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
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