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NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.1 - 2025-26

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Maths Class 12 Chapter 7 Questions and Answers - Free PDF Download

In NCERT Solutions Class 12 Maths Chapter 7 Exercise 7 1, you’ll dive into the basics of integrals, learning how to find antiderivatives (indefinite integrals) step by step. This part of calculus is all about reversing differentiation, making it key for topics like area, volume, and even real-life problems in science and engineering.

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If you ever feel confused about when to use formulas or how to start an integral, these solutions are here to make things simple. Vedantu’s easy explanations and downloadable PDFs can help you clear your doubts and practise confidently. For a quick look at what else you’ll study, you can check the Class 12 Maths syllabus.


With clear steps in these NCERT Solutions, you’ll gain a solid grip on the chapter. Don’t forget—this chapter carries 9 marks in your CBSE exam, so practising with these solutions can really boost your score.


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Access NCERT Solutions for Maths Class 12 Chapter 7 Integrals

Exercise 7.1

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

sin $2x$

Ans:  We use the method of inspection as follows:

$\frac{d}{dx}\left( \cos 2x \right)=-2\sin 2x\Rightarrow -\frac{1}{2}\frac{d}{dx}\left( \cos 2x \right) $ 

$\therefore \sin 2x=\frac{d}{dx}\left( -\frac{1}{2}\cos 2x \right) $ 

Thus, the anti-derivative of sin $2x$ is $-\frac{1}{2}\cos 2x$.

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

cos $3x$

Ans: We use the method of inspection as follows:

$\frac{d}{dx}\left( \sin 3x \right)=3\cos 3x\Rightarrow \frac{1}{3}\frac{d}{dx}\left( \sin 3x \right) $ 

$\therefore \cos 3x=\frac{d}{dx}\left( \frac{1}{3}\sin 3x \right) $ 

Thus, the anti - derivative of cos $3x$ is $\frac{1}{3}\sin 3x$.

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

${{e}^{2x}}$

Ans: We use the method of inspection as follows:

$\frac{d}{dx}\left( {{e}^{2x}} \right)\Rightarrow 2{{e}^{2x}}=\frac{1}{2}\frac{d}{dx}\left( {{e}^{2x}} \right) $ 

$\therefore {{e}^{2x}}=\frac{d}{dx}\left( \frac{1}{2}{{e}^{2x}} \right) $ 

Thus, the anti-derivative of ${{e}^{2x}}$ is $\frac{1}{2}{{e}^{2x}}$.

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

${{\left( ax+b \right)}^{2}}$

Ans: We use the method of inspection as follows:

$\frac{d}{dx}{{\left( ax+b \right)}^{3}}=3a{{\left( ax+b \right)}^{2}} $ 

$\Rightarrow {{\left( ax+b \right)}^{2}}=\frac{1}{3a}\frac{d}{dx}{{\left( ax+b \right)}^{3}} $ 

$\therefore {{\left( ax+b \right)}^{2}}=\frac{d}{dx}\left( \frac{1}{3a}{{\left( ax+b \right)}^{3}} \right) $ 

Thus, the anti-derivative of ${{\left( ax+b \right)}^{2}}$ is $\frac{1}{3a}{{\left( ax+b \right)}^{3}}$.

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

sin $2x-4{{e}^{3x}}$

Ans: We use the method of inspection as follows:

$\frac{d}{dx}\left( -\frac{1}{2}\cos 2x-\frac{4}{3}{{e}^{3x}} \right)=\left( \sin 2x-4{{e}^{3x}} \right)$

Thus, the anti-derivative of $\left( \sin 2x-4{{e}^{3x}} \right)$ is $\left( -\frac{1}{2}\cos 2x-\frac{4}{3}{{e}^{3x}} \right)$.

