NCERT Solutions for Class 12 Maths Chapter 7 Integrals

NCERT Solutions for Class 12 Maths Chapter 7 – Free PDF Download

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

The concepts of integrals given in this chapter are detailed and in easy-to-understand manner. These NCERT Solutions for Class 12 Maths integrals are very simple and can help the students understand the problem-solving method very easily. Students can reach for these NCERT Solutions for Class 12 Maths Chapter 7 and download them 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 the 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) x² + 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) = 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. 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. 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 (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 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 the 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 the 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 the 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 the 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  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 e^x

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. 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 = sec² 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 Exercises 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 the 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 Exercises 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 Exercises 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) π (D) 1

Solution:

(C) π

Explanation:

(A) 2 (B) 3/4 (C) 0 (D) –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, the correct option is (D).

(A) 1 (B) 0 (C) –1 (D) π

Solution:

(B) 0

Explanation:

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 included in CBSE Syllabus 2023-24, which 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 thoroughly understand the concept of Integrals. Chapter 7 of NCERT Solutions for Class 12 Maths discusses the following:

1. Integration is the inverse process of differentiation. In differential calculus, we are given a function, and we have to find the derivative or differential of this function. But in 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 a collection of a 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 × integral of the second function – integral of {differential coefficient of the first function × integral of the second function}.
8. First fundamental theorem of integral calculus.
9. Second fundamental theorem of integral calculus.

Students can utilise the NCERT Solutions for Class 12 Maths Chapter 7 to solve difficult problems, clear doubts, prepare for exams and revise. Referring to these solutions will boost confidence among students to take the Class 12 Maths board exam.

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

Learning the NCERT Solutions of chapter Integrals enables 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.

Disclaimer – 

Dropped Topics – Points (xi)–(xiii) in the List of Derivatives, 7.2.1 Geometrical Interpretation of Indefinite Integral, 7.2.3 Comparison between Differentiation and Integration, 7.6.3 Type of Integral, 7.7.1 Definite Integral as the Limit of a Sum, Ques. 19, 32, 40 and 44 Point 2 in the Summary, (xiv) and (xv) in Some Standard Integrals