NCERT Class 12 Mathematics Chapter 7 Integrals

By Dev Kirtonia / NCERT Class 12 Mathematics

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NCERT Class 12 Mathematics Chapter 7 Integrals

Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 12 Mathematics Chapter 7 Integrals Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 12 Mathematics Chapter 7 Integrals Solutions for All Subject, You can practice these here.

Integrals

Chapter – 7

Exercise 7.1

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

1. sin 2 x

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2. cos 3 x

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3. e²ˣ

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4. (ax + b)²

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5. sin 2x – 4 e³ˣ

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Find the following integrals in Exercises 6 to 20:

6. ∫ (4 e³ˣ + 1)dx

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8. ∫(ax² + bx + c) dx

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9. ∫ (2x² + eˣ ) dx

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14. ∫ (1 – x)√x dx

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15. ∫ √x(3x² + 2x + 3) dx

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16. ∫ (2x – 3 cos x + eˣ) dx

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17. ∫ (2x² – 3sin x + 5√x) dx

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18. ∫ sec x (sec x + tan x) dx

Ans: ∫ sec x (sec x + tan x) dx = ∫ (sec² x + sec x tan x) dx

= ∫ sec² x dx + ∫ sec x tan x dx

= tan x + sec x + C

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= ∫ 2sec² x dx – 3 ∫ tan x sec x dx

= 2 tan x – 3sec(x) + C

Choose the correct answer in Exercises 21 and 22

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Thus, the correct option is C.

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Thus, the correct option is A.

EXERCISE 7.2

Integrate the functions in Exercises 1 to 37:

Ans: Put 1 + x² = t

Therefore, 2xdx = dt

= log|1 + x²| + C

= log(1 + x²) + C

Ans:

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4. sin x sin (cos x).

Ans: Put cos x = t

Therefore, – sin xd x = dt

∫ sin x sin(cos x) dx = – ∫ sin tdt = – [- cos t] + C

= cos t + C

= cos(cos x) + C

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

Ans:

6. √ax + b.

Ans: Put ax + b = t

Therefore,

7. x√x + 2.

Ans: Put, x + 2 = t

∴ dx = dt

⇒ ∫ x √x + 2 = ∫ (t – 2) √t dt

8. x√1 + 2x²

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9. (4x + 2)√x² + x + 1

Ans: Put, x² + x + 1 = t

∴ (2x + 1) dx = dt

∫ (4x + 2) √x² + x + 1dx

= ∫ 2√t dt

= 2 ∫ √t dt

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16. e²ˣ ⁺ ³

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Dividing Nr and Dr by eˣ, we get

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21. tan² (2x – 3)

Ans: tan² (2x – 3) = sec² (2x – 3) – 1

Put, 2x – 3 = t

∴ 2 dx = dt

⇒ ∫ tan² (2x – 3) dx = ∫ [sec² (2x – 3) – 1] dx

22. sec² (7 – 4x)

Ans: Put, 7 – 4x = t

∴ – 4 dx = dt

Ans: Put, sin¯¹ x = t

Ans:

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Ans: Let √x = t

= 2sin t + C

= 2sin √x + C

27. √sin 2x cos 2x

Ans: Put, sin 2x = t

So, 2cos 2x dx = dt

Ans: Put, 1 + sin x = t

∴ cos xdx = dt

= 2√t + C

= 2√1 + sin x + C

29. cos x log sin x

Ans:

Ans: Put, 1 + cos x = t

∴ – sin xdx = dt

= – log|t| + C

= – log|1+cos x| + C

Ans: Put, 1 + cos x = t

∴ – sin xdx = dt

Ans:

Ans:

Ans:

Ans:

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Ans: Put, x⁴ = t

∴ 4x³ dx = dt

Let tan⁻¹ t = u

Choose the correct answer in Exercises 38 and 39.

(A) 10ˣ – x¹⁰ + C

(B) 10ˣ + x¹⁰ + C

(D) log(10ˣ + x¹⁰) + C

Ans: Put, x¹⁰ + 10ˣ = t

= log t + C

= log(10ˣ + x¹⁰) + C

Thus, the correct option is D.

