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NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.9 - 2025-26

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Integrals Questions and Answers - Free PDF Download

In NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9, you’ll dive into the world of definite integrals and learn how to solve them using the substitution method. This method is very useful when the usual ways seem tricky or confusing, and it helps make tough integral questions much easier to handle. If you ever wondered how to quickly crack definite integrals, this exercise is perfect for building that confidence.

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Vedantu's solutions give you step-by-step guidance for every question, so you can follow along without getting lost. You can also get the chapter’s NCERT Solutions as a free PDF—handy for last-minute revision or offline study whenever you need. If you want to see the full list of what’s in store for Class 12 Maths, do check out the latest CBSE Class 12 Maths syllabus.


This chapter is part of Calculus and is important for your final exams, covering concepts that will also help you in competitive exams. Feeling stuck on substitutions or how to change the limits? You'll find clear explanations and worked examples here, so practicing these will help you master the technique in no time. You can always refer to more NCERT Solutions for Class 12 Maths whenever you need extra help!


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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

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Class 12 Maths Chapter 7 Exercise 7.1 - 22 Questions & Solutions (21 Short Answers, 1 MCQs)

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FAQs on NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.9 - 2025-26

1. What is the primary topic covered in the NCERT Solutions for Class 12 Maths Chapter 7, Exercise 7.9?

The NCERT Solutions for Class 12 Maths Chapter 7, Exercise 7.9, primarily focus on the Evaluation of Definite Integrals by Substitution. This exercise provides step-by-step methods for solving definite integrals where a direct integration is complex, in line with the CBSE 2025-26 syllabus.

2. How is the substitution method used to solve definite integrals in Exercise 7.9?

The substitution method involves a series of steps to simplify and solve the integral:

  • Choose a substitution: Identify a part of the integrand to substitute with a new variable, say u.
  • Find the differential: Calculate the differential du in terms of dx.
  • Change the limits: Convert the original limits of integration (which are for x) to the corresponding limits for the new variable u.
  • Integrate: Solve the simplified integral with respect to u using the new limits.
  • Evaluate: Calculate the final value by finding the difference of the function at the upper and lower limits.

3. How many questions are there in Exercise 7.9 of NCERT Class 12 Maths, and what is their main purpose?

Exercise 7.9 contains 10 questions. The main purpose of these questions is to provide students with thorough practice in applying the substitution method to a variety of functions, including trigonometric, algebraic, and inverse trigonometric functions, to build problem-solving proficiency.

4. Why is it essential to change the limits of integration when using the substitution method for definite integrals?

It is essential because the original limits are defined for the variable x. When you change the variable of integration from x to u, the entire integral, including its bounds, must be expressed in terms of u. Applying the old x-limits to the new u-based integral would be mathematically incorrect and would lead to the wrong answer. The new limits correspond to the values of u when x is at its original lower and upper bounds.

5. What is a common mistake students make while solving problems from NCERT Exercise 7.9?

A very common mistake is forgetting to change the limits of integration after substituting the variable. Students often perform the integration with the new variable 'u' but incorrectly apply the original 'x' limits for evaluation. Another frequent error is making a mistake while finding the new differential (e.g., dx in terms of du).

6. How do you solve a problem like Question 7 from Ex 7.9, ∫dx/(x² + 2x + 5) from -1 to 1?

To solve this, you use the method of completing the square and then substitution. The steps are:
1. Rewrite the denominator: x² + 2x + 5 can be written as (x² + 2x + 1) + 4, which simplifies to (x + 1)² + 2².
2. The integral becomes ∫dx/((x + 1)² + 2²).
3. Substitute: Let u = x + 1. Then du = dx.
4. Change limits: When x = -1, u = 0. When x = 1, u = 2.
5. The new integral is ∫du/(u² + 2²) from 0 to 2.
6. This is a standard integral form, which evaluates to (1/2)tan⁻¹(u/2).
7. Apply the new limits [0, 2] to get the final answer, which is π/8.

7. Can all definite integrals in Chapter 7 be solved using the substitution method from Exercise 7.9?

No, the substitution method is not a universal solution. It is effective only when the integrand contains a function and its derivative (or a function that can be simplified via substitution). For other types of integrals, you must use other methods taught in Chapter 7, such as integration by parts, integration using trigonometric identities, or integration by partial fractions.

8. How does mastering the substitution method in Exercise 7.9 help in preparing for board exams?

Mastering the substitution method is crucial as it is a foundational technique frequently tested in the CBSE board exams. It improves problem-solving speed and accuracy. Furthermore, this method is a prerequisite for understanding more advanced topics like finding the area under curves (Chapter 8) and solving differential equations (Chapter 9), which carry significant weightage in exams.