What is Class 12 Mathematics Chapter 7 (Integrals) about?

This chapter treats integration as the reverse of differentiation and builds methods to find antiderivatives: standard formulas, substitution, trigonometric identities, partial fractions and integration by parts, including special forms like integrals of e^x[f(x) + f′(x)] and square roots of quadratics. It then defines the definite integral through the fundamental theorems of calculus and uses substitution and properties such as P0–P7 to evaluate definite integrals efficiently.

Are these solutions based on the new NCERT syllabus?

Yes. The answers follow the current NCERT textbook for Class 12 Mathematics and the latest CBSE syllabus, so the question numbers and wording match the book you use in class.

How should I use these solutions to prepare for exams?

Attempt each question on your own first, then compare your working with the answer here. Note the steps or keywords you missed, and revise the chapter notes before taking a chapter-wise MCQ test to check your understanding.

NCERT Solutions for Class 12 Mathematics Chapter 7 Integrals

Q: What do these NCERT solutions for Class 12 Mathematics Chapter 7 (Integrals) cover?

All 266 questions from the latest NCERT textbook for this chapter are answered step by step, organised as Exercise 7.1, Exercise 7.2, Exercise 7.3, Exercise 7.4, Exercise 7.5, Exercise 7.6, Exercise 7.7 and Exercise 7.8.

Q: Are the NCERT solutions for Class 12 Mathematics Chapter 7 (Integrals) free?

Yes. All Circle of Learning NCERT solutions and chapter notes are free to read online, with no sign-up needed.

Exercise 7.1

Find an anti derivative (or integral) of the following functions by the method of inspection (Q1–5); find the integrals in Q6–20.

sin 2x

Answer

d/dx(cos 2x) = –2 sin 2x
⇒ anti derivative = –½ cos 2x + C

cos 3x

Answer

d/dx(sin 3x) = 3 cos 3x
⇒ anti derivative = ⅓ sin 3x + C

e^2x

Answer

d/dx(e^2x) = 2e^2x
⇒ anti derivative = ½ e^2x + C

(ax + b)²

Answer

d/dx (ax + b)³ = 3a(ax + b)²
⇒ anti derivative = (ax + b)³/(3a) + C

sin 2x – 4e^3x

Answer

Using Q1 and d/dx(e^3x) = 3e^3x
⇒ –½ cos 2x – (4/3) e^3x + C

∫(4e^3x + 1) dx

Answer

= 4·e^3x/3 + x + C = (4/3)e^3x + x + C

∫x²(1 – 1/x²) dx

Answer

x²(1 – 1/x²) = x² – 1
⇒ ∫(x² – 1) dx = x³/3 – x + C

∫(ax² + bx + c) dx

Answer

= ax³/3 + bx²/2 + cx + C

∫(2x² + e^x) dx

Answer

= 2x³/3 + e^x + C

Q10

∫(√x – 1/√x)² dx

Answer

(√x – 1/√x)² = x – 2 + 1/x
⇒ x²/2 – 2x + log|x| + C

Q11

∫(x³ + 5x² – 4)/x² dx

Answer

= ∫(x + 5 – 4x⁻²) dx
⇒ x²/2 + 5x + 4/x + C

Q12

∫(x³ + 3x + 4)/√x dx

Answer

= ∫(x^5/2 + 3x^1/2 + 4x^–1/2) dx
⇒ (2/7)x^7/2 + 2x^3/2 + 8√x + C

Q13

∫(x³ – x² + x – 1)/(x – 1) dx

Answer

x³ – x² + x – 1 = x²(x – 1) + (x – 1) = (x – 1)(x² + 1)
⇒ ∫(x² + 1) dx = x³/3 + x + C

Q14

∫(1 – x)√x dx

Answer

= ∫(x^1/2 – x^3/2) dx
⇒ (2/3)x^3/2 – (2/5)x^5/2 + C

Q15

∫√x(3x² + 2x + 3) dx

Answer

= ∫(3x^5/2 + 2x^3/2 + 3x^1/2) dx
⇒ (6/7)x^7/2 + (4/5)x^5/2 + 2x^3/2 + C

Q16

∫(2x – 3 cos x + e^x) dx

Answer

= x² – 3 sin x + e^x + C

Q17

∫(2x² – 3 sin x + 5√x) dx

Answer

= (2/3)x³ + 3 cos x + (10/3)x^3/2 + C

Q18

∫sec x (sec x + tan x) dx

Answer

= ∫(sec²x + sec x tan x) dx
⇒ tan x + sec x + C

Q19

∫sec²x/cosec²x dx

Answer

sec²x/cosec²x = tan²x = sec²x – 1
⇒ tan x – x + C

Q20

∫(2 – 3 sin x)/cos²x dx

Answer

= ∫(2 sec²x – 3 sec x tan x) dx
⇒ 2 tan x – 3 sec x + C

Q21

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

Answer

∫(x^1/2 + x^–1/2) dx = (2/3)x^3/2 + 2x^1/2 + C
⇒ (C)

Q22

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

Answer

f(x) = ∫(4x³ – 3x⁻⁴) dx = x⁴ + 1/x³ + C
⇒ f(2) = 16 + 1/8 + C = 0
⇒ C = –129/8
⇒ f(x) = x⁴ + 1/x³ – 129/8
⇒ (A)

Exercise 7.2

Integrate the functions in Exercises 1 to 37.

2x/(1 + x²)

Answer

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

(log x)²/x

Answer

Put log x = t, dx/x = dt
⇒ ∫t² dt = t³/3
⇒ (log x)³/3 + C

1/(x + x log x)

Answer

= 1/[x(1 + log x)]; put 1 + log x = t, dx/x = dt
⇒ ∫dt/t
⇒ log|1 + log x| + C

sin x sin (cos x)

Answer

Put cos x = t, –sin x dx = dt
⇒ –∫sin t dt = cos t
⇒ cos (cos x) + C

sin (ax + b) cos (ax + b)

Answer

= ½ sin 2(ax + b)
⇒ ½ · [–cos 2(ax + b)]/(2a)
⇒ –cos 2(ax + b)/(4a) + C

√(ax + b)

Answer

Put ax + b = t, a dx = dt
⇒ (1/a)∫t^1/2 dt
⇒ (2/(3a))(ax + b)^3/2 + C

x√(x + 2)

Answer

Put x + 2 = t, dx = dt
⇒ ∫(t – 2)t^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

x√(1 + 2x²)

Answer

Put 1 + 2x² = t, 4x dx = dt
⇒ ¼∫t^1/2 dt = (1/6)t^3/2
⇒ (1/6)(1 + 2x²)^3/2 + C

(4x + 2)√(x² + x + 1)

Answer

Put x² + x + 1 = t, (2x + 1) dx = dt
⇒ 2∫t^1/2 dt = (4/3)t^3/2
⇒ (4/3)(x² + x + 1)^3/2 + C

Q10

1/(x – √x)

Answer

= 1/[√x(√x – 1)]; put √x – 1 = t, dx/(2√x) = dt
⇒ 2∫dt/t
⇒ 2 log|√x – 1| + C

Q11

x/√(x + 4), x > 0

Answer

Put x + 4 = t, dx = dt
⇒ ∫(t – 4)/√t dt = (2/3)t^3/2 – 8t^1/2
⇒ (2/3)√(x + 4)(x – 8) + C

Q12

(x³ – 1)^1/3 x⁵

Answer

Put x³ – 1 = t, 3x² dx = dt, x³ = t + 1
⇒ ⅓∫(t + 1)t^1/3 dt = ⅓[(3/7)t^7/3 + (3/4)t^4/3]
⇒ (1/7)(x³ – 1)^7/3 + ¼(x³ – 1)^4/3 + C

Q13

x²/(2 + 3x³)³

Answer

Put 2 + 3x³ = t, 9x² dx = dt
⇒ (1/9)∫t⁻³ dt = –1/(18t²)
⇒ –1/[18(2 + 3x³)²] + C

Q14

1/[x(log x)^m], x > 0, m ≠ 1

Answer

Put log x = t, dx/x = dt
⇒ ∫t^–m dt = t^1–m/(1 – m)
⇒ (log x)^1–m/(1 – m) + C

Q15

x/(9 – 4x²)