6. $\int{\left( 4{{e}^{3x}}+1 \right)dx}$

Ans: 

$\int{\left( 4{{e}^{3x}}+1 \right)dx} $ 

$=4\int{{{e}^{3x}}dx}+\int{1}dx $ 

$=4\left( \frac{{{e}^{3x}}}{3} \right)+x+C $ 

$=\frac{4}{3}{{e}^{3x}}+x+C $

7. $\int{{{x}^{2}}\left( 1-\frac{1}{{{x}^{2}}} \right)dx}$

Ans: 

$\int{{{x}^{2}}\left( 1-\frac{1}{{{x}^{2}}} \right)dx} $ 

$=\int{\left( {{x}^{2}}-1 \right)dx} $ 

$=\frac{{{x}^{3}}}{3}-x+C $

8. $\int{\left( a{{x}^{2}}+bx+c \right)dx}$

Ans:

$\int{\left( a{{x}^{2}}+bx+c \right)}dx $ 

$=a\int{{{x}^{2}}dx+b\int{xdx+c\int{1.dx}}} $ 

$=a\left( \frac{{{x}^{3}}}{3} \right)+b\left( \frac{{{x}^{2}}}{2} \right)+cx+D $ 

$=\frac{a{{x}^{3}}}{3}+\frac{b{{x}^{2}}}{2}+cx+D $

9. $\int{\left( 2{{x}^{2}}+{{e}^{x}} \right)dx}$

Ans:  

$\int{\left( 2{{x}^{2}}+{{e}^{x}} \right)dx} $ 

$=2\int{{{x}^{2}}dx+\int{{{e}^{x}}dx}} $ 

$=2\left( \frac{{{x}^{3}}}{3} \right)+{{e}^{x}}+C $ 

$=\frac{2}{3}{{x}^{3}}+{{e}^{x}}+C $

10. $\int{{{\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)}^{2}}dx}$

Ans:

$\int{{{\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)}^{2}}}dx $ 

$=\int{\left( x+\frac{1}{x}-2 \right)dx} $ 

$=\int{xdx}+\int{\frac{1}{x}dx}-2\int{1.dx} $ 

$=\frac{{{x}^{2}}}{2}+\log \left| x \right|-2x+C $

11. $\int{\frac{{{x}^{3}}+5{{x}^{2}}-4}{{{x}^{2}}}dx}$

Ans: $\int{\frac{{{x}^{3}}+5{{x}^{2}}-4}{{{x}^{2}}}dx}$

$=\int{\left( x+5-4{{x}^{-2}} \right)dx} $ 

$=\int{xdx}+5\int{1.dx}-4\int{{{x}^{-2}}dx} $ 

$=\frac{{{x}^{2}}}{2}+5x+\frac{4}{x}+C $

12. $\int{\frac{{{x}^{3}}+3x+4}{\sqrt{x}}dx}$

Ans:$\int{\frac{{{x}^{3}}+3x+4}{\sqrt{x}}dx}$

$=\int{\left( {{x}^{\frac{5}{2}}}+3{{x}^{\frac{1}{2}}}+4{{x}^{-\frac{1}{2}}} \right)dx} $ 

$=\frac{{{x}^{\frac{7}{2}}}}{\frac{7}{2}}+\frac{3\left( {{x}^{\frac{3}{2}}} \right)}{\frac{3}{2}}+\frac{4\left( {{x}^{\frac{1}{2}}} \right)}{\frac{1}{2}}+C $ 

$=\frac{2}{7}{{x}^{\frac{7}{2}}}+2{{x}^{\frac{3}{2}}}+8\sqrt{x}+C $

13. $\int{\frac{{{x}^{3}}-{{x}^{2}}+x-1}{x-1}dx}$

Ans: $\int{\frac{{{x}^{3}}-{{x}^{2}}+x-1}{x-1}dx}$

$=\int{\frac{{{x}^{2}}(x-1)+x-1}{x-1}dx}$

$=\int{\frac{(x-1)({{x}^{2}}+1)}{x-1}}$

We obtain, on dividing:

$=\int{\left( {{x}^{2}}+1 \right)dx} $ 

$=\int{{{x}^{2}}dx}+\int{1.dx} $ 

$=\frac{{{x}^{3}}}{3}+x+C $

14. $\int{\left( 1-x \right)}\sqrt{x}dx$

Ans: $\int{\left( 1-x \right)}\sqrt{x}dx$

$=\int{\left( \sqrt{x}-{{x}^{\frac{3}{2}}} \right)dx} $ 

$=\int{{{x}^{\frac{1}{2}}}dx-\int{{{x}^{\frac{3}{2}}}dx}} $ 

$=\frac{2}{3}{{x}^{\frac{3}{2}}}-\frac{2}{5}{{x}^{\frac{5}{2}}}+C $

15. $\int{\sqrt{x}\left( 3{{x}^{2}}+2x+3 \right)dx}$

Ans: $\int{\sqrt{x}\left( 3{{x}^{2}}+2x+3 \right)dx}$

$=3\int{\left( {{x}^{\frac{5}{2}}}+2{{x}^{\frac{3}{2}}}+3{{x}^{\frac{1}{2}}} \right)} $  $=3\int{{{x}^{\frac{5}{2}}}dx+2\int{{{x}^{\frac{3}{2}}}dx+3\int{{{x}^{\frac{1}{2}}}}dx}} $  $=\frac{6}{7}{{x}^{\frac{7}{2}}}+\frac{4}{5}{{x}^{\frac{5}{2}}}+2{{x}^{\frac{3}{2}}}+C $

