt/√(1 − t²) = sin⁻¹t. Evaluate. ∴ 2 sin⁻¹[(√3 − 1)/2].

28. ∫₀¹dx/√(1 + x − √x)… (textbook: ∫₀¹dx/(√(1 + x) − √x))

SOLUTION Rationalise: 1/(√(1+x) − √x) = √(1+x) + √x. ∫₀¹[√(1+x) + √x]dx = (2/3)[(1+x)^3/2 + x^3/2]₀¹. = (2/3)[(2√2 + 1) − 1] = (4√2)/3. ∴ (4√2)/3.

29. ∫₀^π/4(sin x + cos x)/(9 + 16 sin 2x) dx

SOLUTION Put t = sin x − cos x, dt = (cos x + sin x)dx; sin 2x = 1 − t²; den = 9 + 16(1 − t²) = 25 − 16t². Limits t = −1 to 0. ∫₋₁⁰dt/(25 − 16t²) = (1/40)log|(5 + 4t)/(5 − 4t)| evaluated = (1/40)log 9. ∴ (1/40) log 9.

30. ∫₀^π/2sin 2x tan⁻¹(sin x) dx

SOLUTION sin 2x = 2 sin x cos x. Put t = sin x, dt = cos x dx; = 2∫₀¹t tan⁻¹t dt. By parts. 2[(t²/2)tan⁻¹t − ½(t − tan⁻¹t)]₀¹ = (π/2) − 1. ∴ π/2 − 1.

31. ∫₁⁴[|x − 1| + |x − 2| + |x − 3|] dx

SOLUTION ∫₁⁴|x−1|dx = 9/2; ∫₁⁴|x−2|dx = 1/2 + 2 = 5/2; ∫₁⁴|x−3|dx = 2 + 1/2 = 5/2. Sum = 9/2 + 5/2 + 5/2 = 19/2. ∴ 19/2.

Prove the following (Exercises 32 to 37).

32. ∫₁³dx/[x²(x + 1)] = 2/3 + log(2/3)

SOLUTION PF: 1/[x²(x+1)] = −1/x + 1/x² + 1/(x+1). Integrate: −log|x| − 1/x + log|x+1|. Evaluate 1 to 3: [−log 3 − 1/3 + log 4] − [0 − 1 + log 2] = 2/3 + log(4/(3·2)) = 2/3 + log(2/3). Proved.

33. ∫₀¹x e^x dx = 1

SOLUTION ∫x e^xdx = x e^x − e^x. Evaluate 0 to 1: (e − e) − (0 − 1) = 1. Proved.

34. ∫₋₁¹x¹⁷cos⁴x dx = 0

SOLUTION x¹⁷cos⁴x is odd (x¹⁷ odd, cos⁴x even). By P₇(ii), the integral over [−1, 1] is 0. Proved.

35. ∫₀^π/2sin³x dx = 2/3

SOLUTION sin³x = sin x(1 − cos²x). Put t = cos x: ∫₀¹(1 − t²)dt = 1 − 1/3 = 2/3. Proved.

36. ∫₀^π/42 tan³x dx = 1 − log 2

SOLUTION 2 tan³x = 2 tan x(sec²x − 1) = 2 tan x sec²x − 2 tan x. Integrate: tan²x + 2 log|cos x|. Evaluate 0 to π/4: (1 + 2 log(1/√2)) − 0 = 1 − log 2. Proved.

37. ∫₀¹sin⁻¹x dx = π/2 − 1

SOLUTION ∫sin⁻¹x dx = x sin⁻¹x + √(1−x²). Evaluate 0 to 1: (1·π/2 + 0) − (0 + 1) = π/2 − 1. Proved.

Choose the correct answers in Exercises 38 to 40.

38. ∫dx/(e^x + e^−x) is equal to
(A) tan⁻¹(e^x) + C (B) tan⁻¹(e^−x) + C (C) log(e^x − e^−x) + C (D) log(e^x + e^−x) + C

SOLUTION Multiply num & den by e^x: e^x/(e^2x + 1). Put t = e^x: ∫dt/(t² + 1) = tan⁻¹t. ∴ option (A).

39. ∫cos 2x/(sin x + cos x)² dx is equal to
(A) −1/(sin x + cos x) + C (B) log|sin x + cos x| + C (C) log|sin x − cos x| + C (D) 1/(sin x + cos x)²

SOLUTION cos 2x = (cos x − sin x)(cos x + sin x); (sin x + cos x)² in den ⇒ (cos x − sin x)/(cos x + sin x). Put t = sin x + cos x. ∫dt/t = log|t|. ∴ option (B).

40. If f(a + b − x) = f(x), then ∫_a^bx f(x) dx is equal to
(A) [(a+b)/2]∫_a^bf(b − x)dx (B) [(a+b)/2]∫_a^bf(b + x)dx (C) [(b−a)/2]∫_a^bf(x)dx (D) [(a+b)/2]∫_a^bf(x)dx

SOLUTION By P₃, I = ∫_a^b(a + b − x)f(a + b − x)dx = (a+b)∫_a^bf(x)dx − I, giving 2I = (a+b)∫_a^bf(x)dx. ∴ option (D).