Answer

Put 9 – 4x² = t, –8x dx = dt
⇒ –⅛∫dt/t
⇒ –⅛ log|9 – 4x²| + C

Q16

e^{2x + 3}

Answer

Put 2x + 3 = t → ½∫e^t dt → ½e^{2x + 3} + C

Q17

x/e^x²

Answer

= xe^–x²; put x² = t, 2x dx = dt
⇒ ½∫e^–t dt
⇒ –½e^–x² + C

Q18

e^tan⁻¹x/(1 + x²)

Answer

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

Q19

(e^2x – 1)/(e^2x + 1)

Answer

Divide by e^x: (e^x – e^–x)/(e^x + e^–x); put e^x + e^–x = t
⇒ ∫dt/t
⇒ log(e^x + e^–x) + C

Q20

(e^2x – e^–2x)/(e^2x + e^–2x)

Answer

Put e^2x + e^–2x = t, 2(e^2x – e^–2x) dx = dt
⇒ ½∫dt/t
⇒ ½ log(e^2x + e^–2x) + C

Q21

tan²(2x – 3)

Answer

= sec²(2x – 3) – 1
⇒ ½ tan(2x – 3) – x + C

Q22

sec²(7 – 4x)

Answer

Put 7 – 4x = t, –4 dx = dt
⇒ –¼∫sec²t dt
⇒ –¼ tan(7 – 4x) + C

Q23

sin⁻¹x/√(1 – x²)

Answer

Put sin⁻¹x = t, dx/√(1 – x²) = dt
⇒ ∫t dt
⇒ ½(sin⁻¹x)² + C

Q24

(2 cos x – 3 sin x)/(6 cos x + 4 sin x)

Answer

Put 6 cos x + 4 sin x = t
⇒ (4 cos x – 6 sin x) dx = dt, i.e. (2 cos x – 3 sin x) dx = dt/2
⇒ ½∫dt/t
⇒ ½ log|6 cos x + 4 sin x| + C

Q25

1/[cos²x (1 – tan x)²]

Answer

Put 1 – tan x = t, –sec²x dx = dt
⇒ –∫t⁻² dt = 1/t
⇒ 1/(1 – tan x) + C

Q26

cos √x/√x

Answer

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

Q27

√(sin 2x) cos 2x

Answer

Put sin 2x = t, 2 cos 2x dx = dt
⇒ ½∫t^1/2 dt
⇒ ⅓(sin 2x)^3/2 + C

Q28

cos x/√(1 + sin x)

Answer

Put 1 + sin x = t, cos x dx = dt
⇒ ∫t^–1/2 dt
⇒ 2√(1 + sin x) + C

Q29

cot x log sin x

Answer

Put log sin x = t, cot x dx = dt
⇒ ∫t dt
⇒ ½(log sin x)² + C

Q30

sin x/(1 + cos x)

Answer

Put 1 + cos x = t, –sin x dx = dt
⇒ –∫dt/t
⇒ –log|1 + cos x| + C

Q31

sin x/(1 + cos x)²

Answer

Put 1 + cos x = t, –sin x dx = dt
⇒ –∫t⁻² dt = 1/t
⇒ 1/(1 + cos x) + C

Q32

1/(1 + cot x)

Answer

= sin x/(sin x + cos x) = ½[1 – (cos x – sin x)/(sin x + cos x)]
⇒ x/2 – ½ log|sin x + cos x| + C

Q33

1/(1 – tan x)

Answer

= cos x/(cos x – sin x) = ½[1 + (cos x + sin x)/(cos x – sin x)]; numerator = –d(cos x – sin x)
⇒ x/2 – ½ log|cos x – sin x| + C

Q34

√(tan x)/(sin x cos x)

Answer

= √(tan x) sec²x/tan x = sec²x/√(tan x); put tan x = t
⇒ ∫t^–1/2 dt
⇒ 2√(tan x) + C

Q35

(1 + log x)²/x

Answer

Put 1 + log x = t, dx/x = dt
⇒ ∫t² dt
⇒ (1 + log x)³/3 + C

Q36

(x + 1)(x + log x)²/x

Answer

= (1 + 1/x)(x + log x)²; put x + log x = t, (1 + 1/x) dx = dt
⇒ ∫t² dt
⇒ (x + log x)³/3 + C

Q37

x³ sin(tan⁻¹x⁴)/(1 + x⁸)

Answer

Put x⁴ = t, 4x³ dx = dt
⇒ ¼∫sin(tan⁻¹t)/(1 + t²) dt; put tan⁻¹t = u
⇒ ¼∫sin u du = –¼ cos u
⇒ –¼ cos(tan⁻¹x⁴) + C

Q38

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

Answer

Put x¹⁰ + 10^x = t
⇒ (10x⁹ + 10^x log 10) dx = dt
⇒ ∫dt/t = log(10^x + x¹⁰) + C
⇒ (D)

Q39

∫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

Answer

1/(sin²x cos²x) = (sin²x + cos²x)/(sin²x cos²x) = sec²x + cosec²x
⇒ tan x – cot x + C
⇒ (B)

Exercise 7.3

Find the integrals of the functions in Exercises 1 to 22.

sin²(2x + 5)

Answer

= [1 – cos(4x + 10)]/2
⇒ x/2 – sin(4x + 10)/8 + C

sin 3x cos 4x

Answer

= ½[sin 7x + sin(–x)] = ½(sin 7x – sin x)
⇒ –cos 7x/14 + cos x/2 + C

cos 2x cos 4x cos 6x

Answer

cos 2x cos 4x = ½(cos 6x + cos 2x)
⇒ product = ½(cos²6x + cos 2x cos 6x) = ¼(1 + cos 12x + cos 8x + cos 4x)
⇒ ¼[x + sin 12x/12 + sin 8x/8 + sin 4x/4] + C

sin³(2x + 1)

Answer

= sin(2x + 1)[1 – cos²(2x + 1)]; put cos(2x + 1) = t, –2 sin(2x + 1) dx = dt
⇒ –½∫(1 – t²) dt
⇒ –½ cos(2x + 1) + (1/6)cos³(2x + 1) + C

sin³x cos³x

Answer

= sin³x(1 – sin²x) cos x; put sin x = t
⇒ ∫(t³ – t⁵) dt
⇒ sin⁴x/4 – sin⁶x/6 + C

sin x sin 2x sin 3x

Answer

sin x sin 3x = ½(cos 2x – cos 4x)
⇒ ½(sin 2x cos 2x – sin 2x cos 4x) = ¼(sin 4x – sin 6x + sin 2x)
⇒ –cos 2x/8 – cos 4x/16 + cos 6x/24 + C

sin 4x sin 8x

Answer

= ½(cos 4x – cos 12x)
⇒ sin 4x/8 – sin 12x/24 + C

(1 – cos x)/(1 + cos x)

Answer

= 2 sin²(x/2)/2 cos²(x/2) = tan²(x/2) = sec²(x/2) – 1
⇒ 2 tan(x/2) – x + C

cos x/(1 + cos x)

Answer

= 1 – 1/(1 + cos x) = 1 – ½sec²(x/2)
⇒ x – tan(x/2) + C

Q10

sin⁴x

Answer

= ¼(1 – cos 2x)² = ¼[1 – 2 cos 2x + (1 + cos 4x)/2] = 3/8 – ½cos 2x + ⅛cos 4x
⇒ 3x/8 – sin 2x/4 + sin 4x/32 + C

Q11

cos⁴2x

Answer

= ¼(1 + cos 4x)² = 3/8 + ½cos 4x + ⅛cos 8x
⇒ 3x/8 + sin 4x/8 + sin 8x/64 + C

Q12

sin²x/(1 + cos x)