16. $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$

Ans: $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$          

$=2\int{xdx-3\int{\cos xdx}+\int{{{e}^{x}}dx}} $ 

$=\frac{2{{x}^{2}}}{2}-3\left( \sin x \right)+{{e}^{x}}+C $ 

$={{x}^{2}}-3\sin x+{{e}^{x}}+C $

17. $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

Ans: $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

  $=2\int{{{x}^{2}}dx-3\int{\sin xdx+5\int{{{x}^{\frac{1}{2}}}}dx}} $ 

 $=\frac{2{{x}^{3}}}{3}-3\left( -\cos x \right)+5\left( \frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}} \right)+C $ 

 $=\frac{2}{3}{{x}^{3}}+3\cos x+\frac{10}{3}{{x}^{\frac{3}{2}}}+C $

18. $\int{\sec x\left( \sec x+\tan x \right)dx}$

Ans: $\int{\sec x\left( \sec x+\tan x \right)dx}$

$=\int{\left( {{\sec }^{2}}x+\sec x\tan x \right)dx} $ 

$=\int{{{\sec }^{2}}xdx+\int{\sec x\tan xdx}} $ 

$=\tan x+\sec x+C $

19. $\int{\frac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

Ans: $\int{\frac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

$=\int{\frac{\frac{1}{{{\cos }^{2}}x}}{\frac{1}{{{\sin }^{2}}x}}dx} $ 

$=\int{\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}dx} $ 

$=\int{{{\tan }^{2}}xdx} $ 

$=\int{{{\sec }^{2}}xdx}-\int{1dx} $ 

$=\tan x-x+C $

20. $\int{\frac{2-3\sin x}{{{\cos }^{2}}x}dx}$

Ans: $\int{\frac{2-3\sin x}{{{\cos }^{2}}x}dx}$

$=\int{\left( \frac{2}{{{\cos }^{2}}x}-\frac{3\sin x}{{{\cos }^{2}}x} \right)dx} $ 

$=\int{2{{\sec }^{2}}xdx}-3\int{\tan x\sec xdx} $ 

$=2\tan x-3\sec x+C $ 

21. The anti – derivative of $\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)$ equals

  1. $\frac{1}{3}{{x}^{\frac{1}{3}}}+2{{x}^{\frac{1}{2}}}+C$

  2. $\frac{2}{3}{{x}^{\frac{2}{3}}}+\frac{1}{2}{{x}^{2}}+C$

  3. $\frac{2}{3}{{x}^{\frac{3}{2}}}+2{{x}^{\frac{1}{2}}}+C$

  4. $\frac{3}{2}{{x}^{\frac{3}{2}}}+\frac{1}{2}{{x}^{\frac{1}{2}}}+C$

Ans:

  $\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right) $  $=\int{{{x}^{\frac{1}{2}}}dx}+\int{{{x}^{-\frac{1}{2}}}}dx=\frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}}+\frac{{{x}^{\frac{1}{2}}}}{\frac{1}{2}}+C $ 

 $=\frac{2}{3}{{x}^{\frac{3}{2}}}+2{{x}^{\frac{1}{2}}}+C $ 

Thus, the correct answer is C.

