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# NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.9

Last updated date: 24th Jul 2024
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## NCERT Solutions for Class 12 Maths Chapter 7 (Ex 7.9) Integrals

The topic of "Evaluation of Definite Integrals by Substitution" is the main focus of the NCERT Solutions for Maths Exercise 7.9 in Class 12 Chapter 7 - Integrals by Vedantu. This exercise follows a fundamental mathematics concept: the substitution method for solving integrals. Students improve a full understanding of the method by practicing these tasks.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 7 (Ex 7.9) Integrals
2. Glance on NCERT Solutions Maths Chapter 7 Exercise 7.9 Class 12 | Vedantu
3. Formulas Used in Class 12 Chapter 7 Exercise 7.9
4. Access NCERT Solutions for Maths Class 12 Chapter 7 - Integrals
4.1Exercise 7.9
5. Class 12 Maths Chapter 7: Exercises Breakdown
6. CBSE Class 12 Maths Chapter 7 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Exercise 7.9 contains 10 questions designed to give students plenty of practice with substitution in definite integrals from NCERT Solutions for Class 12 Maths. It is important that one focus on learning the substitution method and using it to solve different kinds of problems. The goal of these solutions is to help students become better at solving problems and to increase their exam confidence by using CBSE Class 12 Maths Syllabus.

## Glance on NCERT Solutions Maths Chapter 7 Exercise 7.9 Class 12 | Vedantu

• Exercise 7.9 Class 12, focuses on the evaluation of definite integrals using the substitution method.

• Definite Integral: This is the calculation of the integral of a function over a defined interval [a, b], which represents the total area under the curve between these points.

• Substitution Method: This technique involves simplifying integrals by substituting part of the integrand with a new variable.

• Integration by Substitution: This process entails replacing a complex expression with a single variable to make integration easier.

• Limits of Integration: These limits define the interval for evaluating the definite integral and are adjusted based on the substitution.

• Change of Variable: This method involves replacing the original variable with a new variable to simplify the integration.

• Integral Bounds Adjustment: Following substitution, the original integration limits are changed to match the new variable.

• Antiderivative: This is the function that results in the original integrand when differentiated, found through the substitution method in definite integrals.

• There are 10 fully solved questions in Chapter 7 Exercise 7.9 Class 12 Integrals.

## Formulas Used in Class 12 Chapter 7 Exercise 7.9

• Basic Substitution Formula: $\int f\left ( g(x)\right ){g}'(x)dx=\int f(u)du$

• Integration by Substitution: $\int f(x)dx=\int f(g(t))\cdot {g}'(t)dt$

• Definite Integral Formula: $\int_{a}^{b}f(x) dx$

Competitive Exams after 12th Science

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

### Exercise 7.9

Evaluate the integrals in Exercises 1 to 8 using substitution.

1. $\int\limits_0^1 {\dfrac{x}{{{x^2} + 1}}} dx$.

Ans: To simplify the question Let us suppose ${x^2} + 1 = t$

Derivative of it will be, $2xdx = dt$

When $x = 0$, then $t = 1$ again when $x = 1$ then $t = 2$.

$\therefore \int\limits_0^1 {\dfrac{x}{{{x^2} + 1}}} dx = \dfrac{1}{2}\int\limits_1^2 {\dfrac{{dt}}{t}}$

$= \dfrac{1}{2}[\log \left| t \right|]_1^2$

$= \dfrac{1}{2}[\log 2 - \log 1]$

$= \dfrac{1}{2}\log 2(\because \log 1 = 0)$

Thus, the answer is $\dfrac{1}{2}\log 2$.

2. $\int\limits_0^{\dfrac{\pi }{2}} {\sqrt {\sin \phi {{\cos }^5}\phi d\phi } }$

Ans:To simplify the question, let us suppose

$\sin \phi = t \Rightarrow \cos \phi d\phi = dt$

When, $\phi = 0,t = 0$and when $\phi = \dfrac{\pi }{2},t = 1$

$\therefore I = \int\limits_0^1 {\sqrt t {{(1 - {t^2})}^2}dt}$

$= \int\limits_0^1 {\sqrt t {{(1 - {t^2})}^2}dt}$

$= \int\limits_0^1 {\sqrt t {{(1 - {t^2})}^2}dt}$

$= \int\limits_0^1 {{t^{\dfrac{1}{2}}}(1 + {t^4} - 2{t^2})dt}$

$= \int\limits_0^1 {[{t^{\dfrac{1}{2}}} + {t^{\dfrac{9}{2}}} - 2{t^{\dfrac{5}{2}}}]dt}$