Answer

= (1 – cos x)(1 + cos x)/(1 + cos x) = 1 – cos x
⇒ x – sin x + C

Q13

(cos 2x – cos 2α)/(cos x – cos α)

Answer

= 2(cos²x – cos²α)/(cos x – cos α) = 2(cos x + cos α)
⇒ 2 sin x + 2x cos α + C

Q14

(cos x – sin x)/(1 + sin 2x)

Answer

1 + sin 2x = (sin x + cos x)²; put sin x + cos x = t, (cos x – sin x) dx = dt
⇒ ∫t⁻² dt
⇒ –1/(sin x + cos x) + C

Q15

tan³2x sec 2x

Answer

= (sec²2x – 1) sec 2x tan 2x; put sec 2x = t, 2 sec 2x tan 2x dx = dt
⇒ ½∫(t² – 1) dt
⇒ sec³2x/6 – sec 2x/2 + C

Q16

tan⁴x

Answer

= tan²x(sec²x – 1) = tan²x sec²x – sec²x + 1
⇒ tan³x/3 – tan x + x + C

Q17

(sin³x + cos³x)/(sin²x cos²x)

Answer

= sin x/cos²x + cos x/sin²x = sec x tan x + cosec x cot x
⇒ sec x – cosec x + C

Q18

(cos 2x + 2 sin²x)/cos²x

Answer

cos 2x + 2 sin²x = 1
⇒ ∫sec²x dx = tan x + C

Q19

1/(sin x cos³x)

Answer

= sec⁴x/tan x = (1 + tan²x) sec²x/tan x; put tan x = t
⇒ ∫(1/t + t) dt
⇒ log|tan x| + ½tan²x + C

Q20

cos 2x/(cos x + sin x)²

Answer

= (cos x – sin x)(cos x + sin x)/(cos x + sin x)² = (cos x – sin x)/(cos x + sin x)
⇒ log|cos x + sin x| + C

Q21

sin⁻¹(cos x)

Answer

sin⁻¹(cos x) = sin⁻¹[sin(π/2 – x)] = π/2 – x (0 ≤ x ≤ π)
⇒ πx/2 – x²/2 + C

Q22

1/[cos(x – a) cos(x – b)]

Answer

Q23

∫(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

Answer

= ∫(sec²x – cosec²x) dx = tan x + cot x + C
⇒ (A)

Q24

∫e^x(1 + x)/cos²(e^xx) dx equals (A) –cot(ex^x) + C (B) tan(xe^x) + C (C) tan(e^x) + C (D) cot(e^x) + C

Answer

Put xe^x = t, e^x(1 + x) dx = dt
⇒ ∫sec²t dt = tan(xe^x) + C
⇒ (B)

Exercise 7.4

Integrate the functions in Exercises 1 to 23.

3x²/(x⁶ + 1)

Answer

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

1/√(1 + 4x²)

Answer

Put 2x = t
⇒ ½∫dt/√(1 + t²) = ½ log|t + √(1 + t²)|
⇒ ½ log|2x + √(1 + 4x²)| + C

1/√[(2 – x)² + 1]

Answer

Put 2 – x = t, dx = –dt
⇒ –∫dt/√(t² + 1)
⇒ –log|2 – x + √(x² – 4x + 5)| + C

1/√(9 – 25x²)

Answer

Put 5x = t
⇒ (1/5)∫dt/√(9 – t²)
⇒ (1/5) sin⁻¹(5x/3) + C

3x/(1 + 2x⁴)

Answer

Put √2 x² = t, 2√2 x dx = dt
⇒ (3/(2√2))∫dt/(1 + t²)
⇒ (3/(2√2)) tan⁻¹(√2 x²) + C

x²/(1 – x⁶)

Answer

Put x³ = t, 3x² dx = dt
⇒ ⅓∫dt/(1 – t²) = (1/6) log|(1 + t)/(1 – t)|
⇒ (1/6) log|(1 + x³)/(1 – x³)| + C

(x – 1)/√(x² – 1)

Answer

= x/√(x² – 1) – 1/√(x² – 1)
⇒ √(x² – 1) – log|x + √(x² – 1)| + C

x²/√(x⁶ + a⁶)

Answer

Put x³ = t, 3x² dx = dt
⇒ ⅓∫dt/√(t² + a⁶)
⇒ ⅓ log|x³ + √(x⁶ + a⁶)| + C

sec²x/√(tan²x + 4)

Answer

Put tan x = t
⇒ ∫dt/√(t² + 4)
⇒ log|tan x + √(tan²x + 4)| + C

Q10

1/√(x² + 2x + 2)

Answer

= 1/√[(x + 1)² + 1]
⇒ log|x + 1 + √(x² + 2x + 2)| + C

Q11

1/(9x² + 6x + 5)

Answer

= 1/[(3x + 1)² + 2²]; put 3x + 1 = t
⇒ ⅓ · ½ tan⁻¹(t/2)
⇒ (1/6) tan⁻¹[(3x + 1)/2] + C

Q12

1/√(7 – 6x – x²)

Answer

7 – 6x – x² = 16 – (x + 3)²
⇒ sin⁻¹[(x + 3)/4] + C

Q13

1/√[(x – 1)(x – 2)]

Answer

(x – 1)(x – 2) = x² – 3x + 2 = (x – 3/2)² – (½)²
⇒ log|x – 3/2 + √(x² – 3x + 2)| + C

Q14

1/√(8 + 3x – x²)

Answer

8 + 3x – x² = 41/4 – (x – 3/2)²
⇒ sin⁻¹[(2x – 3)/√41] + C

Q15

1/√[(x – a)(x – b)]

Answer

(x – a)(x – b) = [x – (a + b)/2]² – [(a – b)/2]²
⇒ log|x – (a + b)/2 + √[(x – a)(x – b)]| + C

Q16

(4x + 1)/√(2x² + x – 3)

Answer

4x + 1 = d/dx(2x² + x – 3); put 2x² + x – 3 = t
⇒ ∫t^–1/2 dt
⇒ 2√(2x² + x – 3) + C

Q17

(x + 2)/√(x² – 1)

Answer

= x/√(x² – 1) + 2/√(x² – 1)
⇒ √(x² – 1) + 2 log|x + √(x² – 1)| + C

Q18

(5x – 2)/(1 + 2x + 3x²)

Answer

5x – 2 = (5/6)(6x + 2) – 11/3
⇒ (5/6) log|3x² + 2x + 1| – (11/3)∫dx/(3x² + 2x + 1); 3x² + 2x + 1 = 3[(x + ⅓)² + 2/9]
⇒ (5/6) log|3x² + 2x + 1| – (11/(3√2)) tan⁻¹[(3x + 1)/√2] + C

Q19

(6x + 7)/√[(x – 5)(x – 4)]

Answer

(x – 5)(x – 4) = x² – 9x + 20; 6x + 7 = 3(2x – 9) + 34
⇒ 3 · 2√(x² – 9x + 20) + 34∫dx/√[(x – 9/2)² – ¼]
⇒ 6√(x² – 9x + 20) + 34 log|x – 9/2 + √(x² – 9x + 20)| + C

Q20

(x + 2)/√(4x – x²)

Answer

x + 2 = –½(4 – 2x) + 4; 4x – x² = 4 – (x – 2)²
⇒ –√(4x – x²) + 4 sin⁻¹[(x – 2)/2] + C

Q21

(x + 2)/√(x² + 2x + 3)

Answer

x + 2 = ½(2x + 2) + 1; x² + 2x + 3 = (x + 1)² + 2
⇒ √(x² + 2x + 3) + log|x + 1 + √(x² + 2x + 3)| + C

Q22

(x + 3)/(x² – 2x – 5)

Answer

x + 3 = ½(2x – 2) + 4; x² – 2x – 5 = (x – 1)² – 6
⇒ ½ log|x² – 2x – 5| + (2/√6) log|(x – 1 – √6)/(x – 1 + √6)| + C

Q23

(5x + 3)/√(x² + 4x + 10)

Answer

5x + 3 = (5/2)(2x + 4) – 7; x² + 4x + 10 = (x + 2)² + 6
⇒ 5√(x² + 4x + 10) – 7 log|x + 2 + √(x² + 4x + 10)| + C

Q24

∫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

Answer

= ∫dx/[(x + 1)² + 1] = tan⁻¹(x + 1) + C
⇒ (B)

Q25

∫dx/√(9x – 4x²) equals (A) (1/9) sin⁻¹[(9x – 8)/8] + C (B) ½ sin⁻¹[(8x – 9)/9] + C (C) ⅓ sin⁻¹[(9x – 8)/8] + C (D) ½ sin⁻¹[(9x – 8)/9] + C

Answer

9x – 4x² = 4[(9/8)² – (x – 9/8)²]
⇒ ½ sin⁻¹[(x – 9/8)/(9/8)] = ½ sin⁻¹[(8x – 9)/9] + C
⇒ (B)

Exercise 7.5

Integrate the rational functions in Exercises 1 to 21.

