**NCERT Solutions for Class 12 Maths Chapter 7 Integrals** have been designed by subject experts at BYJUâ€™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.

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### 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 â€“ 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 â€“ 4 e^{3x}

The anti-derivative of (sin 2x â€“ 4 e^{3x}) is the function of x whose derivative of (sin 2x â€“ 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} â€“ 3 / x^{4} such that f (2) = 0. Then f (x) is**

**(A) x ^{4} + 1 / x^{3} â€“ 129 / 8**

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

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

**(D) x ^{3} + 1 / x^{4} â€“ 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. âˆšax + b**

**Solution:**

**7. x âˆšx + 2**

**Solution:**

**8. x âˆš1 + 2x ^{2}**

**Solution:**

**9. (4x + 2) âˆšx ^{2} + x + 1**

**Solution:**

**10. 1 / (x â€“ âˆšx)**

**Solution:**

**11. x / (âˆšx + 4), x > 0**

**Solution:**

**12. (x ^{3} â€“ 1)^{1 / 3} x^{5}**

**Solution:**

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

**Solution:**

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

**Solution:**

**15. x / (9 â€“ 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

â€“ sin x dx = dt

By integrating both sides

So we get

= â€“ log |t| + C

Substituting the value of t

= â€“ log |1 + cos x| + C

**31.**

**Solution:**

Take 1 + cos x = t

By differentiation

â€“ 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 â€“ cot x + C**

**(C) tan x cot x + C**

**(D) tan x â€“ 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 = Â½ {cos(A + B) + cos(A â€“ 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 â€“ 1,

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

By multiplying, we get,

= (sec^{2}2x Ã— tan2xsec2x) â€“ (tan2xsec2x)

Integrating both sides,

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

**Solution:-**

By splitting the given function, we have,

tan^{4}x = tan^{2}x Ã— tan^{2}x

Then,

From trigonometric identity, tan^{2} x = sec^{2} x â€“ 1

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

By multiplying, we get,

= sec^{2}x tan^{2}x â€“ tan^{2}x

Again by using trigonometric identity, tan^{2} x = sec^{2} x â€“ 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 â€¦ [equation (i)]

Then, substitute â€˜tâ€™ in place of cosx

**Solution:-**

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

**Solution:-**

**Solution:-**

Let us assume that, (xe^{x})Â = t

Differentiating both sides we get,

((e^{x} Ã— x) + (e^{x} Ã— 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 â€“ 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 â€“ 3 = t

By differentiation

(4x + 1) dx = dt

Integrating both sides

We get

= 2 âˆšt + 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 = Â½

B = 2

Using equation (1) we get

**18.**

**Solution:**

**19.**

**Solution:**

**20.**

**Solution:**

Consider

It can be written as

x + 2 = A (4 â€“ 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 â€“ 1) (x + 2) + B (x + 2) + C (x â€“ 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 â€“ 1 = A (x + 2) + B â€¦.. (1)

Now by equating the coefficient of x and constant terms

A = 3

2A + B = â€“ 1

Solving the equations

B = â€“ 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 Î¸ we get dx = sec^{2} Î¸ dÎ¸

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

**Solution:**

**24. âˆ«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:**

**Solution:**

**Choose the correct answer in Exercises 10 to 11**

**Solution:**

**Solution:**

EXERCISE 7.8 PAGE NO: 334

**Evaluate the following definite integrals as limit of sums.**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

Exercise 7.9 Page No: 338

**Evaluate the definite integrals in Exercises 1 to 20.**

**1. **

**Solution: **

**2. **

**Solution: **

**3. **

**Solution: **

**4. **

**Solution: **

**5. **

**Solution: **

**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: **

**21.**

**Solution: **

Hence, option (D) is correct.

**22. **

**Solution: **

Hence, option (C) is correct.

