STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

$x = \cos \theta $ रखने पर,

$\sin \left[ {2{{\tan }^{ - 1}}\sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} } \right]$

$ = \sin \,\,\left[ {2{{\tan }^{ - 1}}\left( {\cot \frac{\theta }{2}} \right)} \right]$

$ = \sin \left[ {2{{\tan }^{ - 1}}\left[ {\tan \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right]} \right] $

$= \sin \left[ {2\,\left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right]$

$ = \sin (\pi - \theta ) = \sin \theta $

$= \sqrt {1 - {{\cos }^2}\theta } = \sqrt {1 - {x^2}} $

अब, $\int_0^1 {\sin \left( {2{{\tan }^{ - 1}}\sqrt {\frac{{1 + x}}{{1 - x}}} } \right)} \,dx = \int_0^1 {\sqrt {1 - {x^2}} \,dx} $

$ = \left[ {\frac{1}{2}x\sqrt {1 - {x^2}} } \right]_0^1 + \frac{1}{2}[{\sin ^{ - 1}}x]_0^1 = \frac{\pi }{4}.$