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

Answer

x/[(x + 1)(x + 2)] = –1/(x + 1) + 2/(x + 2)
⇒ –log|x + 1| + 2 log|x + 2| = log[(x + 2)²/|x + 1|] + C

1/(x² – 9)

Answer

= (1/6)[1/(x – 3) – 1/(x + 3)]
⇒ (1/6) log|(x – 3)/(x + 3)| + C

(3x – 1)/[(x – 1)(x – 2)(x – 3)]

Answer

= A/(x – 1) + B/(x – 2) + C/(x – 3); x = 1
⇒ A = 1, x = 2
⇒ B = –5, x = 3
⇒ C = 4
⇒ log|x – 1| – 5 log|x – 2| + 4 log|x – 3| + C

x/[(x – 1)(x – 2)(x – 3)]

Answer

A = ½, B = –2, C = 3/2 (by putting x = 1, 2, 3)
⇒ ½ log|x – 1| – 2 log|x – 2| + (3/2) log|x – 3| + C

2x/(x² + 3x + 2)

Answer

= 2x/[(x + 1)(x + 2)] = –2/(x + 1) + 4/(x + 2)
⇒ 4 log|x + 2| – 2 log|x + 1| + C

(1 – x²)/[x(1 – 2x)]

Answer

Improper: (1 – x²)/(x – 2x²) = ½ + (x/2 – 1)/[x(2x – 1)] = ½ + 1/x – (3/2)/(2x – 1)
⇒ x/2 + log|x| – (3/4) log|1 – 2x| + C

x/[(x² + 1)(x – 1)]

Answer

= A/(x – 1) + (Bx + C)/(x² + 1); A = ½, B = –½, C = ½
⇒ ½ log|x – 1| – ¼ log(x² + 1) + ½ tan⁻¹x + C

x/[(x – 1)²(x + 2)]

Answer

= A/(x – 1) + B/(x – 1)² + C/(x + 2); B = ⅓, C = –2/9, A = 2/9
⇒ (2/9) log|(x – 1)/(x + 2)| – 1/[3(x – 1)] + C

(3x + 5)/(x³ – x² – x + 1)

Answer

x³ – x² – x + 1 = (x – 1)²(x + 1); = A/(x – 1) + B/(x – 1)² + C/(x + 1) with B = 4, C = ½, A = –½
⇒ ½ log|(x + 1)/(x – 1)| – 4/(x – 1) + C

Q10

(2x – 3)/[(x² – 1)(2x + 3)]

Answer

= A/(x – 1) + B/(x + 1) + C/(2x + 3); A = –1/10, B = 5/2, C = –24/5
⇒ –(1/10) log|x – 1| + (5/2) log|x + 1| – (12/5) log|2x + 3| + C

Q11

5x/[(x + 1)(x² – 4)]

Answer

= 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

Q12

(x³ + x + 1)/(x² – 1)

Answer

= x + (2x + 1)/(x² – 1) = x + (3/2)/(x – 1) + (1/2)/(x + 1)
⇒ x²/2 + (3/2) log|x – 1| + ½ log|x + 1| + C

Q13

2/[(1 – x)(1 + x²)]

Answer

= A/(1 – x) + (Bx + C)/(1 + x²); A = 1, B = 1, C = 1
⇒ –log|1 – x| + ½ log(1 + x²) + tan⁻¹x + C

Q14

(3x – 1)/(x + 2)²

Answer

3x – 1 = 3(x + 2) – 7
⇒ 3/(x + 2) – 7/(x + 2)²
⇒ 3 log|x + 2| + 7/(x + 2) + C

Q15

1/(x⁴ – 1)

Answer

= 1/[(x – 1)(x + 1)(x² + 1)] = ¼/(x – 1) – ¼/(x + 1) – ½/(x² + 1)
⇒ ¼ log|(x – 1)/(x + 1)| – ½ tan⁻¹x + C

Q16

1/[x(xⁿ + 1)] [Hint: multiply numerator and denominator by x^n–1 and put xⁿ = t]

Answer

= x^n–1/[xⁿ(xⁿ + 1)]; xⁿ = t, nx^n–1 dx = dt
⇒ (1/n)∫dt/[t(t + 1)] = (1/n) log|t/(t + 1)|
⇒ (1/n) log|xⁿ/(xⁿ + 1)| + C

Q17

cos x/[(1 – sin x)(2 – sin x)] [Hint: Put sin x = t]

Answer

sin x = t
⇒ ∫dt/[(1 – t)(2 – t)] = ∫[1/(1 – t) – 1/(2 – t)] dt = log|(2 – t)/(1 – t)|
⇒ log|(2 – sin x)/(1 – sin x)| + C

Q18

(x² + 1)(x² + 2)/[(x² + 3)(x² + 4)]

Answer

With y = x²: (y + 1)(y + 2)/[(y + 3)(y + 4)] = 1 – (4y + 10)/[(y + 3)(y + 4)] = 1 + 2/(y + 3) – 6/(y + 4)
⇒ x + (2/√3) tan⁻¹(x/√3) – 3 tan⁻¹(x/2) + C

Q19

2x/[(x² + 1)(x² + 3)]

Answer

Put x² = t
⇒ ∫dt/[(t + 1)(t + 3)] = ½∫[1/(t + 1) – 1/(t + 3)] dt
⇒ ½ log[(x² + 1)/(x² + 3)] + C

Q20

1/[x(x⁴ – 1)]

Answer

= x³/[x⁴(x⁴ – 1)]; put x⁴ = t
⇒ ¼∫dt/[t(t – 1)] = ¼ log|(t – 1)/t|
⇒ ¼ log|(x⁴ – 1)/x⁴| + C

Q21

1/(e^x – 1) [Hint: Put e^x = t]

Answer

e^x = t, dx = dt/t
⇒ ∫dt/[t(t – 1)] = log|(t – 1)/t|
⇒ log|(e^x – 1)/e^x| + C

Q22

∫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

Answer

x/[(x – 1)(x – 2)] = –1/(x – 1) + 2/(x – 2)
⇒ 2 log|x – 2| – log|x – 1| = log|(x – 2)²/(x – 1)| + C
⇒ (B)

Q23

∫dx/[x(x² + 1)] equals (A) log|x| – ½ log(x² + 1) + C (B) log|x| + ½ log(x² + 1) + C (C) –log|x| + ½ log(x² + 1) + C (D) ½ log|x| + log(x² + 1) + C

Answer

1/[x(x² + 1)] = 1/x – x/(x² + 1)
⇒ log|x| – ½ log(x² + 1) + C
⇒ (A)

Exercise 7.6

Integrate the functions in Exercises 1 to 22.