(A) tan x + cot x + C

(B) tan x – cot x + C

(D) tan x – cot 2x + C

Ans: Put,

= ∫ sec² xdx + ∫ cos ec² dx

= tan x- cot x + C

Thus, the correct option is B.

EXERCISE 7.3

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

1. sin² (2x + 5)

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2. sin 3x cos 4x

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3. cos 2x cos 4x cos 6x

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4. sin³ (2x + 1)

Ans: Put, I = ∫ sin³ (2x + 1)

⇒ ∫ sin³ (2x + 1) dx = ∫ sin² (2x + 1) sin(2x + 1) dx

= ∫ (1 – cos² (2x + 1)) sin(2x + 1) dx

Let cos(2x + 1) = t

⇒ – 2sin(2x + 1) dx = dt

5. sin³ x cos³ x

Ans: Let I = ∫ sin³ x cos³ xdx

= ∫ cos³ x sin² x sin xdx

= ∫ cos³ x (1 – cos² x) sin xdx

Let cosx = t

⇒ – sin xdx = dt

⇒ I = – ∫ t³ (1 – t²) dt

6. sin x sin 2x sin 3x

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7. sin 4x sin 8x

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10. sin⁴ x

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11. cos⁴ 2x

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15. tan³ 2x sec 2x

Ans: tan³ 2x sec 2x = tan² 2x tan 2x sec 2x

= (sec² 2x – 1) tan 2x sec 2x

= sec² 2x tan 2x sec 2x – tan 2x sec 2x

∴ ∫ tan³ 2x sec 2 xdx = ∫ sec ² 2x tan 2x sec 2x – ∫ tan 2x sec 2x

16. tan⁴ x

Ans: tan⁴ x

= tan² x tan² x

= (sec² x – 1) tan² x

= sec² x tan² x – tan² x

= sec² x tan² x – (sec² x – 1)

= sec² x tan² x – sec² x + 1

∴ ∫ tan⁴ xdx = ∫ sec² x tan² xdx – ∫ sec² xdx + ∫ 1dx

= ∫ sec² x tan² xdx – tan x + x + C …(1)

Consider sec² x tan² xdx

Let tan x = t ⇒ sec² xdx = dt

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21. sin⁻¹ (cos x)

Ans: sin⁻¹ (cos x)

Let cos x = t

Then, sin x = √ 1 – t²

⇒ (- sin x) dx = dt

Let sin⁻¹ t = u

Substituting in equation (1), we get

Ans:

Choose the correct answer in Exercises 23 and 24.

(A) tan x + cot x + C

(B) tan x + cosec x + C

(D) tan x + sec x + C

Ans:

= ∫ (sec² x – cosec² x) dx

= tan x + cot x + C

Thus, the correct option is A.

(A) – cot (exˣ) + C

(B) tan (xeˣ) + C

(D) cot(eˣ) + C

Ans:

Put, eˣ x = t

⇒ (eˣ x + eˣ .1) dx = dt

eˣ (x + 1) dx = dt

= ∫ sec² tdt

= tan t + C

= tan(eˣ x)+C

Thus, the correct answer is B.

EXERCISE 7.4

Integrate the functions in Exercises 1 to 23.

Ans: Put, x³ = t

∴ 3x² dx = dt

= tan⁻¹ t + C

= tan⁻¹ (x³) + C

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Ans: Put, 2 – x = t

⇒ – dx = dt

Ans: Put, 5x = t

∴ 5dx = dt

Ans: Let √2x² = t

∴ 2 √2xdx = dt

Ans: Put, x³ = t

∴ 3x² dx = dt

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Ans: Put, tan x = t

∴ sec² xdx = dt

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Ans: 7 – 6x – x² can be written as 7 – (x² + 6x + 9 – 9)

Thus,

7 – (x² + 6x + 9 – 9)

= 16 – (x² + 6x + 9)