$= [\dfrac{{{t^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}} + \dfrac{{{t^{\dfrac{{11}}{2}}}}}{{\dfrac{{11}}{2}}} + \dfrac{{2{t^{\dfrac{7}{2}}}}}{{\dfrac{7}{2}}}]_0^1$

$= \dfrac{2}{3} + \dfrac{2}{{11}} - \dfrac{4}{7}$

$= \dfrac{{154 + 42 - 132}}{{231}}$

$= \dfrac{{64}}{{231}}$

3.$\int\limits_0^1 {{{\sin }^{ - 1}}} (\dfrac{{2x}}{{1 + {x^2}}})dx$

Ans: To simplify the question, let us suppose$x = \tan \theta$

After differentiating both side we get, $dx = {\sec ^2}\theta d\theta$

$I = \int\limits_0^{\dfrac{\pi }{4}} {{{\sin }^{ - 1}}(\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }})} {\sec ^2}\theta d\theta$

$= \int\limits_0^{\dfrac{\pi }{4}} {{{\sin }^{ - 1}}(\sin 2\theta )} {\sec ^2}\theta d\theta$

$= \int\limits_0^{\dfrac{\pi }{4}} {2\theta } {\sec ^2}\theta d\theta$

$= 2\int\limits_0^{\dfrac{\pi }{4}} \theta {\sec ^2}\theta d\theta$

Taking θ as first function and ${\sec ^2}\theta$ as second function and integrating by parts, we obtain

$I = 2[\theta \int {{{\sec }^2}} \theta d\theta - \int {\{ (\dfrac{d}{{dx}}\theta )\int {{{\sec }^2}\theta d\theta } \} d\theta } ]_0^{\dfrac{\pi }{4}}$

$= 2[\theta \tan \theta - \int {\tan \theta d\theta } ]_0^{\dfrac{\pi }{4}}$

$= 2[\theta \tan \theta + \log \left| {\cos \theta } \right|]_0^{\dfrac{\pi }{4}}$

$= 2[\dfrac{\pi }{4}\tan \dfrac{\pi }{4} + \log \left| {\cos \dfrac{\pi }{4}} \right| - \log \left| {\cos 0} \right|]$

$= 2[\dfrac{\pi }{4} + \log (\dfrac{1}{{\sqrt 2 }}) - \log 1]$

$= 2[\dfrac{\pi }{4} - \dfrac{1}{2}\log 2]$

$= \dfrac{\pi }{2} - \log 2$

4. $\int\limits_0^2 {x\sqrt {x + 2} } (Put(x + 2 = {t^2}))$

Ans: To simplify the question let us suppose

$x + 2 = {t^2}$

After differentiating both side we get,

$dx = 2tdt$

When $x = 0$, then t=$t = \sqrt 2$and when $x = 2$ , $t = 2$

$\int\limits_0^2 {x\sqrt {x + 2} dx}$

$= \int\limits_{\sqrt 2 }^2 {({t^2} - 2)\sqrt {{t^2}} 2dt}$

$= 2\int\limits_{\sqrt 2 }^2 {({t^3} - 2t)dt}$

$= 2[\dfrac{{{t^4}}}{4} - {t^2}]_{\sqrt 2 }^2$

$= 2[(4 - 4) - (1 - 2)]$

$= 2$

5. $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx}$

Ans: To simplify the question, let us suppose

$\cos x = t$

After differentiating both side we get, $- \sin xdx = dt$

$\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx} = - \int\limits_1^0 {\dfrac{{dt}}{{1 + {t^2}}}}$

$= - [{\tan ^{ - 1}}0 - {\tan ^{ - 1}}1]$

$= - [ - \dfrac{\pi }{4}]$

$= \dfrac{\pi }{4}$

6. $\int\limits_0^2 {\dfrac{{dx}}{{x + 4 - {x^2}}}}$

Ans: To simplify the question it is written as $\int\limits_0^2 {\dfrac{{dx}}{{x + 4 - {x^2}}} = \int\limits_0^2 {\dfrac{{dx}}{{ - ({x^2} - x - 4)}}} }$