x sin x

Answer

By parts (u = x): –x cos x + ∫cos x dx
⇒ –x cos x + sin x + C

x sin 3x

Answer

u = x, v = –cos 3x/3
⇒ –x cos 3x/3 + ∫cos 3x/3 dx
⇒ –(x cos 3x)/3 + (sin 3x)/9 + C

x²e^x

Answer

x²e^x – ∫2xe^x dx = x²e^x – 2(xe^x – e^x)
⇒ e^x(x² – 2x + 2) + C

x log x

Answer

u = log x, v = x²/2
⇒ (x²/2) log x – ∫(x/2) dx
⇒ (x²/2) log x – x²/4 + C

x log 2x

Answer

u = log 2x (du = dx/x), v = x²/2
⇒ (x²/2) log 2x – x²/4 + C

x² log x

Answer

u = log x, v = x³/3
⇒ (x³/3) log x – ∫(x²/3) dx
⇒ (x³/3) log x – x³/9 + C

x sin⁻¹x

Answer

(x²/2) sin⁻¹x – ½∫x²/√(1 – x²) dx; x²/√(1 – x²) = 1/√(1 – x²) – √(1 – x²)
⇒ ∫ = ½ sin⁻¹x – ½x√(1 – x²)
⇒ [(2x² – 1)/4] sin⁻¹x + (x/4)√(1 – x²) + C

x tan⁻¹x

Answer

(x²/2) tan⁻¹x – ½∫x²/(1 + x²) dx = (x²/2) tan⁻¹x – ½(x – tan⁻¹x)
⇒ [(x² + 1)/2] tan⁻¹x – x/2 + C

x cos⁻¹x

Answer

(x²/2) cos⁻¹x + ½∫x²/√(1 – x²) dx = (x²/2) cos⁻¹x + ¼ sin⁻¹x – (x/4)√(1 – x²); using sin⁻¹x = π/2 – cos⁻¹x
⇒ [(2x² – 1)/4] cos⁻¹x – (x/4)√(1 – x²) + C

Q10

(sin⁻¹x)²

Answer

Take 1 as second function: x(sin⁻¹x)² – 2∫x sin⁻¹x/√(1 – x²) dx; ∫x sin⁻¹x/√(1 – x²) dx = –√(1 – x²) sin⁻¹x + x
⇒ x(sin⁻¹x)² + 2√(1 – x²) sin⁻¹x – 2x + C

Q11

x cos⁻¹x/√(1 – x²)

Answer

u = cos⁻¹x, dv = x dx/√(1 – x²)
⇒ v = –√(1 – x²)
⇒ –√(1 – x²) cos⁻¹x – ∫dx
⇒ –[√(1 – x²) cos⁻¹x + x] + C

Q12

x sec²x

Answer

x tan x – ∫tan x dx
⇒ x tan x + log|cos x| + C

Q13

tan⁻¹x

Answer

x tan⁻¹x – ∫x/(1 + x²) dx
⇒ x tan⁻¹x – ½ log(1 + x²) + C

Q14

x(log x)²

Answer

(x²/2)(log x)² – ∫x log x dx
⇒ (x²/2)(log x)² – (x²/2) log x + x²/4 + C

Q15

(x² + 1) log x

Answer

(x³/3 + x) log x – ∫(x²/3 + 1) dx
⇒ (x³/3 + x) log x – x³/9 – x + C

Q16

e^x(sin x + cos x)

Answer

Form e^x[f(x) + f′(x)] with f(x) = sin x
⇒ e^x sin x + C

Q17

xe^x/(1 + x)²

Answer

= e^x[1/(1 + x) – 1/(1 + x)²], f(x) = 1/(1 + x)
⇒ e^x/(1 + x) + C

Q18

e^x(1 + sin x)/(1 + cos x)

Answer

= e^x[½sec²(x/2) + tan(x/2)], f(x) = tan(x/2)
⇒ e^x tan(x/2) + C

Q19

e^x(1/x – 1/x²)

Answer

f(x) = 1/x, f′(x) = –1/x² → e^x/x + C

Q20

(x – 3)e^x/(x – 1)³

Answer

= e^x[1/(x – 1)² – 2/(x – 1)³], f(x) = 1/(x – 1)²
⇒ e^x/(x – 1)² + C

Q21

e^2x sin x

Answer

I = –e^2x cos x + 2∫e^2x cos x dx = –e^2x cos x + 2[e^2x sin x – 2I]
⇒ 5I = e^2x(2 sin x – cos x)
⇒ I = e^2x(2 sin x – cos x)/5 + C

Q22

sin⁻¹[2x/(1 + x²)]

Answer

Put x = tan θ
⇒ sin⁻¹(sin 2θ) = 2θ = 2 tan⁻¹x
⇒ 2∫tan⁻¹x dx = 2x tan⁻¹x – log(1 + x²) + C

Q23

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

Answer

Put x³ = t, 3x² dx = dt
⇒ ⅓∫e^t dt = ⅓e^x³ + C
⇒ (A)

Q24

∫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

Answer

= ∫e^x(sec x + sec x tan x) dx, f(x) = sec x
⇒ e^x sec x + C
⇒ (B)

Exercise 7.7

Integrate the functions in Exercises 1 to 9.

√(4 – x²)

Answer

Using ∫√(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

√(1 – 4x²)

Answer

Put 2x = t
⇒ ½∫√(1 – t²) dt
⇒ (x/2)√(1 – 4x²) + ¼ sin⁻¹(2x) + C

√(x² + 4x + 6)

Answer

= √[(x + 2)² + 2]
⇒ [(x + 2)/2]√(x² + 4x + 6) + log|x + 2 + √(x² + 4x + 6)| + C

√(x² + 4x + 1)

Answer

= √[(x + 2)² – 3]
⇒ [(x + 2)/2]√(x² + 4x + 1) – (3/2) log|x + 2 + √(x² + 4x + 1)| + C

√(1 – 4x – x²)

Answer

= √[5 – (x + 2)²]
⇒ [(x + 2)/2]√(1 – 4x – x²) + (5/2) sin⁻¹[(x + 2)/√5] + C

√(x² + 4x – 5)

Answer

= √[(x + 2)² – 9]
⇒ [(x + 2)/2]√(x² + 4x – 5) – (9/2) log|x + 2 + √(x² + 4x – 5)| + C

√(1 + 3x – x²)

Answer

= √[13/4 – (x – 3/2)²]
⇒ [(2x – 3)/4]√(1 + 3x – x²) + (13/8) sin⁻¹[(2x – 3)/√13] + C

√(x² + 3x)

Answer

= √[(x + 3/2)² – 9/4]
⇒ [(2x + 3)/4]√(x² + 3x) – (9/8) log|x + 3/2 + √(x² + 3x)| + C

√(1 + x²/9)

Answer

= ⅓√(x² + 9)
⇒ ⅓[(x/2)√(x² + 9) + (9/2) log|x + √(x² + 9)|] = (x/6)√(x² + 9) + (3/2) log|x + √(x² + 9)| + C

Q10

∫√(1 + x²) dx is equal to (A) (x/2)√(1 + x²) + ½ log|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) + ½x² log|x + √(1 + x²)| + C

Answer

Standard result with a = 1
⇒ (x/2)√(1 + x²) + ½ log|x + √(1 + x²)| + C
⇒ (A)

Q11

∫√(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

Answer

x² – 8x + 7 = (x – 4)² – 3²
⇒ ½(x – 4)√(x² – 8x + 7) – (9/2) log|x – 4 + √(x² – 8x + 7)| + C
⇒ (D)

Exercise 7.8

Evaluate the definite integrals in Exercises 1 to 20.