Exercise 7.10 Page No: 338

**Evaluate the integrals in Exercise 1 to 8 by substitution.**

**1. **

**Solution: **

**2. **

**Solution: **

**3. **

**Solution: **

**4. **

**Solution: **

**5. **

**Solution: **

**6. **

**Solution: **

**7. **

**Solution: **

**8. **

**Solution: **

**Choose the correct answer in Exercise 9 and 10.**

**9.**

** **

**Solution: **

**10.**

** **

**Solution: **

Exercise 7.11 Page No: 347

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

Solution:

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Choose the correct answer in Exercises 20 and 21.**

(A) 0 (B) 2 (C) Ï€ (D) 1

**Solution:**

(C) Ï€

**Explanation:**

(A) 2 (B) 3/4 (C) 0 (D) â€“2

**Solution:**

(C) 0

**Explanation:**

Miscellaneous Exercise Page No: 352

**Integrate the functions in Exercises 1 to 24.**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Evaluate the definite integrals in Exercises 25 to 33.**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Prove the following (Exercises 34 to 39)**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

**Solution:**

Hence the proof.

**Solution:**

**Solution:**

**Choose the correct answers in Exercises 41 to 44.**

**Solution:**

(A) tan^{-1} (e^{x}) + C

**Explanation:**

**Solution:**

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

**Explanation:**

**Solution:**

**Explanation:**

Hence, correct option is (D).

(A) 1 (B) 0 (C) â€“1 (D) Ï€

**Solution:**

(B) 0

**Explanation:**

The major concepts of Maths covered in the Chapter 7- Integrals of NCERT Solutions for Class 12 includes:

7.1 Introduction

7.2 Integration as an Inverse Process of Differentiation

7.2.1 Geometrical interpretation of indefinite integral

7.2.2 Some properties of indefinite integral

7.2.3 Comparison between differentiation and integration

7.3 Methods of Integration

7.3.1 Integration by substitution

7.3.2 Integration using trigonometric identities

7.4 Integrals of Some Particular Functions

7.5 Integration by Partial Fractions

7.6 Integration by Parts

7.7 Definite Integral

7.7.1 Definite integral as the limit of a sum

7.8 Fundamental Theorem of Calculus

7.8.1 Area function

7.8.2 First fundamental theorem of integral calculus

7.8.3 Second fundamental theorem of integral calculus

7.9 Evaluation of Definite Integrals by Substitution

7.10 Some Properties of Definite Integrals

Exercise 7.1 Solutions 22 Questions

Exercise 7.2 Solutions 39 Questions

Exercise 7.3 Solutions 24 Questions

Exercise 7.4 Solutions 25 Questions

Exercise 7.5 Solutions 23 Questions

Exercise 7.6 Solutions 24 Questions

Exercise 7.7 Solutions 11 Questions

Exercise 7.8 Solutions 6 Questions

Exercise 7.9 Solutions 22 Questions

Exercise 7.10 Solutions 10 Questions

Exercise 7.11 Solutions 21 Questions

Miscellaneous Exercise On Chapter 7 Solutions 44 Questions

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

The chapter integrals belongs to the unit Calculus, that adds up to 35 marks of the total marks. There are 11 exercises along with a miscellaneous exercise in this chapter to help the students get thorough with the concept of Integrals. The Chapter 7 of NCERT Solutions for Class 12 Maths discusses the following:

- Integration is the inverse process of differentiation. In the differential calculus, we are given a function and we have to find the derivative or differential of this function, but in the integral calculus, we are to find a function whose differential is given. Thus, integration is a process which is the inverse of differentiation.
- From the geometric point of view, an indefinite integral is collection of family of curves, each of which is obtained by translating one of the curves parallel to itself upwards or downwards along the y-axis.
- Some properties of indefinite integrals.
- Some standard integrals
- Integration by partial fractions
- Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integrals. The method in which we change the variable to some other variable is called the method of substitution. When the integrand involves some trigonometric functions, we use some well known identities to find the integrals.
- Integration by parts: the integral of the product of two functions = first function Ã— integral of the second function â€“ integral of {differential coefficient of the first function Ã— integral of the second function}.
- First fundamental theorem of integral calculus.
- Second fundamental theorem of integral calculus.

### Key Features of NCERT Solutions for Class 12 Maths Chapter 7- Integrals

Learning the chapter Integrals enables the students to study the following:

Integration as the inverse process of differentiation, Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them.

Definite integrals as a limit of a sum, Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.