Thus,

$\int\limits_0^2 {\dfrac{{dx}}{{x + 4 - {x^2}}} = \int\limits_0^2 {\dfrac{{dx}}{{ - ({x^2} - x - 4)}}} }$

$= \int\limits_0^2 {\dfrac{{dx}}{{ - ({x^2} - x + \dfrac{1}{4} - \dfrac{1}{4} - 4)}}}$

$= \int\limits_0^2 {\dfrac{{dx}}{{ - [{{(x - \dfrac{1}{2})}^2} - \dfrac{{17}}{4}]}}}$

$= \int\limits_0^2 {\dfrac{{dx}}{{{{(\dfrac{{\sqrt {17} }}{2})}^2} - {{(x - \dfrac{1}{2})}^2}}}}$

Let $x - \dfrac{1}{2} = t$

So, $dx = dt$

When $x = 0,t = - \dfrac{1}{2}$and when $x = 2,t = \dfrac{3}{2}$

$\int\limits_0^2 {\dfrac{{dx}}{{2(\dfrac{{\sqrt {17} }}{2}) - {{(x - \dfrac{1}{2})}^2}}}}$

$= \int\limits_{\dfrac{{ - 1}}{2}}^{\dfrac{3}{2}} {\dfrac{{dt}}{{{{(\dfrac{{\sqrt {17} }}{2})}^2} - {t^2}}}}$

$= [\dfrac{1}{{2(\dfrac{{\sqrt {17} }}{2})}}\log \dfrac{{\dfrac{{\sqrt {17} }}{2} + t}}{{\dfrac{{\sqrt {17} }}{2} - t}}]_{\dfrac{{ - 1}}{2}}^{\dfrac{3}{2}}$

$= \dfrac{1}{{\sqrt {17} }}[\log \dfrac{{\dfrac{{\sqrt {17} }}{2} + \dfrac{3}{2}}}{{\dfrac{{\sqrt {17} }}{2} - \dfrac{3}{2}}} - \dfrac{{\log \dfrac{{\sqrt {17} }}{2} - \dfrac{1}{2}}}{{\log \dfrac{{\sqrt {17} }}{2} + \dfrac{1}{2}}}]$

$= \dfrac{1}{{\sqrt {17} }}[\log \dfrac{{\sqrt {17} + 3}}{{\sqrt {17} - 3}} - \log \dfrac{{\sqrt {17} - 1}}{{\sqrt {17} + 1}}]$

$= \dfrac{1}{{\sqrt {17} }}[\log \dfrac{{\sqrt {17} + 3}}{{\sqrt {17} - 3}} \times \dfrac{{\sqrt {17} - 1}}{{\sqrt {17} + 1}}]$

$= \dfrac{1}{{\sqrt {17} }}[\log \dfrac{{\sqrt {17} + 3 + 4\sqrt {17} }}{{\sqrt {17} + 3 - 4\sqrt {17} }}]$

$= \dfrac{1}{{\sqrt {17} }}[\log \dfrac{{20 + 4\sqrt {17} }}{{20 - 4\sqrt {17} }}]$

$= \dfrac{1}{{\sqrt {17} }}\log (\dfrac{{5 + \sqrt {17} }}{{5 - \sqrt {17} }})$

$= \dfrac{1}{{\sqrt {17} }}\log [\dfrac{{(5 + \sqrt {17} )(5 + \sqrt {17} )}}{{25 - 17}}]$

