Chapter 7: Integrals

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 $\dfrac{1}{2}\log 2$.

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 $

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

 $= \dfrac{{154 + 42 - 132}}{{231}}$

$= \dfrac{{64}}{{231}}$ 

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 $ 

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 $ 

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} $

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}] $ 

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} $ 

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.

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.

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)$x\sin x$.