= 16 – (x + 3)²

= (4)² – (x + 3)²

Ans: (x – 1)(x – 2) can be written as x² – 3x + 2

Thus,

x² – 3x + 2

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Thus,

Ans: (x – a)(x – b) = x² – (a + b) x + ab

Thus,

x² – (a + b) x + ab

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⇒ 4x + 1 = A(4x + 1) + B

⇒ 4x + 1 = 4Ax + A + B

Equating the coefficients of x and constant term on both sides, we get

4A = 4 ⇒ A =1

A + B =1 ⇒ B = 0

Let 2x² + x – 3 = t

∴ (4x + 1) dx = dt

= 2√t + C

= 2√2x² + x – 3) + C

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Ans: Let

(x + 3) = A(2x – 2) + B

Equating the coefficients of x and constant term on both sides, we get

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Choose the correct answer in Exercises 24 and 25.

(A) x tan⁻¹ (x + 1) + C

(B) tan⁻¹ (x + 1) + C

(D) tan⁻¹ x + C

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Hence, the correct option is B.

EXERCISE 7.5

Integrate the rational functions in Exercises 1 to 21.

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Ans: Let

1 = A(x – 3) + B(x + 3)

Equating the coefficients of x and constants term, we get

A + B = 0

– 3A + 3B = 1

On solving, we get

Ans:

Ans: Let

x = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2) …(1)

Equating the coefficients of x² , x and constant terms, we get

A + B + C = 0

– 5A – 4B – 3C = 1

6A + 4B + 2C = 0

Solving these equations, we get

Ans:

Ans: It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (1 – x²)by x(1 – 2x), we get

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3x + 5 = A(x – 1)(x + 1) + B(x + 1) + C (x – 1)²

3x + 5 = A(x – 1)(x + 1) + B(x + 1) + C(x² – 2x + 1) …(1)

Equating the coefficients of x²,x and constant term, we get

A + C = 0

B – 2C = 3

– A + B + C = 5

On solving these equations, we get

Ans:

⇒ (2x – 3) = A(x – 1)(2x – 3) + B(x + 1)(2x + 3) + C(x + 1)(x – 1)

⇒ (2x – 3) = A(2x²+ x – 3) + B(2x² + 5x + 3) + C(x² – 1)

⇒ (2x – 3) = (2A + 2B + C) x² + (A + 5B) x + (- 3A + 3B – C)

Equating the coefficients of x², x and constant term, we get

2A + 2B + C = 0

A + 5B = 2

– 3A + 3B – C = – 3

On solving, we get

Ans:

5x = A(x + 2)(x – 2) + B(x + 1)(x – 2) + C(x + 1)(x + 2) …(1)

Equating the coefficients of x²,x and constant term, we get

A + B + C = 0

– B + 3C = 5

– 4A – 2B + 2C = 0

On solving, we get

Ans: On dividing (x³ + x + 1) by x² – 1 , we get

2x + 1 = A(x – 1) + B(x + 1) …(1)

Equating the coefficients of x and constant term, we get

A + B = 2

– A + B = 1

On solving, we get

Ans: Let

2 = A(1 + x²) + (Bx + C)(1 – x)

2 = A + A x² + Bx – B x² + C – Cx

Equating the coefficients of x², x and constant term, we get

A – B = 0

B – C = 0

A + C = 2

On solving these equations, we get

A = 1, B = 1 and C = 1

Ans: Let

⇒ 3x – 1 = A(x + 2) + B

Equating the coefficient of x and constant term, we get

A = 3

2A + B = -1 ⇒ B = -7

Ans:

l = A(x – 1)(1 + x²) + B(x + 1)(1 + x²) + (Cx + D)(x² – 1)

l = A(x³ + x – x² – 1) + B(x³ + x + x² + 1) + C x³ + D x² – Cx – D

I = (A + B + C) x³ + (- A + B + D) x² + (A + B – C) x + (- A + B – D)

Equating the coefficients of x³, x², x and constant term, we get

A + B + C = 0

– A + B + D = 0

A + B – C = 0

– A + B – D = 1

On solving, we get

[Hint: multiply numerator and denominator by xⁿ⁻¹ and put xⁿ = t]

