**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. e ^{2x}**

**4. (ax + b) ^{2}**

**5. sin 2x – 4 e ^{3x}**

**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. e^{2x}

The anti-derivative of e^{2x} is the function of x whose derivative is e^{2x}

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 e^{3x}

The anti-derivative of (sin 2x – 4 e^{3x}) is the function of x whose derivative of (sin 2x – 4e^{3x})

**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) x ^{1 / 3 }+ (2) x^{1 / 2} + C**

**(B) (2 / 3) x ^{2 / 3} + (1 / 2) x^{2} + C**

**(C) (2 / 3) x ^{3 / 2} + (2) x^{1 / 2} + C**

**(D) (3 / 2) x ^{3 / 2} + (1 / 2) x^{1 / 2} + C**

**Solution:**

**22. If d / dx f (x) = 4x ^{3} – 3 / x^{4} such that f (2) = 0. Then f (x) is**

**(A) x ^{4} + 1 / x^{3} – 129 / 8**

**(B) x ^{3} + 1 / x^{4} + 129 / 8**

**(C) x ^{4} + 1 / x^{3} + 129 / 8**

**(D) x ^{3} + 1 / x^{4} – 129 / 8**

**Solution:**

Exercise 7.2 page no: 304

**Integrate the functions in Exercises 1 to 37:**

**1. 2x / 1 + x ^{2}**

**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 + 2x ^{2}**

**Solution:**

**9. (4x + 2) √x ^{2} + x + 1**

**Solution:**

**10. 1 / (x – √x)**

**Solution:**

**11. x / (√x + 4), x > 0**

**Solution:**

**12. (x ^{3} – 1)^{1 / 3} x^{5}**

**Solution:**

**13. x ^{2} / (2 + 3x^{3})^{3}**

**Solution:**

**14. 1 / x (log x) ^{m}, x > 0, m ≠ 1**

**Solution:**

**15. x / (9 – 4x ^{2})**

**Solution:**

**16. e ^{2x + 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. sin ^{2} (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. sin ^{3} (2x + 1)**

**Solution:-**

**5. sin ^{3} x cos^{3} x**

**Solution:-**

**6. sin x sin 2x sin 3x**

**Solution:-**

**7. sin 4x sin 8x**

**Solution:-**

**Solution:-**

**Solution:-**

**10. sin ^{4} x**

**Solution:-**

On simplifying, we get,

**11. cos ^{4} 2x**

**Solution:-**

**Solution:-**

**Solution:-**

**Solution:-**

**15. tan ^{3} 2x sec 2x**

**Solution:-**

By splitting the given function, we have,

tan^{3}2x sec2x = tan^{2}2x tan2x sec2x

From the standard trigonometric identity, tan^{2} 2x = sec^{2} 2x – 1,

= (sec^{2}2x -1) tan2x sec2x

By multiplying, we get,

= (sec^{2}2x × tan2xsec2x) – (tan2xsec2x)

Integrating both sides,

**16. tan ^{4} x**

**Solution:-**

By splitting the given function, we have,

tan^{4}x = tan^{2}x × tan^{2}x

Then,

From trigonometric identity, tan^{2} x = sec^{2} x – 1

= (sec^{2}x -1) tan^{2}x

By multiplying, we get,

= sec^{2}x tan^{2}x – tan^{2}x

Again by using trigonometric identity, tan^{2} x = sec^{2} x – 1

= sec^{2}x tan^{2}x- (sec^{2}x-1)

= sec^{2}x tan^{2}x- sec^{2}x+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, (xe^{x}) = t

Differentiating both sides we get,

((e^{x} × x) + (e^{x} × 1)) dx = dt

e^{x }(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 x^{3} = t

We get 3 x^{2} 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 (2x^{2} + 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 2x^{2} + 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 x^{2} 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. x ^{2} e^{x}**

**Solution:**

**4. x log x**

**Solution:**

**5. x log 2x**

**Solution:**

**6. x ^{2} 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 sec ^{2} x**

**Solution:**

**13. tan ^{-1 }x**

**Solution:**

**14. x (log x) ^{2}**

**Solution:**

**15. (x ^{2} + 1) log x**

**Solution:**

**16. e ^{x} (sin x + cos x)**

**Solution:**

**17. **

**Solution:**

**18.**

**Solution:**

**19.**

**Solution:**

**20.**

**Solution:**

It is given that

**21. e ^{2x} sin x**

**Solution:**

**22. **

**Solution:**

Take x = tan θ we get dx = sec^{2} θ dθ

**Choose the correct answer in Exercises 23 and 24.**

**Solution:**

**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:**

EXERCISE 7.7 PAGE NO: 330

**Integrate the functions in exercise 1 to 9**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**