$= \dfrac{1}{{\sqrt {17} }}\log [\dfrac{{(25 + 17 + 10\sqrt {17} }}{8}]$

$= \dfrac{1}{{\sqrt {17} }}\log [\dfrac{{(42 + 10\sqrt {17} }}{8}]$

$= \dfrac{1}{{\sqrt {17} }}\log [\dfrac{{(21 + 5\sqrt {17} }}{4}]$

7. $\int\limits_{ - 1}^1 {\dfrac{{dx}}{{{x^2} + 2x + 5}}}$

Ans: To simplify the question it is written as,

$\int\limits_{ - 1}^1 {\dfrac{{dx}}{{{x^2} + 2x + 1 + 4}}}$

$= \int\limits_{ - 1}^1 {\dfrac{{dx}}{{({x^2} + 2x + 1) + 4}}}$

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

Let $x + 1 = t$

Differentiating both side we get, $dx = dt$

When $x = - 1,t = 0$and when $x = 1,t = 2$

$\int\limits_{ - 1}^1 {\dfrac{{dx}}{{{{(x + 1)}^2} + {2^2}}}}$

$= \int\limits_0^2 {\dfrac{{dx}}{{{t^2} + {2^2}}}}$

$= [\dfrac{1}{2}{\tan ^{ - 1}}\dfrac{t}{2}]_0^2$

$= \dfrac{1}{2}{\tan ^{ - 1}}1 - \dfrac{1}{2}{\tan ^{ - 1}}0$

$= \dfrac{1}{2}(\dfrac{\pi }{4})$

$= \dfrac{\pi }{8}$

8. $\int\limits_1^2 {(\dfrac{1}{x} - \dfrac{1}{{2{x^2}}}){e^{2x}}dx}$

Ans: Let $2x = t$

After differentiating both side we get, $2dx = dt$

When $x = 1$, $t = 2$ and $x = 2$, $t = 4$

$\int\limits_1^2 {(\dfrac{1}{x} - \dfrac{1}{{2{x^2}}}){e^{2x}}dx}$

$= \dfrac{1}{2}\int\limits_2^4 {(\dfrac{2}{{{t^2}}} - \dfrac{2}{{{t^2}}})} {e^t}dt$

Let, $\dfrac{1}{t} = f(t)$

$f'(t) = - \dfrac{1}{{{t^2}}}$

$\int\limits_2^4 {(\dfrac{1}{x} - \dfrac{1}{{2{x^2}}}){e^{2x}}dx = \dfrac{1}{2}\int\limits_2^4 {(\dfrac{2}{t} - \dfrac{2}{{{t^2}}}){e^t}dt} }$

$= [{e^t}\dfrac{1}{t}]_2^4$

$= [\dfrac{{{e^t}}}{t}]_2^4$

$= \dfrac{{{e^4}}}{t} - \dfrac{{{e^2}}}{t}$

$= \dfrac{{{e^2}({e^2} - 2)}}{4}$

Choose the Correct Answer in Question 9 and 10.

9.The value of the integral $\int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{(x - {x^3})}^{\dfrac{1}{3}}}}}{{{x^4}}}dx}$ is,

Ans: Let $I = \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{(x - {x^3})}^{\dfrac{1}{3}}}}}{{{x^4}}}dx}$

Also, to simplify the question let, $x = \sin \theta$

After differentiating both side we get $dx = \cos \theta d\theta$

When,$x = \dfrac{1}{3},\theta = {\sin ^{ - 1}}(\dfrac{1}{3})$ and when$x = 1,\theta = \dfrac{\pi }{2}$

$\Rightarrow I = \int\limits_{{{\sin }^{ - 1}}(\dfrac{1}{3})}^{\dfrac{\pi }{2}} {\dfrac{{\sin \theta - {{\sin }^3}\theta }}{{{{\sin }^4}\theta }}} \cos \theta d\theta$

$= \int\limits_{{{\sin }^{ - 1}}(\dfrac{1}{3})}^{\dfrac{\pi }{2}} {\dfrac{{{{(\sin \theta )}^{\dfrac{1}{3}}}{{(1 - {{\sin }^2}\theta )}^{\dfrac{1}{3}}}}}{{{{\sin }^4}\theta }}} \cos \theta d\theta$

$= \int\limits_{{{\sin }^{ - 1}}(\dfrac{1}{3})}^{\dfrac{\pi }{2}} {\dfrac{{{{(\sin \theta )}^{\dfrac{1}{3}}}{{(\cos \theta )}^{\dfrac{2}{3}}}}}{{{{\sin }^4}\theta }}} \cos \theta d\theta$

$= \int\limits_{{{\sin }^{ - 1}}(\dfrac{1}{3})}^{\dfrac{\pi }{2}} {\dfrac{{{{(\sin \theta )}^{\dfrac{1}{3}}}{{(\cos \theta )}^{\dfrac{2}{3}}}}}{{{{\sin }^2}\theta {{\sin }^2}\theta }}} \cos \theta d\theta$

