NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9

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.

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