22. If $\frac{d}{dx}f\left( x \right)=4{{x}^{3}}-\frac{3}{{{x}^{4}}}$ such that $f\left( 2 \right)=0$ then $f\left( x \right)$ is

  1. ${{x}^{4}}+\frac{1}{{{x}^{3}}}-\frac{129}{8}$

  2. ${{x}^{3}}+\frac{1}{{{x}^{4}}}+\frac{129}{8}$

  3. ${{x}^{4}}+\frac{1}{{{x}^{3}}}+\frac{129}{8}$

  4. ${{x}^{3}}+\frac{1}{{{x}^{4}}}-\frac{129}{8}$

Ans: Given, $\frac{d}{dx}f\left( x \right)=4{{x}^{3}}-\frac{3}{{{x}^{4}}}$

  Anti-derivative of $4{{x}^{3}}-\frac{3}{{{x}^{4}}}=f\left( x \right)$

 $\therefore f\left( x \right)=\int{4{{x}^{3}}-\frac{3}{{{x}^{4}}}=f\left( x \right)} $ 

 $f\left( x \right)=4\int{{{x}^{3}}dx-3\int{\left( {{x}^{-4}} \right)}dx} $ 

$f\left( x \right)=4\left( \frac{{{x}^{4}}}{4} \right)-3\left( \frac{{{x}^{-3}}}{-3} \right)+C $ 

$f\left( x \right)={{x}^{4}}+\frac{1}{{{x}^{3}}}+C $ 

$Also, $ 

$f\left( 2 \right)=0 $ 

$\therefore f\left( 2 \right)={{\left( 2 \right)}^{4}}+\frac{1}{{{\left( 2 \right)}^{3}}}+C=0 $ 

$\Rightarrow 16+\frac{1}{8}+C=0 $ 

$\Rightarrow C=\frac{-129}{8} $ 

$\therefore f\left( x \right)={{x}^{4}}+\frac{1}{{{x}^{3}}}-\frac{129}{8} $ 

Thus, the correct answer is A.

Conclusion

In conclusion, Exercise 7.1 Class 12 Maths is a crucial stepping stone in mastering the concept of integrals, a fundamental aspect of calculus. By focusing on the basics of indefinite integrals and their standard formulas, Ex 7.1 Class 12 helps students build a strong foundation. Completing these problems not only enhances problem-solving skills but also prepares students for more advanced integration techniques in subsequent exercises and chapters.


Practice and ensure a thorough understanding, as the concepts learned here are integral to various applications in mathematics and beyond. Keep practicing, and students will find themselves well-prepared for more complex calculus challenges ahead!


Class 12 Maths Chapter 7: Exercises Breakdown

S.No.

Chapter 7 - Integrals Exercises in PDF Format

1

Class 12 Maths Chapter 7 Exercise 7.2 - 39 Questions & Solutions (37 Short Answers, 2 MCQs)

2

Class 12 Maths Chapter 7 Exercise 7.3 - 24 Questions & Solutions (22 Short Answers, 2 MCQs)

3

Class 12 Maths Chapter 7 Exercise 7.4 - 25 Questions & Solutions (23 Short Answers, 2 MCQs)

4

Class 12 Maths Chapter 7 Exercise 7.5 - 23 Questions & Solutions (21 Short Answers, 2 MCQs)

5

Class 12 Maths Chapter 7 Exercise 7.6 - 24 Questions & Solutions (22 Short Answers, 2 MCQs)

6

Class 12 Maths Chapter 7 Exercise 7.7 - 11 Questions & Solutions (9 Short Answers, 2 MCQs)

7

Class 12 Maths Chapter 7 Exercise 7.8 - 6 Questions & Solutions (6 Short Answers)

8

Class 12 Maths Chapter 7 Exercise 7.9 - 22 Questions & Solutions (20 Short Answers, 2 MCQs)

9

Class 12 Maths Chapter 7 Exercise 7.10 - 10 Questions & Solutions (8 Short Answers, 2 MCQs)

10

Class 12 Maths Chapter 7 Miscellaneous Exercise - 40 Questions & Solutions



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FAQs on NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.1 - 2025-26

1. What is the correct step-by-step method to solve problems in NCERT Class 12 Maths Chapter 7, Integrals?

To solve integration problems from Chapter 7 as per the CBSE pattern, follow this standard methodology:

  • Identify the Integrand: First, carefully examine the function to be integrated to determine its type (e.g., algebraic, trigonometric, exponential).
  • Choose the Right Method: Select the appropriate integration technique. For basic functions, use standard formulas. For complex functions, use integration by substitution, by parts, or by partial fractions.
  • Apply the Method: Execute the chosen method step-by-step, showing all transformations and substitutions clearly.
  • Integrate: Perform the integration using the relevant formulas.
  • Add the Constant of Integration: For indefinite integrals, always add the constant '+ C' to your final answer to represent the entire family of antiderivatives.

