# NCERT Solutions for Class 12 Maths Chapter 7 Integrals

NCERT Solutions for Class 12 Maths Chapter 7 Integrals have been designed by subject experts at CoolGyan’S to help the students in their exam preparations. The Class 12 NCERT Maths Book contains the concept of integrals in chapter 7. In this chapter, students learn about integral calculus (definite and indefinite), their properties and much more. This topic is extremely important for both CBSE board exam and for competitive exams.

The concepts of integrals are given in this chapter in a detailed and easy to understand manner. These Class 12 NCERT Solutions for integrals are very simple and can help the students in understanding the problem solving method very easily. Students can reach for these NCERT Solutions and download it for free to practise them offline as well. These solved questions help the students understand the method of applying the concepts of Integrals in each problem.

## Download PDF of NCERT Solutions for Class 12 Maths Chapter 7- Integrals

### Access NCERT Solutions for Class 12 Maths Chapter 7- Integrals

Exercise 7.1 page no: 299

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

1. sin 2x

2. cos 3x

3. e2x

4. (ax + b)2

5. sin 2x – 4 e3x

Solution:

1. sin 2x

The anti-derivative of sin 2x is a function of x whose derivative is sin 2x

We know that,

2. cos 3x

The anti-derivative of cos 3x is a function of x whose derivative is cos 3x

We know that,

3. e2x

The anti-derivative of e2x is the function of x whose derivative is e2x

We know that,

4. (ax + b)2

The anti-derivative of (ax + b) 2 is the function of x whose derivative is (ax + b)2

5. sin 2x – 4 e3x

The anti-derivative of (sin 2x – 4 e3x) is the function of x whose derivative of (sin 2x – 4e3x)

Find the following integrals in Exercises 6 to 20:

6.

Solution:

7.

Solution:

8.

Solution:

9.

Solution:

10.

Solution:

11.

Solution:

12.

Solution:

13.

Solution:

14.

Solution:

15.

Solution:

16.

Solution:

17.

Solution:

18.

Solution:

19.

Solution:

20.

Solution:

Choose the correct answer in Exercises 21 and 22

21. The anti-derivative of
equals

(A) (1 / 3) x1 / 3 + (2) x1 / 2 + C

(B) (2 / 3) x2 / 3 + (1 / 2) x2 + C

(C) (2 / 3) x3 / 2 + (2) x1 / 2 + C

(D) (3 / 2) x3 / 2 + (1 / 2) x1 / 2 + C

Solution:

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:

Exercise 7.2 page no: 304

Integrate the functions in Exercises 1 to 37:

1. 2x / 1 + x2

Solution:

2. (log x)2 / x

Solution:

3. 1 / (x + x log x)

Solution:

4. sin x sin (cos x)

Solution:

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

Solution:

6. √ax + b

Solution:

7. x √x + 2

Solution:

8. x √1 + 2x2

Solution:

9. (4x + 2) √x2 + x + 1

Solution:

10. 1 / (x – √x)

Solution:

11. x / (√x + 4), x > 0

Solution:

12. (x3 – 1)1 / 3 x5

Solution:

13. x 2 / (2 + 3x3)3

Solution:

14. 1 / x (log x) m, x > 0, m ≠ 1

Solution:

15. x / (9 – 4x2)

Solution:

16. e2x + 3

Solution:

17.

Solution:

18.

Solution:

19.

Solution:

20.

Solution:

21.

Solution:

22.

Solution:

23.

Solution:

24.

Solution:

25.

Solution:

26.

Solution:

27.

Solution:

28.

Solution:

29.

Solution:

Take

log sin x = t

By differentiation we get

So we get

cot x dx = dt

Integrating both sides

30.

Solution:

Take 1 + cos x = t

By differentiation

– sin x dx = dt

By integrating both sides

So we get

= – log |t| + C

Substituting the value of t

= – log |1 + cos x| + C

31.

Solution:

Take 1 + cos x = t

By differentiation

– sin x dx = dt

32.

Solution:

It is given that

33.

Solution:

34.

Solution:

35.

Solution:

36.

Solution:

37.

Solution:

Choose the correct answer in Exercises 38 and 39.

Solution:

(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:

Exercise 7.3 Page: 307

1. sin2 (2x + 5)

Solution:-

2. sin 3x cos 4x

Solution:-

3. cos 2x cos 4x cos 6x

Solution:-

By standard trigonometric identity cosA cosB = ½ {cos(A + B) + cos(A – B)}

4. sin3 (2x + 1)

Solution:-

5. sin3 x cos3 x

Solution:-

6. sin x sin 2x sin 3x

Solution:-

7. sin 4x sin 8x

Solution:-

Solution:-

Solution:-

10. sin4 x

Solution:-

On simplifying, we get,

11. cos4 2x

Solution:-

Solution:-

Solution:-

Solution:-

15. tan3 2x sec 2x

Solution:-

By splitting the given function, we have,

tan32x sec2x = tan22x tan2x sec2x

From the standard trigonometric identity, tan2 2x = sec2 2x – 1,

= (sec22x -1) tan2x sec2x

By multiplying, we get,

= (sec22x × tan2xsec2x) – (tan2xsec2x)

Integrating both sides,

16. tan4 x

Solution:-

By splitting the given function, we have,

tan4x = tan2x × tan2x

Then,

From trigonometric identity, tan2 x = sec2 x – 1

= (sec2x -1) tan2x

By multiplying, we get,

= sec2x tan2x – tan2x

Again by using trigonometric identity, tan2 x = sec2 x – 1

= sec2x tan2x- (sec2x-1)

= sec2x tan2x- sec2x+1

Now, integrating on both sides we get,

Solution:-

Solution:-

Solution:-

Solution:-

21. sin-1 (cos x)

Solution:-

Given, sin-1(cosx)

Let us assume cosx = t … [equation (i)]

Then, substitute ‘t’ in place of cosx

Solution:-

Choose the correct answer in Exercises 23 and 24.