$= \int\limits_{{{\sin }^{ - 1}}(\dfrac{1}{3})}^{\dfrac{\pi }{2}} {\dfrac{{{{(\cos \theta )}^{\dfrac{5}{3}}}}}{{{{(\sin \theta )}^{\dfrac{5}{3}}}}}} \cos e{c^2}\theta d\theta$

$= \int\limits_{{{\sin }^{ - 1}}(\dfrac{1}{3})}^{\dfrac{\pi }{2}} {{{(\cot \theta )}^{\dfrac{5}{3}}}} \cos e{c^2}\theta d\theta$

Let $\cot \theta = t$

After differentiating both side we get, $- \cos e{c^2}\theta d\theta = dt$

When $\theta = {\sin ^{ - 1}}(\dfrac{1}{3}),t = 2\sqrt 2$and when $\theta = \dfrac{\pi }{2},t = 0$

$\therefore I = - \int\limits_{2\sqrt 2 }^0 {{{(t)}^{\dfrac{5}{3}}}dt}$

$= - [\dfrac{3}{8}{(t)^{\dfrac{8}{3}}}]_{2\sqrt 2 }^0$

$= - \dfrac{3}{8}[ - {(2\sqrt 2 )^{\dfrac{8}{3}}}]_{2\sqrt 2 }^0$

$= \dfrac{3}{8}[{(\sqrt 8 )^{\dfrac{8}{3}}}]$

$= \dfrac{3}{8}[{(8)^{\dfrac{4}{3}}}]$

$= \dfrac{3}{8}[16]$

$= 3 \times 2$

$= 6$

Hence, the correct Answer is option(a) 6.

10. If $f(x) = \int\limits_0^x {t\sin tdt,}$ then $f'(x)$ is

1. $\cos x + x\sin x$

2. $x\sin x$

3. $x\cos x$

4. $\sin x + x\cos x$

Ans: $f(x) = \int\limits_0^x {t\sin tdt}$

Integrating by parts, we obtain

$f(x) = t\int\limits_0^x {\sin tdt} - \int\limits_0^x {\{ (\dfrac{d}{{dx}}t)\int {\sin tdt} \} dt}$

$= [t( - \cos t)]_0^x - \int\limits_0^x {( - \cot t)dt}$

$= [ - t\cos t + \sin x]_0^x$

$= [ - x\cos x + \sin t]_0^x$

$= - x\cos x + \sin x$

$\Rightarrow f'(x) = - [\{ x( - \sin x)\} + \cos x] + \cos x$

$= x\sin x - \cos x + \cos x$

$= x\sin x$

Hence, the correct answer is b)$x\sin x$.

## Conclusion

The NCERT Solutions for Maths Exercise 7.9 in Class 12 Chapter 7 - Integrals by Vedantu helps you learn definite integrals using substitution. This exercise is important for understanding how to use substitution to solve integrals. By practicing these solutions, you can improve your problem-solving skills and feel more confident for exams. Focus on understanding each step and practice the questions thoroughly to get a strong grasp of the concepts.

## Class 12 Maths Chapter 7: Exercises Breakdown

 Exercise Number of Questions Exercise 7.1 22 Questions & Solutions (21 Short Answers, 1 MCQs) Exercise 7.2 39 Questions & Solutions (37 Short Answers, 2 MCQs) Exercise 7.3 24 Questions & Solutions (22 Short Answers, 2 MCQs) Exercise 7.4 25 Questions & Solutions (23 Short Answers, 2 MCQs) Exercise 7.5 23 Questions & Solutions (21 Short Answers, 2 MCQs) Exercise 7.6 24 Questions & Solutions (22 Short Answers, 2 MCQs) Exercise 7.7 11 Questions & Solutions (9 Short Answers, 2 MCQs) Exercise 7.8 22 Questions & Solutions (20 Short Answers, 2 MCQs) Miscellaneous Exercise 40 Questions & Solutions

## CBSE Class 12 Maths Chapter 7 Other Study Materials

 S.No. Important Links for Chapter 7 Integrals 1 Class 12 Integrals Important Questions 2 Class 12 Integrals Revision Notes 3 Class 12 Integrals Important Formulas 4 Class 12 Integrals NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.9

1. What is the meaning of NCERT solutions for Exercise 7.9 Class 12 Maths Chapter 7 Integrals?

NCERT solutions for Exercise 7.9 Class 12 Maths Chapter 7 integrals answers all the questions of exercise 7.9 of the NCERT textbook. These NCERT solutions are taken from the topic “Evaluation of definite integral by substitution”. These solutions consist of questions in which integrals have been evaluated by using substitution. When there are some difficulties in solving these questions of NCERT EX 7.9, we need these solutions. It contains explanations of each concept of solutions.

