# NCERT Solutions for Class 12 Maths Chapter 7 Integrals

## NCERT Solutions for Class 12 Maths Chapter 7 â€“ Free PDF Download

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

The concepts of integrals are given in this chapter is detailed and easy to understand manner. TheseÂ NCERT Solutions forÂ Class 12 Maths integrals are very simple and can help the students in understanding the problem solving method very easily. Students can reach for these NCERT Solutions for Class 12Â Maths Chapter 7Â 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

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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. e2x

4. (ax + b)2

5. sin 2x â€“ 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 â€“ 4 e3x

The anti-derivative of (sin 2x â€“ 4 e3x) is the function of x whose derivative of (sin 2x â€“ 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 â€“ 3 / x4 such that f (2) = 0. Then f (x) is

(A) x4 + 1 / x3 â€“ 129 / 8

(B) x3 + 1 / x4 + 129 / 8

(C) x4 + 1 / x3 + 129 / 8

(D) x3 + 1 / x4 â€“ 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. âˆšax + b

Solution:

7. x âˆšx + 2

Solution:

8. x âˆš1 + 2x2

Solution:

9. (4x + 2) âˆšx2 + x + 1

Solution:

10. 1 / (x â€“ âˆšx)

Solution:

11. x / (âˆšx + 4), x > 0

Solution:

12. (x3 â€“ 1)1 / 3 x5

Solution:

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

Solution:

14. 1 / x (log x) m, x > 0, m â‰  1

Solution:

15. x / (9 â€“ 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

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. sin2 (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. 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 â€“ 1,

= (sec22x -1) tan2x sec2x

By multiplying, we get,

= (sec22x Ã— tan2xsec2x) â€“ (tan2xsec2x)

Integrating both sides,

16. tan4 x

Solution:-

By splitting the given function, we have,

tan4x = tan2x Ã— tan2x

Then,

From trigonometric identity, tan2 x = sec2 x â€“ 1

= (sec2x -1) tan2x

By multiplying, we get,

= sec2x tan2x â€“ tan2x

Again by using trigonometric identity, tan2 x = sec2 x â€“ 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 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, (xex)Â = t

Differentiating both sides we get,

((ex Ã— x) + (ex Ã— 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 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 2x2 + 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 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 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. 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 Î¸ we get dx = sec2 Î¸ dÎ¸

Choose the correct answer in Exercises 23 and 24.

Solution:

24. âˆ«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:

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 (ex) + 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:

 Also AccessÂ NCERT Exemplar for Class 12 Maths Chapter 7 CBSE Notes for Class 12 Maths Chapter 7

## Class 12 Maths NCERT Solutions Chapter 7 Integrals

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

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

### Access exercise-wise NCERT Solutions for Class 12 Maths Chapter 7 from the links below:

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 include in term â€“ II of CBSE Syllabus 2021-22, 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 SolutionsÂ for Class 12 Maths discusses the following:

1. 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.
2. 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.
3. Some properties of indefinite integrals.
4. Some standard integrals
5. Integration by partial fractions
6. 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.
7. 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}.
8. First fundamental theorem of integral calculus.
9. Second fundamental theorem of integral calculus.

Students can utilize the NCERT Solutions for Class 12 Maths Chapter 7 to solve difficult problems, doubt clearance, exam preparations and revisions. Referring to these solutions will boost confidence among students to take the Class 12 Maths second term exam.

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

Learning the NCERT Solutions of 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.

## Frequently Asked Questions on NCERT Solutions for Class 12 Maths Chapter 7

### What is the meaning of integrals in NCERT Solutions for Class 12 Maths Chapter 7?

The area under a curve is often referred to as integral in Calculus. It can be represented on a graph, as a mathematical function. List of important formulas for determining the integration of various functions is covered in this chapter. Integral is also known as antiderivative and we also know that integration formulas of some functions are the reverse of their differentiation formulas. This chapter is also considered to be one of the important chapters from the exam point of view.

### Is Chapter 7 of NCERT Solutions for Class 12 Maths difficult to understand?

No, Chapter 7 of NCERT Solutions for Class 12 Maths is not difficult to understand. It is one of the interesting topics in Class 12 which would be continued in higher levels of education also. Good knowledge of integral formulas will make it easier for the students to solve the basic sums of integration efficiently. Proper knowledge of derivatives is necessary to understand the concepts of integral calculus without any difficulty.

### Where can I download the NCERT Solutions for Class 12 Maths Chapter 7 PDF online?

You can find the NCERT Solutions for Class 12 Maths Chapter 7 on BYJUâ€™S. The solutions are prepared by the highly experienced faculty having vast conceptual knowledge. These are considered to be one of the best-rated solutions available online. All the problems from the NCERT textbook are solved in a stepwise manner based on the latest syllabus for 2021-22 and the exam pattern designed by the CBSE board.