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[Hint : Put sin x = t]

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4x² + 10 = (Ax + B)(x² + 4) + (Cx + D)(x² + 3)

4x² + 10 = Ax³ + 4Ax + Bx² + 4B + Cx³ + 3Cx + Dx² + 3D

4x² + 10 = (A + C) x³ + (B + D) x² + (4A + 3C) x + (4B + 3D)

Equating the coefficients of x³, x² , x and constant term, we get

A + C = 0

B + D = 4

4A + 3C = 0

4B + 3D = 10

On solving these equations, we get

A = 0, B = – 2, C = 0 and D = 6

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Multiplying Nr and Dr by x³, we get

Ans: Put eˣ = t ⇒ eˣ dx = dt

1 = A(t – 1) + Bt …(1)

Equating the coefficients of t and constant, we get

A = – 1 and B = 1

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EXERCISE 7.5

Integrate the functions in Exercises 1 to 22.

1. x sin x.

Ans: Let I = ∫ x sin xdx

Taking u = x and v = sin x and integrating by parts,

I = x ∫ sin xdx – ∫ {d/dx (x)) ∫ sin xdx} dx

= x(- cos x) – ∫ 1. (- cos x) dx

= – x cos x + sin x + C

2. x sin 3x

Ans: Let I = ∫ x sin 3 xdx

Taking u = x and v = sin 3x and integrating by parts,

3. x² eˣ

Ans: Let I = ∫ x² eˣ dx

Taking u = x² and v = eˣ and integrating by parts, we get

= x²eˣ – ∫ 2x.eˣ dx

=x²eˣ – 2 ∫ x.eˣ

Again using integration by parts, we get

= x²eˣ – 2xeˣ + 2eˣ + C

= eˣ(x² – 2x + 2) + C

4. x log x

Ans: Let I = ∫ x log xdx

Taking u = log x and v = x and integrating by parts, we get

5. x log 2x

Ans: Let I = ∫ x log 2xdx

Taking u = log 2x and v = x and integrating by parts, we get

6. x² log x

Ans: Let I = ∫ x² log xdx

Taking u = log x and v = x² and integrating by parts, we get

7. x sin⁻¹ x

Ans: Let I = ∫ x sin⁻¹ xdx

Taking u = sin⁻¹ x and v = x and integrating by parts, we get

8. x tan⁻¹ x

Ans: Let I = ∫ x tan⁻¹ xdx

Taking tann⁻¹ x and v = x and integrating by parts, we get

9. x cos⁻¹ x

Ans: Let I = ∫ x cos⁻¹ xdx

Taking u = cos⁻¹ x and v = x and integrating by parts, we get

10. (sin⁻¹ x)²

Ans: Let I = ∫ (sin⁻¹ x)².1 dx

Taking u = (sin⁻¹ x)² and v = 1 and integrating by parts, we get

Ans:

12. x sec² x

Ans: Let I = ∫ x sec² xdx

Taking u = x and v = sec² x and integrating by parts, we get

= x tan x – ∫ 1.tan xdx

= x tan x + log |cosx| + C

13. tan⁻¹ x

Ans: Let I = ∫1.tan⁻¹ xdx

Taking u = tan⁻¹ x and v = 1 and integrating by parts, we get

14. x (log x)²

Ans: Let I = ∫ x (log x)² dx

Taking u = (log x)² and v = x and integrating by parts, we get

Again, using integration by parts, we get

15. (x² + 1) log x

Ans: Let I = ∫(x² + 1) log xdx = ∫ x² log xdx + ∫ log xdx

Let I = I₁ + I₂ ……….(1)

Where, I₁ = ∫ x² log xdx and I₂ = ∫ log xdx

I₁ = ∫ x² log xdx

Taking u = log x and v = x² and integrating by parts, we get

I₂ = ∫ log xdx

Taking u = log x and v = 1 and integrating by parts,

Using equations (2) and (3) in (1),

16. eˣ (sin x + cos x)

