Mathematics

# $\displaystyle \int_{0}^{1}\sin \left(2\tan^{-1}\sqrt{\dfrac{1+x}{1-x}}\right)dx=$

##### ANSWER

$\dfrac{\pi}{4}$

##### SOLUTION
$\int_0^1 {\sin \left( {2{{\tan }^{ - 1}}\sqrt {\cfrac{{1 + x}}{{1 - x}}} } \right)dx}$
$x = \cos \theta$ $\Rightarrow dx = - \sin \theta d\theta$
$= - \int_{\cfrac{\pi }{2}}^0 {\sin \left( {2{{\tan }^{ - 1}}\sqrt {\cfrac{{1 + \cos \theta }}{{1 - \cos \theta }}} } \right)\sin \theta d\theta }$
$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {2{{\tan }^{ - 1}}\sqrt {\cfrac{{2{{\cos }^2}\cfrac{\theta }{2}}}{{2{{\sin }^2}\cfrac{\theta }{2}}}} } \right)\sin \theta d\theta }$         (Since $1 + \cos 2\theta = 2{\cos ^2}\theta ,1 - \cos 2\theta = 2{\sin ^2}\theta$)
$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {2{{\tan }^{ - 1}}\left( {\cot \cfrac{\theta }{2}} \right)} \right)} \sin \theta d\theta$
$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {2{{\tan }^{ - 1}}\left( {\tan \left( {\cfrac{\pi }{2} - \cfrac{\theta }{2}} \right)} \right)} \right)\sin \theta d\theta }$       (Since $\tan \left( {\cfrac{\pi }{2} - \theta } \right) = \cot \theta$)
$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {2 \times \left( {\cfrac{\pi }{2} - \cfrac{\theta }{2}} \right)} \right)} \sin \theta d\theta$
$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {\pi - \theta } \right)\sin \theta d\theta }$
$= \int_0^{\cfrac{\pi }{2}} {{{\sin }^2}\theta d\theta }$
$= \cfrac{1}{2}\int_0^{\cfrac{\pi }{2}} {2{{\sin }^2}\theta d\theta }$
$= \cfrac{1}{2}\int_0^{\cfrac{\pi }{2}} {\left( {1 - \cos 2\theta } \right)d\theta }$
$= \cfrac{1}{2}\left( {\theta - \cfrac{{\sin 2\theta }}{2}} \right)_0^{\cfrac{\pi }{2}}$
$= \cfrac{1}{2}\left( {\cfrac{\pi }{2}} \right)$
$= \cfrac{\pi }{4}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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