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

Exercise 7.9

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

CBSE Class 12 Maths Chapter 7 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

Related Links for NCERT Class 12 Maths in Hindi

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

Important Related Links for NCERT Class 12 Maths