Ans: Let I = ∫eˣ (sin x + cos x) dx

Let ∫ (x) = sin x

f’(x) = cos x

I = ∫eˣ {f(x) + f'(x)} dx

Since, ∫ eˣ {f(x) + f'(x)} dx

= eˣ f (x) + c

∴ I = eˣ sin x + C

Ans:

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It is known that, ∫ eˣ {f(x) + f ‘(x)} dx

= eˣ f (x) + C

From equation (1), we get

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It is known that,

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Ans: Let I = e²ˣ sin xdx………. (1)

Taking u = sin x and v = e²ˣ and integrating by parts, we get

Again, using integration by parts, we get

Ans:

Using integration by parts, we get

Choose the correct answer in Exercises 23 and 24.

Ans:

Thus, the correct option is A.

24. ∫ eˣ sec x (1+ tan x) dx equals

(A) eˣ cos x + C

(B) eˣ sec x + C

(D) eˣ tan x + C

Ans: ∫ eˣ sec x (1 + tan x) dx

Consider, I = ∫ eˣ sec x (1 + tan x)dx = ∫ eˣ (sec x + sec x tan x)dx

Let sec x = f (x) sec x tan x = f'(x)

It is known that, ∫ eˣ {f (x) + f'(x)} dx = eˣ f(x) + C

∴ I = eˣ sec x + C

Thus, the correct option is B.

EXERCISE 7.7

Integrate the functions in Exercises 1 to 9

1. √ 4 – x²

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2. √1 – 4x²

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3. √x² + 4x + 6

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4. √x² + 4x + 1

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5. √1 – 4x – x²

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6. √x² + 4x – 5

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7. √1 + 3x – x²

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8. √x² + 3x

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Choose the correct answer in Exercises 10 to 11.

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EXERCISE 7.8

Evaluate the definite integrals in Exercises 1 to 20.

Ans:

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Ans: Let

Ans:

Ans:

Ans: Let

∫ eˣ dx = eˣ = F(x)

Using second fundamental theorem of calculus, we get

I = F(5) – F(4)

= e⁵ – e⁴

= e⁴ (e – 1)

Ans: Let

∫ tan xdx = – log|cos x| = F(x)

Using second fundamental theorem of calculus, we get

Ans:

Ans:

Using second fundamental theorem of calculus, we get

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Ans:

Ans:

Ans:

Ans: Let

Ans:

Ans:

= 2Ax + (4A + B)

Equating the coefficients of x and constant term, we get

A = 10 and B = – 25

Let x² + 4x + 3 = t

⇒ (2x + 4) dx = dt

Substituting the value I₁ in (1), we get

Ans:

Ans:

∫ cos xdx = sin x = F(x)

Using second fundamental theorem of calculus, we get

I = F(π) – F(0)

= sin π – sin 0

= 0

Ans:

Ans: Let

Choose the correct answer in Exercises 21 and 22.

Ans:

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EXERCISE 7.9

Evaluate the integrals in Exercises 1 to 8 using substitution.

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Choose the correct answer in Exercises 9 and 10.

9. The value of the integral

(A) 6

(B) 0

(D) 4

Ans:

(A) cos x + x sin x

(B) x sin x

(D) sin x + x cos x

Ans:

EXERCISE 7.10

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

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Multiplying and Dividing by (1-sinx)

⇒ 2 I = π[(tan(π) – tan(0)) – (sec(π) – sec(0))]

⇒ 2I = π[2]

⇒ I = π

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Ans:

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19. Show that f (x) dx, if f and g are defined as f(x) = f(a – x) and g(x) + g(a – x) = 4

Ans:

Choose the correct answer in Exercises 20 and 21.

(A) 0

(B) 2

(D) 1

Ans:

(A) 2

(B) ¾

(D) -2

Ans:

Thus, the correct option is C.

Miscellaneous Exercise on Chapter 7

Integrate the functions in Exercises 1 to 23.