∫_–1¹ (x + 1) dx

Answer

[x²/2 + x]_–1¹ = (½ + 1) – (½ – 1) = 2

∫₂³ (1/x) dx

Answer

[log x]₂³ = log 3 – log 2 = log(3/2)

∫₁² (4x³ – 5x² + 6x + 9) dx

Answer

[x⁴ – 5x³/3 + 3x² + 9x]₁² = (16 – 40/3 + 12 + 18) – (1 – 5/3 + 3 + 9) = 98/3 – 34/3 = 64/3

∫₀^π/4 sin 2x dx

Answer

[–cos 2x/2]₀^π/4 = –½(0 – 1) = ½

∫₀^π/2 cos 2x dx

Answer

[sin 2x/2]₀^π/2 = ½(sin π – 0) = 0

∫₄⁵ e^x dx

Answer

[e^x]₄⁵ = e⁵ – e⁴ = e⁴(e – 1)

∫₀^π/4 tan x dx

Answer

[log|sec x|]₀^π/4 = log √2 – 0 = ½ log 2

∫_π/6^π/4 cosec x dx

Answer

[log|cosec x – cot x|]_π/6^π/4 = log(√2 – 1) – log(2 – √3) = log[(√2 – 1)/(2 – √3)]

∫₀¹ dx/√(1 – x²)

Answer

[sin⁻¹x]₀¹ = π/2

Q10

∫₀¹ dx/(1 + x²)

Answer

[tan⁻¹x]₀¹ = π/4

Q11

∫₂³ dx/(x² – 1)

Answer

½[log|(x – 1)/(x + 1)|]₂³ = ½[log(½) – log(⅓)] = ½ log(3/2)

Q12

∫₀^π/2 cos²x dx

Answer

= ½∫₀^π/2(1 + cos 2x) dx = ½[x + sin 2x/2]₀^π/2 = π/4

Q13

∫₂³ x dx/(x² + 1)

Answer

½[log(x² + 1)]₂³ = ½(log 10 – log 5) = ½ log 2

Q14

∫₀¹ (2x + 3)/(5x² + 1) dx

Answer

= (1/5)∫10x/(5x² + 1) dx + 3∫dx/(5x² + 1)
⇒ [(1/5) log(5x² + 1) + (3/√5) tan⁻¹(√5 x)]₀¹ = (1/5) log 6 + (3/√5) tan⁻¹√5

Q15

∫₀¹ xe^x² dx

Answer

Put x² = t → ½∫₀¹ e^t dt = ½(e – 1)

Q16

∫₁² 5x²/(x² + 4x + 3) dx

Answer

5x²/(x² + 4x + 3) = 5 – 5(4x + 3)/[(x + 1)(x + 3)] = 5 + (5/2)/(x + 1) – (45/2)/(x + 3)
⇒ [5x + (5/2) log(x + 1) – (45/2) log(x + 3)]₁² = 5 – (5/2)[9 log(5/4) – log(3/2)]

Q17

∫₀^π/4 (2 sec²x + x³ + 2) dx

Answer

[2 tan x + x⁴/4 + 2x]₀^π/4 = 2 + π⁴/1024 + π/2

Q18

∫₀^π (sin²(x/2) – cos²(x/2)) dx

Answer

= –∫₀^π cos x dx = –[sin x]₀^π = 0

Q19

∫₀² (6x + 3)/(x² + 4) dx

Answer

= [3 log(x² + 4) + (3/2) tan⁻¹(x/2)]₀² = 3 log 2 + (3/2)(π/4) = 3 log 2 + 3π/8

Q20

∫₀¹ (xe^x + sin(πx/4)) dx

Answer

[e^x(x – 1) – (4/π) cos(πx/4)]₀¹ = (0 – 2√2/π) – (–1 – 4/π) = 1 + 4/π – 2√2/π

Q21

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

Answer

tan⁻¹√3 – tan⁻¹1 = π/3 – π/4 = π/12
⇒ (D)

Q22

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

Answer

= [(1/6) tan⁻¹(3x/2)]₀^2/3 = (1/6)(π/4) = π/24
⇒ (C)

Exercise 7.9

Evaluate the integrals in Exercises 1 to 8 using substitution.

∫₀¹ x/(x² + 1) dx

Answer

Put x² + 1 = t (t: 1
⇒ 2)
⇒ ½∫₁² dt/t = ½ log 2

∫₀^π/2 √(sin φ) cos⁵φ dφ

Answer

Put sin φ = t (0
⇒ 1), cos⁴φ = (1 – t²)²
⇒ ∫₀¹ (t^1/2 – 2t^5/2 + t^9/2) dt = 2/3 – 4/7 + 2/11 = 64/231

∫₀¹ sin⁻¹[2x/(1 + x²)] dx

Answer

Put x = tan θ (0
⇒ π/4): integrand = 2θ, dx = sec²θ dθ
⇒ 2∫₀^π/4 θ sec²θ dθ = 2[θ tan θ + log cos θ]₀^π/4 = 2[π/4 – ½ log 2] = π/2 – log 2

∫₀² x√(x + 2) dx (Put x + 2 = t²)

Answer

dx = 2t dt, t: √2
⇒ 2
⇒ ∫(t² – 2)·t·2t dt = [2t⁵/5 – 4t³/3]_√2² = 32/15 + 16√2/15 = 16√2(√2 + 1)/15

∫₀^π/2 sin x/(1 + cos²x) dx

Answer

Put cos x = t (1
⇒ 0)
⇒ ∫₀¹ dt/(1 + t²) = π/4

∫₀² dx/(x + 4 – x²)

Answer

x + 4 – x² = (√17/2)² – (x – ½)²
⇒ (1/√17)[log|(√17 – 1 + 2x)/(√17 + 1 – 2x)|]₀² = (1/√17) log[(5 + √17)/(5 – √17)] = (1/√17) log[(21 + 5√17)/4]

∫_–1¹ dx/(x² + 2x + 5)

Answer

= ∫dx/[(x + 1)² + 4] = ½[tan⁻¹((x + 1)/2)]_–1¹ = ½(π/4 – 0) = π/8

∫₁² (1/x – 1/(2x²)) e^2x dx

Answer

Put 2x = t (2
⇒ 4)
⇒ ∫₂⁴ e^t(1/t – 1/t²) dt = [e^t/t]₂⁴ = e⁴/4 – e²/2 = e²(e² – 2)/4

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

Answer

= ∫(1/x² – 1)^1/3/x³ dx; put 1/x² – 1 = t, –2dx/x³ = dt (t: 8
⇒ 0)
⇒ ½∫₀⁸ t^1/3 dt = ½·(3/4)·16 = 6
⇒ (A)

Q10

If f(x) = ∫₀^x t 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

Answer

By the first fundamental theorem, f′(x) = x sin x
⇒ (B)

Exercise 7.10

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

∫₀^π/2 cos²x dx

Answer

By P₄, I = ∫₀^π/2 sin²x dx
⇒ 2I = ∫₀^π/2 1 dx = π/2
⇒ I = π/4

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

Answer

By P₄ (x
⇒ π/2 – x), I = ∫√(cos x)/[√(cos x) + √(sin x)] dx
⇒ 2I = ∫₀^π/2 dx = π/2
⇒ I = π/4

∫₀^π/2 sin^3/2x dx/(sin^3/2x + cos^3/2x)

Answer

By P₄, I = ∫cos^3/2x/(cos^3/2x + sin^3/2x) dx
⇒ 2I = π/2
⇒ I = π/4

∫₀^π/2 cos⁵x dx/(sin⁵x + cos⁵x)

Answer

By P₄, I = ∫sin⁵x/(cos⁵x + sin⁵x) dx
⇒ 2I = π/2
⇒ I = π/4

∫_–5⁵ |x + 2| dx

Answer

= ∫_–5^–2 –(x + 2) dx + ∫_–2⁵ (x + 2) dx = 9/2 + 49/2 = 29

∫₂⁸ |x – 5| dx

Answer

= ∫₂⁵ (5 – x) dx + ∫₅⁸ (x – 5) dx = 9/2 + 9/2 = 9

∫₀¹ x(1 – x)ⁿ dx

Answer

By P₄: I = ∫₀¹ (1 – x)xⁿ dx = 1/(n + 1) – 1/(n + 2) = 1/[(n + 1)(n + 2)]

∫₀^π/4 log(1 + tan x) dx

Answer

By P₄: 1 + tan(π/4 – x) = 2/(1 + tan x)
⇒ I = ∫₀^π/4 [log 2 – log(1 + tan x)] dx
⇒ 2I = (π/4) log 2
⇒ I = (π/8) log 2

∫₀² x√(2 – x) dx

Answer

By P₄: I = ∫₀² (2 – x)√x dx = [(4/3)x^3/2 – (2/5)x^5/2]₀² = 8√2/3 – 8√2/5 = 16√2/15