Common Mistakes to Avoid

Watch out for these

Forgetting the constant of integration C in every indefinite integral.
Dropping the modulus in log answers: it must be log|x|, log|sec x + tan x|, etc.
In substitution, not changing the differential (dx → dt) or, for definite integrals, not changing the limits when you stay in the new variable.
Choosing the wrong “first function” in integration by parts — follow ILATE.
Trying partial fractions on an improper rational function without first dividing.
Mixing up the special-function formulas: x²−a² gives log|(x−a)/(x+a)|, while a²−x² gives log|(a+x)/(a−x)|.
Not exploiting odd/even symmetry (P₇) — an odd integrand over [−a, a] is simply 0.

Practice MCQs & Assertion–Reason

1. ∫e^x(sin x + cos x) dx equals

(a) e^x cos x + C (b) e^x sin x + C (c) e^x tan x + C (d) e^x(sin x − cos x) + C

2. ∫sec²x dx equals

(a) sec x + C (b) cot x + C (c) tan x + C (d) −cot x + C

3. ∫dx/(x² + a²) equals

(a) (1/a)tan⁻¹(x/a) + C (b) a tan⁻¹(x/a) + C (c) tan⁻¹(ax) + C (d) (1/a)sin⁻¹(x/a) + C

4. ∫tan x dx equals

5. ∫₀^π/2cos²x dx equals

(a) π/2 (b) π/4 (c) π (d) 1

6. The value of ∫_−a^af(x) dx when f is odd is

(a) 2∫₀^af(x)dx (b) 0 (c) a (d) f(a) − f(−a)

7. ∫e^x[f(x) + f′(x)] dx equals

(a) e^xf′(x) + C (b) e^xf(x) + C (c) e^x + f(x) + C (d) f(x) + C

8. ∫1/x dx equals

(a) −1/x² + C (b) log|x| + C (c) x log x + C (d) 1 + C

9. By the Second Fundamental Theorem, ∫_a^bf(x) dx equals

(a) F(a) − F(b) (b) F(b) − F(a) (c) F(b) + F(a) (d) F′(b) − F′(a)

10. ∫dx/√(a² − x²) equals

(a) log|x + √(a²−x²)| + C (b) sin⁻¹(x/a) + C (c) (1/a)tan⁻¹(x/a) + C (d) cos⁻¹(x/a) + C

Answer key: 1-(b), 2-(c), 3-(a), 4-(b), 5-(b), 6-(b), 7-(b), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: ∫₋₁¹x³ dx = 0.

Reason: For an odd function f, ∫_−a^af(x) dx = 0.

A-R 2. Assertion: ∫sec x dx = log|sec x + tan x| + C.

Reason: It is obtained by multiplying numerator and denominator by (sec x + tan x) and substituting.

A-R 3. Assertion: To integrate a product of two functions we always use partial fractions.

Reason: Integration by parts uses the rule ∫u v dx = u∫v dx − ∫(u′∫v dx) dx.

A-R 4. Assertion: ∫e^x(1/x − 1/x²) dx = e^x/x + C.

Reason: ∫e^x[f(x) + f′(x)] dx = e^xf(x) + C, with f(x) = 1/x.

A-R 5. Assertion: ∫₀^π/2sin x/(sin x + cos x) dx = π/4.

Reason: ∫₀^af(x) dx = ∫₀^af(a − x) dx (property P₄).

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

Integration is the inverse of differentiation; ∫f(x) dx = F(x) + C where F′ = f.
Learn the standard integrals (powers, e^x, a^x, trig, inverse-trig forms) by heart.
Substitution: pick t whose derivative is in the integrand; change limits for definite integrals.
Partial fractions handle proper rational functions; reduce improper ones by division first.
Integration by parts: ∫u v dx = u∫v dx − ∫(u′∫v dx) dx; first function by ILATE; remember ∫e^x[f + f′] = e^xf.
Second Fundamental Theorem: ∫_a^bf = F(b) − F(a).
Definite-integral properties P₀–P₇ (especially P₄ and the even/odd rule P₇) often shorten the work dramatically.

How to score full marks in this chapter

Show the substitution clearly (state “Put t = …, dt = …”), always carry the constant C, and keep modulus signs in logarithms. For definite integrals, decide early whether a symmetry property (P₄ or odd/even) will save time. In integration by parts, write the ILATE choice before you start, and in “prove that” questions finish with a clear “Hence proved.” Practising the standard special-function and radical formulas until they are automatic is the single biggest scoring boost.

Frequently Asked Questions

What does Class 12 Maths Chapter 7 Integrals cover?

It covers anti-derivatives and indefinite integrals, the methods of substitution, partial fractions and integration by parts, integrals of special functions, definite integrals, the Fundamental Theorem of Calculus, evaluation by substitution, and the properties of definite integrals.

How many exercises are there in Chapter 7?

There are ten exercises (7.1 to 7.10) plus a Miscellaneous Exercise. Every numbered question of each, including the MCQs and the “prove that” items, is solved on this page.

What is the ILATE rule for integration by parts?

ILATE tells you which factor to take as the first function (the one to differentiate): Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential — in that order of priority.

Are these Class 12 Maths Chapter 7 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics Part II textbook for the 2026–27 session, with answers verified against the book’s answer key.

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