Ans:

⇒ 1 = A(1 – x²) + Bx(1 + x) + Cx(1 – x)

⇒ 1 = A – Ax² + Bx + Bx² + Cx – Cx²

Equating the coefficients of x², x and constant terms, we get

– A + B – C = 0

B + C = 0

A = 1

On solving these equations, we get

A = 1

Ans:

Ans:

Ans:

Ans:

Ans: Consider,

⇒ 5x = A(x² + 9) + (Bx + C)(x + 1)

⇒ 5x = Ax² + 9A + Bx² + Bx + Cx + C

Equating the coefficients of x²,x and constant term, we get

A + B = 0

B + C = 5

9A + C = 0

On solving these equations, we get

A = – 1/2

B = 1/2

C = 9/2

From equation (1), we get

Ans:

= t cos a + sin a log|sint| + C₁

= (x – a) cos a + sin a log|sin(x – a)| + C₁

= x cos a + sin a log|sin(x – a)| – a cos a + C₁

= sin a log|sin(x – a)| + x cos a + C

Ans:

Ans:

Ans:

Ans:

Multiplying and dividing by sin(a – b), we get

Ans:

Ans:

Ans:

⇒ 1 = (Ax + B)(x² + 4) + (Cx + D)(x² + 1)

⇒ 1 = Ax³ + 4Ax + Bx² + 4B + Cx³ + Cx + Dx² + D

Equating the coefficients of x³, x²,x and constant term, we get

A + C = 0

B + D = 0

4A + C = 0

4B + D = 1

On solving these equations, we get

A = 0

B = 1/3

C = 0

D = – 1/3

From equation (1), we get

15. cos³ x eˡᵒᵍ ˢⁱⁿˣ

Ans: cos³ xeˡᵒᵍ ˢⁱⁿ ˣ = cos³ x × sin x

Let cos x = t ⇒ -sin xdx = dt

⇒- ∫ cos³ xeˡᵒᵍ ˢⁱⁿ ˣ dx = ∫ cos³ x sin xdx

= – ∫ t³ dt

16. e³ ˡᵒᵍˣ (x⁴ + 1)⁻¹

Ans:

Let x⁴ +1= t ⇒ 4x³ dx = dt

17. f’ (ax + b) [f (ax + b)]ⁿ

Ans:

Ans:

Put, cos α + cot x sin α = t ⇒ – cos ec² x sin α dx = dt

Ans:

= θ + √1-cos² θ. cos θ – 2√1 – cos² θ + C

= cos⁻¹ √x + √1-x. √x – 2√1 – x + C

= – 2√1 – x + cos⁻¹ √x + √x(1 – x) + C

= – 2√1 – x + cos⁻¹ √x + √x – x² + C

Ans:

= ∫ (sec² x + tan x) eˣ

Let f(x) = tan x ⇒ f’ (x) = sec² x

∴ I = ∫ (f(x) + f’ (x)) eˣ dx

= eˣ f(x)+C

= eˣ tan x +C

Ans: Let

⇒ x² + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x² + 2x + 1)

⇒ x² + x + 1 = A(x² + 3x + 2) + B(x + 2) + C(x² + 2x + 1)

⇒ x² + x + 1 = (A + C) x² + (3A + B + 2C) x + (2A + 2B + C)

Equating the coefficients of x² ,x and constant term, we get

A + C = 1

3A + B + 2C = 1

2A + 2B + C = 1

On solving these equations, we get

A = – 2

B = 1

C = 3

From equation (1), we get

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⇒ 1 = Ax(x + 1) + B(x + 1) + C(x²)

⇒ 1 = Ax² + Ax + Bx + B + Cx²

Equating the coefficients of x², x and constant terms, we get

A + C = 0

A + B = 0

B = 1

On solving these equations, we get

A = – 1

C = 1

B = 1

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(A) tan⁻¹ (eˣ) + C.

(B) tan⁻¹ (e⁻ˣ) + C.

(D) log(eˣ + e⁻ˣ) + C.

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(B) log|sin x + cos x| + C

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∴ I = ∫ dt/t

= log|t|+C

= log|cosx + sinx|+ C

Thus, the correct option is B.

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