Solution:-

Solution:-

Let us assume that, (xex) = t

Differentiating both sides we get,

((ex × x) + (ex × 1)) dx = dt

ex (x + 1) = dt

Applying integrals,

Exercise 7.4 Page: 315

Integrate the functions in Exercises 1 to 23.

Solution:-

2.

Solution:

Substituting the value of t

3.

Solution:

4.

Solution:

5.

Solution:

6.

Solution:

Take x3 = t

We get 3 x2 dx = dt

Integrating both sides

On further calculation

7.

Solution:

8.

Solution:

9.

Solution:

10.

Solution:

11.

Solution:

12.

Solution:

Substituting the value of t

13.

Solution:

14.

Solution:

15.

Solution:

16.

Solution:

Consider

4x + 1 = A d/dx (2x2 + x – 3) + B

So we get

4x + 1 = A (4x + 1) + B

On further calculation

4x + 1 = 4 Ax + A + B

By equating the coefficients of x and constant term on both sides

4A = 4

A = 1

A + B = 1

B = 0

Take 2x2 + x – 3 = t

By differentiation

(4x + 1) dx = dt

Integrating both sides

We get

= 2 √t + C

Substituting the value of t

17.

Solution:

Consider

It can be written as

x + 2 = A (2x) + B

Now equating the coefficients of x and constant term on both sides

2A = 1

A = ½

B = 2

Using equation (1) we get

18.

Solution:

19.

Solution:

20.

Solution:

Consider

It can be written as

x + 2 = A (4 – 2x) + B

Now equating the coefficients of x and constant term on both sides

-2A = 1

A = -1/2

4A + B = 2

B = 4

Using equation (1) we get

21.

Solution:

22.

Solution:

23.

Solution:

Consider

It can be written as

5x + 3 = A (2x + 4) + B

Now equating the coefficients of x and constant term on both sides

2A = 5

A = 5/2

4A + B = 3

B = -7

Using equation (1) we get

Choose the correct answer in Exercises 24 and 25.

24.

(A) x tan -1 (x + 1) + C (B) tan -1 (x + 1) + C

(C) (x + 1) tan -1 x + C (D) tan -1 x + C

Solution:

25.

Solution:

Exercise 7.5 page: 322

Integrate the rational functions in Exercises 1 to 21.

1.

Solution:

2.

Solution:

3.

Solution:

4.

Solution:

5.

Solution:

6.

Solution:

7.

Solution:

8.

Solution:

We know that

It can be written as

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

Taking x = 1 we get

B = 1/3

Now by equating the coefficients of x2 and constant terms we get

A + C = 0

-2A + 2B + C = 0

By solving the equations

A = 2/9 and C = -2/9

We get

9.

Solution:

By further calculation

10.

Solution:

11.

Solution:

12.

Solution:

13.

Solution:

14.

Solution:

We know that

It can be written as

3x – 1 = A (x + 2) + B ….. (1)

Now by equating the coefficient of x and constant terms

A = 3

2A + B = – 1

Solving the equations

B = – 7

We get

15.

Solution:

16.

Solution:

17.

Solution:

18.

Solution:

19.

Solution:

20.

Solution:

21.

Solution:

Choose the correct answer in each of the Exercises 22 and 23.

Solution:

Solution:

Exercise 7.6 page: 327

Integrate the functions in Exercises 1 to 22.

1. x sin x

Solution:

2. x sin 3x

Solution:

3. x2 ex

Solution:

4. x log x

Solution:

5. x log 2x

Solution:

6. x2 log x

Solution:

7. x sin -1 x

Solution:

8. x tan -1 x

Solution:

9. x cos -1 x

Solution:

10. (sin -1 x)2

Solution:

11.

Solution:

12. x sec2 x

Solution:

13. tan -1 x

Solution:

14. x (log x)2

Solution:

15. (x2 + 1) log x

Solution:

16. ex (sin x + cos x)

Solution:

17.

Solution:

18.

Solution:

19.

Solution:

20.

Solution:

It is given that

21. e2x sin x

Solution:

22.

Solution:

Take x = tan θ we get dx = sec2 θ dθ

Choose the correct answer in Exercises 23 and 24.

Solution:

24. ∫ex sec x (1 + tan x) dx equals

(A) ex cos x + C (B) ex sec x + C

(C) ex sin x + C (D) ex tan x + C

Solution:

EXERCISE 7.7 PAGE NO: 330

Integrate the functions in exercise 1 to 9

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution: