int_0^1 sin left(2 tan ^{-1} sqrt{frac{1+x}{1 | vedclass

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(B) Let $I = \int_0^1 \sin \left(2 	an ^{-1} \sqrt{\frac{1+x}{1-x}}ight) d x$. Put $x = \cos 	heta$, then $d x = -\sin 	heta d 	heta$. When $x =

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$\frac{\pi}{4}$

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(B) Let $I = \int_0^1 \sin \left(2 	an ^{-1} \sqrt{\frac{1+x}{1-x}}ight) d x$. Put $x = \cos 	heta$, then $d x = -\sin 	heta d 	heta$. When $x = 0$, $	heta = \frac{\pi}{2}$ and when $x = 1$, $	heta = 0$. Substituting these into the integral: $I = \int_{\pi/2}^0 \sin \left(2 	an ^{-1} \sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}}ight) (-\sin 	heta) d 	heta$. Using the identity $\sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}} = \sqrt{\frac{2\cos^2(	heta/2)}{2\sin^2(	heta/2)}} = \cot(	heta/2) = 	an(\frac{\pi}{2} - \frac{	heta}{2})$. $I = \int_0^{\pi/2} \sin \left(2 	an ^{-1} 	an \left(\frac{\pi}{2} - \frac{	heta}{2}ight)ight) \sin 	heta d 	heta$. $I = \int_0^{\pi/2} \sin \left(\pi - 	hetaight) \sin 	heta d 	heta = \int_0^{\pi/2} \sin^2 	heta d 	heta$. Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$: $I = \int_0^{\pi/2} \frac{1 - \cos 2	heta}{2} d 	heta = \frac{1}{2} \left[ 	heta - \frac{\sin 2	heta}{2} ight]_0^{\pi/2}$. $I = \frac{1}{2} \left[ (\frac{\pi}{2} - 0) - (0 - 0) ight] = \frac{\pi}{4}$.

$\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$ is equal to :

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