Q10

∫₀^π/2 (2 log sin x – log sin 2x) dx

Answer

= ∫₀^π/2 (log sin x – log cos x – log 2) dx; by P₄ ∫log sin x = ∫log cos x
⇒ I = –(π/2) log 2 = (π/2) log(½)

Q11

∫_–π/2^π/2 sin²x dx

Answer

sin²x is even
⇒ 2∫₀^π/2 sin²x dx = 2(π/4) = π/2

Q12

∫₀^π x dx/(1 + sin x)

Answer

By P₄: I = ∫(π – x)/(1 + sin x) dx
⇒ 2I = π∫₀^π dx/(1 + sin x) = π∫₀^π (sec²x – sec x tan x) dx = π[tan x – sec x]₀^π = π[1 – (–1)] = 2π
⇒ I = π

Q13

∫_–π/2^π/2 sin⁷x dx

Answer

sin⁷x is an odd function → I = 0

Q14

∫₀^2π cos⁵x dx

Answer

cos⁵(2π – x) = cos⁵x
⇒ I = 2∫₀^π cos⁵x dx; cos⁵(π – x) = –cos⁵x
⇒ ∫₀^π cos⁵x dx = 0 (P₆)
⇒ I = 0

Q15

∫₀^π/2 (sin x – cos x)/(1 + sin x cos x) dx

Answer

By P₄: I = ∫(cos x – sin x)/(1 + sin x cos x) dx = –I
⇒ 2I = 0
⇒ I = 0

Q16

∫₀^π log(1 + cos x) dx

Answer

By P₄: I = ∫₀^π log(1 – cos x) dx
⇒ 2I = ∫₀^π log sin²x dx = 4∫₀^π/2 log sin x dx = 4(–(π/2) log 2)
⇒ I = –π log 2

(Using ∫₀^π/2 log sin x dx = –(π/2) log 2 and P₆ for ∫₀^π log sin x dx = 2∫₀^π/2 log sin x dx.)

Q17

∫₀^a √x dx/(√x + √(a – x))

Answer

By P₄: I = ∫₀^a √(a – x)/(√(a – x) + √x) dx
⇒ 2I = ∫₀^a dx = a
⇒ I = a/2

Q18

∫₀⁴ |x – 1| dx

Answer

= ∫₀¹ (1 – x) dx + ∫₁⁴ (x – 1) dx = ½ + 9/2 = 5

Q19

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

Answer

Let I = ∫₀^a f(x)g(x) dx. By P₄, I = ∫₀^a f(a – x)g(a – x) dx = ∫₀^a f(x)g(a – x) dx.

Adding: 2I = ∫₀^a f(x)[g(x) + g(a – x)] dx = 4∫₀^a f(x) dx
⇒ I = 2∫₀^a f(x) dx. Hence proved.

Q20

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

Answer

x³, x cos x, tan⁵x are odd
⇒ their integrals vanish
⇒ ∫_–π/2^π/2 1 dx = π
⇒ (C)

Q21

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

Answer

By P₄: I = ∫log[(4 + 3 cos x)/(4 + 3 sin x)] dx = –I
⇒ I = 0
⇒ (C)

Miscellaneous Exercise on Chapter 7

Integrate the functions in Exercises 1 to 23.

1/(x – x³)

Answer

= 1/[x(1 – x)(1 + x)] = 1/x + ½/(1 – x) – ½/(1 + x)
⇒ log|x| – ½ log|1 – x| – ½ log|1 + x| = ½ log|x²/(1 – x²)| + C

1/[√(x + a) + √(x + b)]

Answer

Rationalise: [√(x + a) – √(x + b)]/(a – b)
⇒ [2/(3(a – b))][(x + a)^3/2 – (x + b)^3/2] + C

1/[x√(ax – x²)] [Hint: Put x = a/t]

Answer

x = a/t, dx = –(a/t²) dt, √(ax – x²) = (a/t)√(t – 1)
⇒ –(1/a)∫dt/√(t – 1) = –(2/a)√(t – 1)
⇒ –(2/a)√[(a – x)/x] + C

1/[x²(x⁴ + 1)^3/4]

Answer

= x⁻⁵(1 + x⁻⁴)^–3/4; put 1 + x⁻⁴ = t, –4x⁻⁵ dx = dt
⇒ –¼∫t^–3/4 dt = –t^1/4
⇒ –(1 + 1/x⁴)^1/4 + C

1/(x^1/2 + x^1/3) [Hint: put x = t⁶]

Answer

x = t⁶, dx = 6t⁵ dt
⇒ ∫6t³/(t + 1) dt = 6∫[t² – t + 1 – 1/(t + 1)] dt = 2t³ – 3t² + 6t – 6 log|1 + t|
⇒ 2√x – 3x^1/3 + 6x^1/6 – 6 log(1 + x^1/6) + C

5x/[(x + 1)(x² + 9)]

Answer

= A/(x + 1) + (Bx + C)/(x² + 9); A = –½, B = ½, C = 9/2
⇒ –½ log|x + 1| + ¼ log(x² + 9) + (3/2) tan⁻¹(x/3) + C

sin x/sin(x – a)

Answer

sin x = sin[(x – a) + a]
⇒ integrand = cos a + sin a cot(x – a)
⇒ x cos a + sin a log|sin(x – a)| + C

(e^{5 log x} – e^{4 log x})/(e^{3 log x} – e^{2 log x})

Answer

= (x⁵ – x⁴)/(x³ – x²) = x⁴(x – 1)/[x²(x – 1)] = x²
⇒ x³/3 + C

cos x/√(4 – sin²x)

Answer

Put sin x = t
⇒ ∫dt/√(4 – t²)
⇒ sin⁻¹(sin x/2) + C

Q10

(sin⁸x – cos⁸x)/(1 – 2 sin²x cos²x)

Answer

Numerator = (sin⁴x – cos⁴x)(sin⁴x + cos⁴x) and sin⁴x + cos⁴x = 1 – 2 sin²x cos²x
⇒ integrand = sin⁴x – cos⁴x = sin²x – cos²x = –cos 2x
⇒ –½ sin 2x + C

Q11

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

Answer

Multiply and divide by sin(a – b) = sin[(x + a) – (x + b)]
⇒ [tan(x + a) – tan(x + b)]/sin(a – b)
⇒ (1/sin(a – b)) log|cos(x + b)/cos(x + a)| + C

Q12

x³/√(1 – x⁸)

Answer

Put x⁴ = t, 4x³ dx = dt
⇒ ¼∫dt/√(1 – t²)
⇒ ¼ sin⁻¹(x⁴) + C

Q13

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

Answer

Put e^x = t
⇒ ∫dt/[(1 + t)(2 + t)] = ∫[1/(1 + t) – 1/(2 + t)] dt
⇒ log[(1 + e^x)/(2 + e^x)] + C

Q14

1/[(x² + 1)(x² + 4)]

Answer

= ⅓[1/(x² + 1) – 1/(x² + 4)]
⇒ ⅓ tan⁻¹x – (1/6) tan⁻¹(x/2) + C

Q15

cos³x e^{log sin x}

Answer

= cos³x sin x; put cos x = t
⇒ –∫t³ dt
⇒ –cos⁴x/4 + C

Q16

e^{3 log x}(x⁴ + 1)⁻¹

Answer

= x³/(x⁴ + 1); put x⁴ + 1 = t
⇒ ¼∫dt/t
⇒ ¼ log(x⁴ + 1) + C

Q17

f′(ax + b)[f(ax + b)]ⁿ

Answer

Put f(ax + b) = t, a f′(ax + b) dx = dt
⇒ (1/a)∫tⁿ dt
⇒ [f(ax + b)]ⁿ⁺¹/[a(n + 1)] + C, n ≠ –1

Q18

1/√[sin³x sin(x + α)]