2. What is the purpose of the substitution rule in Exercise 7.9 Class 12?

Using substitution to evaluate a definite integral, requires a change to the limit of integration. In the substitution rule, we require a change to the limit of integration. In the substitution rule, we change variables in the integrand, which results in a change of limits of integration. Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently

3. Where can I download the NCERT solutions for Maths Chapter 7 Integrals Exercise 7.9 Class 12?

I can download the NCERT solutions for Maths Chapter 7 Integral Exercise 7.9 Class 12 from vedantu.com. These solutions can be understood easily by the right notes. Solutions of Vedantu class 12 maths chapter 7 integral are more successful in producing the desired result. Higher grades can be found in exams by Vedantu’s class 12 maths chapter 7.9. Students will be able to get a concept of topics by solving these questions.

4. What is meant by definite integration using substitution?

Integration using substitution is to evaluate integrals and antiderivatives, In this method First we find the assumed variables u and du. Whatever select u to be, du has to be du/dx. Then replace given variables say x and dx by assumed variables u and du. Now know the upper limit and lower limit from the x values to the corresponding u values. Then find the values at the lower and upper limits. The difference of which is the final answer.

5. How many questions are there in NCERT solutions for class 12 maths chapter 7 integral Exercise 7.9?

There are 22 questions in Exercise 7.9 of class 12 mathsIntegral. To evaluate these questions by substitution, the steps could be as follows:

1. Assuming  integral without limits, substitute, y = f(x) or x = g(y)

2. Integrate the new reduced integrand concerning new variables without mentioning the constant of integration.

3. Find the values of the answer obtained in step (2), accordingly, change the limits of the integral and the difference of the value at the upper and lower limits.

6. What is Ex 7.9 Class 12 about in Class 12 Maths Chapter 7?

Exercise 7.9 deals with evaluating definite integrals using the substitution method. This method involves changing the variable of integration to simplify the integral. Practicing these problems helps students understand this key technique. It’s a fundamental part of learning calculus.

7. Why is the substitution method used in Ex 7.9 Class 12?

The substitution method is used because it simplifies the process of integration. By substituting a part of the integral with a new variable, the integral becomes easier to solve. This technique is essential for solving complex integrals. Understanding this method helps with advanced calculus topics.

8. How many questions are there in Ex 7.9 Class 12 Maths NCERT Solutions?

Exercise 7.9 contains 10 questions. These questions provide sufficient practice for mastering the substitution method. Each question is designed to help students apply the substitution technique effectively. This practice is crucial for understanding definite integrals.

9. What is the main benefit of practising Ex 7.9 Class 12 Maths NCERT Solutions?

Practicing Exercise 7.9 helps students grasp the substitution method for definite integrals. This improves their problem-solving skills in calculus. It also builds confidence for tackling similar problems in exams. Mastery of this method is important for higher-level math.

10. Are the solutions in Ex 7.9 Class 12 Maths NCERT Solutions detailed?

Yes, the solutions provided by Vedantu are detailed and step-by-step. These detailed explanations help students understand each part of the process. This makes it easier to follow and learn the substitution method. Detailed solutions are crucial for thorough understanding.

11. How does Ex 7.9 Class 12 Maths NCERT Solutions help in exam preparation?

Practicing Exercise 7.9 helps strengthen problem-solving skills. This exercise builds confidence in handling definite integrals. Regular practice of these questions prepares students well for exams. It is an important part of exam preparation in calculus.

12. Is understanding Ex 7.9 Class 12 Maths NCERT Solutions important for future topics?

Yes, understanding Ex 7.9 Class 12 Maths NCERT Solutions is crucial for future calculus topics. The substitution method learned here is a foundational concept. It will be used in more advanced areas of mathematics. Mastery of this method is essential for progressing in calculus.

13. What should students focus on while solving Ex 7.9 Class 12 Maths NCERT Solutions?

Students should focus on identifying the correct substitution to simplify the integral. Following each step carefully is important for accuracy. Practice helps in mastering this technique. Understanding each part of the solution is crucial for learning.