2. What are the fundamental integration formulas I must know for Chapter 7?

Mastering Chapter 7 requires a strong command of the following essential integration formulas from the NCERT syllabus:

  • ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C, where n ≠ -1
  • ∫(1/x) dx = ln|x| + C
  • ∫eˣ dx = eˣ + C
  • ∫aˣ dx = (aˣ / ln a) + C
  • ∫sin(x) dx = -cos(x) + C
  • ∫cos(x) dx = sin(x) + C
  • ∫sec²(x) dx = tan(x) + C
  • ∫csc²(x) dx = -cot(x) + C
  • ∫sec(x)tan(x) dx = sec(x) + C
  • ∫csc(x)cot(x) dx = -csc(x) + C

3. How do I decide when to use integration by substitution, by parts, or by partial fractions?

Choosing the correct integration technique is crucial. Here’s a simple guide:

  • Use Integration by Substitution when the integrand contains a function and its derivative (e.g., ∫f(g(x))g'(x)dx). This method simplifies the integral into a standard form.
  • Use Integration by Parts when the integrand is a product of two different types of functions, like logarithmic, inverse, algebraic, trigonometric, or exponential (follow the ILATE rule). It is used for integrals like ∫x cos(x) dx.
  • Use Integration by Partial Fractions when the integrand is a rational function (a ratio of two polynomials) where the denominator can be factorised.

4. What is the correct approach for solving the Miscellaneous Exercise for Integrals in Chapter 7?

The Miscellaneous Exercise is designed to test your comprehensive understanding of the entire chapter. The best approach is to:

  • Revise all concepts: Before attempting, thoroughly review all integration methods—substitution, by parts, partial fractions, and properties of definite integrals.
  • Identify the core technique: Problems in this exercise often require a combination of techniques or a clever substitution. Analyse the problem carefully to identify the most efficient starting point.
  • Don't give up: These questions are often more challenging and are excellent practice for higher-order thinking skill (HOTS) questions in board exams. Refer to step-by-step solutions to understand the logic if you get stuck.

5. Why is adding the constant of integration, '+ C', necessary in every indefinite integral?

The constant of integration, '+ C', is fundamental because integration is the reverse process of differentiation. The derivative of any constant (like 5, -10, or π) is zero. Therefore, when we find an antiderivative F(x) for a function f(x), we don't know if the original function had a constant term. The '+ C' represents this unknown constant and acknowledges that there is an entire family of functions (e.g., x² + 2, x² - 100) that have the same derivative.

6. What is the main difference between solving an indefinite integral and a definite integral?

The key difference lies in their results and interpretation. An indefinite integral, like ∫f(x)dx, gives a general function or a family of functions, F(x) + C, as its answer. In contrast, a definite integral, like ∫ₐᵇ f(x)dx, gives a single numerical value. This value is calculated using the Fundamental Theorem of Calculus by finding the antiderivative and evaluating it at the upper and lower limits (F(b) - F(a)).

7. I find trigonometric integrals confusing. What is a reliable strategy for integrals involving trigonometric identities?

The primary strategy for solving complex trigonometric integrals is to simplify the integrand before integrating. Your goal is to convert the expression into a form where standard integral formulas can be applied. Key tactics include:

  • Using power-reducing formulas for terms like sin²(x) or cos²(x).
  • Applying product-to-sum formulas for products like sin(ax)cos(bx).
  • Making appropriate substitutions (e.g., u = sin(x) or u = tan(x)) after simplifying.
  • Rewriting the integrand in terms of sin(x) and cos(x) if it involves other trig functions.

8. How do the NCERT Solutions for Class 12 Integrals help in preparing for board exams?

The NCERT Solutions for Class 12 Integrals are essential for board exam preparation because they demonstrate the exact step-by-step methodology that examiners expect. By following these solutions, you learn how to present your answers clearly, apply formulas correctly, and avoid common errors. Mastering them ensures you build a strong foundation that aligns perfectly with the CBSE 2025-26 syllabus, which is crucial for scoring full marks.

9. What is the learning objective of the 'method of inspection' in the first exercise of Integrals?

The 'method of inspection' serves a critical pedagogical purpose beyond just finding an answer. Its main objective is to build a deep, intuitive understanding of the inverse relationship between differentiation and integration. By asking you to guess the antiderivative, it forces you to think backwards: "Which function, when I differentiate it, gives me this result?" This strengthens your mental connection to derivative formulas and solidifies the core concept of an integral as an antiderivative.