Answer

sin³x sin(x + α) = sin⁴x(cos α + cot x sin α)
⇒ integrand = cosec²x/√(cos α + sin α cot x); put cos α + sin α cot x = t, –sin α cosec²x dx = dt
⇒ –(1/sin α)∫t^–1/2 dt
⇒ –(2/sin α)√[sin(x + α)/sin x] + C

Q19

√[(1 – √x)/(1 + √x)]

Answer

Put x = cos²θ, dx = –2 sin θ cos θ dθ; integrand = tan(θ/2)
⇒ ∫–4 sin²(θ/2) cos θ dθ = ∫(–2 cos θ + 1 + cos 2θ) dθ = –2 sin θ + θ + ½ sin 2θ
⇒ –2√(1 – x) + cos⁻¹√x + √(x(1 – x)) + C

Q20

(2 + sin 2x)e^x/(1 + cos 2x)

Answer

= e^x(2 + 2 sin x cos x)/(2 cos²x) = e^x(sec²x + tan x), f(x) = tan x
⇒ e^x tan x + C

Q21

(x² + x + 1)/[(x + 1)²(x + 2)]

Answer

= A/(x + 1) + B/(x + 1)² + C/(x + 2); B = 1, C = 3, A = –2
⇒ –2 log|x + 1| – 1/(x + 1) + 3 log|x + 2| + C

Q22

tan⁻¹√[(1 – x)/(1 + x)]

Answer

Put x = cos θ
⇒ tan⁻¹[tan(θ/2)] = θ/2 = ½ cos⁻¹x
⇒ ½∫cos⁻¹x dx
⇒ ½[x cos⁻¹x – √(1 – x²)] + C

Q23

√(x² + 1)[log(x² + 1) – 2 log x]/x⁴

Answer

= (1/x³)√(1 + 1/x²) log(1 + 1/x²); put 1 + 1/x² = t, –2dx/x³ = dt
⇒ –½∫t^1/2 log t dt = –½[(2/3)t^3/2 log t – (4/9)t^3/2]
⇒ –⅓(1 + 1/x²)^3/2[log(1 + 1/x²) – 2/3] + C

Evaluate the definite integrals in Exercises 24 to 31.

Q24

∫_π/2^π e^x(1 – sin x)/(1 – cos x) dx

Answer

(1 – sin x)/(1 – cos x) = ½cosec²(x/2) – cot(x/2) = f + f′ with f = –cot(x/2)
⇒ [–e^x cot(x/2)]_π/2^π = 0 + e^π/2 = e^π/2

Q25

∫₀^π/4 sin x cos x/(cos⁴x + sin⁴x) dx

Answer

Divide by cos⁴x: tan x sec²x/(1 + tan⁴x); put tan²x = t (0
⇒ 1)
⇒ ½∫₀¹ dt/(1 + t²) = π/8

Q26

∫₀^π/2 cos²x dx/(cos²x + 4 sin²x)

Answer

= ∫dx/(1 + 4 tan²x); put tan x = t
⇒ ∫₀^∞ dt/[(1 + t²)(1 + 4t²)] = ⅓∫₀^∞[4/(1 + 4t²) – 1/(1 + t²)] dt = ⅓(π – π/2) = π/6

Q27

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

Answer

sin 2x = 1 – (sin x – cos x)²; put sin x – cos x = t (–(√3 – 1)/2
⇒ (√3 – 1)/2)
⇒ ∫dt/√(1 – t²) = 2 sin⁻¹[(√3 – 1)/2]

Q28

∫₀¹ dx/[√(1 + x) – √x]

Answer

= ∫₀¹ [√(1 + x) + √x] dx = (2/3)[(1 + x)^3/2 + x^3/2]₀¹ = (2/3)(2√2 + 1 – 1) = 4√2/3

Q29

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

Answer

Put sin x – cos x = t (–1
⇒ 0), sin 2x = 1 – t²
⇒ ∫_–1⁰ dt/(25 – 16t²) = (1/40)[log|(5 + 4t)/(5 – 4t)|]_–1⁰ = (1/40) log 9 = (1/20) log 3

Q30

∫₀^π/2 sin 2x tan⁻¹(sin x) dx

Answer

Put sin x = t
⇒ 2∫₀¹ t tan⁻¹t dt = 2[((t² + 1)/2) tan⁻¹t – t/2]₀¹ = 2(π/4 – ½) = π/2 – 1

Q31

∫₁⁴ [|x – 1| + |x – 2| + |x – 3|] dx

Answer

∫₁⁴|x – 1| = 9/2; ∫₁⁴|x – 2| = ½ + 2 = 5/2; ∫₁⁴|x – 3| = 2 + ½ = 5/2
⇒ total = 19/2

Prove the following (Exercises 32 to 37).

Q32

∫₁³ dx/[x²(x + 1)] = 2/3 + log(2/3)

Answer

1/[x²(x + 1)] = –1/x + 1/x² + 1/(x + 1)
⇒ [log((x + 1)/x) – 1/x]₁³ = (log(4/3) – ⅓) – (log 2 – 1) = 2/3 + log(2/3). Hence proved.

Q33

∫₀¹ xe^x dx = 1

Answer

[e^x(x – 1)]₀¹ = 0 – (–1) = 1. Hence proved.

Q34

∫_–1¹ x¹⁷cos⁴x dx = 0

Answer

f(x) = x¹⁷cos⁴x satisfies f(–x) = –f(x) (odd)
⇒ integral over [–1, 1] = 0 (P₇). Hence proved.

Q35

∫₀^π/2 sin³x dx = 2/3

Answer

= ∫(1 – cos²x) sin x dx = [–cos x + cos³x/3]₀^π/2 = 0 – (–1 + ⅓) = 2/3. Hence proved.

Q36

∫₀^π/4 2 tan³x dx = 1 – log 2

Answer

= 2∫(sec²x – 1) tan x dx = 2[tan²x/2 + log cos x]₀^π/4 = 2[½ – ½ log 2] = 1 – log 2. Hence proved.

Q37

∫₀¹ sin⁻¹x dx = π/2 – 1

Answer

By parts: [x sin⁻¹x + √(1 – x²)]₀¹ = (π/2 + 0) – (0 + 1) = π/2 – 1. Hence proved.

Q38

∫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

Answer

= ∫e^x dx/(e^2x + 1); put e^x = t
⇒ ∫dt/(1 + t²) = tan⁻¹(e^x) + C
⇒ (A)

Q39

∫cos 2x dx/(sin x + cos x)² 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 + C

Answer

= ∫(cos x – sin x)/(cos x + sin x) dx = log|sin x + cos x| + C
⇒ (B)

Q40

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

Answer

By P₃: I = ∫_a^b (a + b – x) f(a + b – x) dx = ∫_a^b (a + b – x) f(x) dx
⇒ 2I = (a + b)∫_a^b f(x) dx
⇒ (D)

Important Extra Questions

Find ∫dx/(x² + 6x + 13).

Answer

x² + 6x + 13 = (x + 3)² + 2²
⇒ ½ tan⁻¹[(x + 3)/2] + C

Find ∫xe^2x dx.

Answer

By parts: (x/2)e^2x – ½∫e^2x dx
⇒ e^2x(2x – 1)/4 + C

Find ∫log x dx.

Answer

Take 1 as second function: x log x – ∫x·(1/x) dx
⇒ x log x – x + C

Evaluate ∫_–1² |x³ – x| dx.

Answer

x³ – x ≥ 0 on [–1, 0] and [1, 2], ≤ 0 on [0, 1]
⇒ ∫_–1⁰(x³ – x) dx + ∫₀¹(x – x³) dx + ∫₁²(x³ – x) dx = ¼ + ¼ + 9/4 = 11/4

Evaluate ∫₀^π x sin x/(1 + cos²x) dx.

Answer

By P₄: 2I = π∫₀^π sin x/(1 + cos²x) dx; put cos x = t
⇒ 2I = π∫_–1¹ dt/(1 + t²) = π(π/2)
⇒ I = π²/4

NCERT Solutions for Class 12 Mathematics Chapter 7: Integrals

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 on Chapter 7

Important Extra Questions

Frequently Asked Questions