MCQ
$\int_{\,0}^{\,1} {\,\sin \left( {2{{\tan }^{ - 1}}\sqrt {\frac{{1 + x}}{{1 - x}}} } \right)\,dx = } $
- A$\pi /6$
- ✓$\pi /4$
- C$\pi /2$
- D$\pi $
Put $x = \cos \theta ,$ then
$\sin \left[ {2{{\tan }^{ - 1}}\sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} } \right]$
$ = \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}} $
Now, $\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}.$
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$\mathrm{F}(\mathrm{x})=\int\limits_{1}^{\mathrm{x}} \mathrm{t}^{2} \mathrm{g}(\mathrm{t}) \mathrm{dt},$ where $\mathrm{g}(\mathrm{t})=\int\limits_{1}^{\mathrm{t}} \mathrm{f}(\mathrm{u}) \mathrm{du}$
Then for the function $\mathrm{F}$, the point $\mathrm{x}=1$ is