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

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

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$\frac{\pi}{4}$

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(B) Let $I = \int_0^1 \sin \left(2 	an ^{-1} \sqrt{\frac{1+x}{1-x}}ight) d x$. Substitute $x = \cos 	heta$,then $d x = -\sin 	heta d 	heta$. When $x = 0$,$	heta = \frac{\pi}{2}$. When $x = 1$,$	heta = 0$. $I = \int_{\pi/2}^0 \sin \left(2 	an ^{-1} \sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}}ight) (-\sin 	heta) d 	heta$. Using $\sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}} = \sqrt{\frac{2\cos^2(	heta/2)}{2\sin^2(	heta/2)}} = \cot(	heta/2) = 	an(\frac{\pi}{2} - \frac{	heta}{2})$. $I = \int_0^{\pi/2} \sin \left(2 	an^{-1} 	an(\frac{\pi}{2} - \frac{	heta}{2})ight) \sin 	heta d 	heta$. $I = \int_0^{\pi/2} \sin(\pi - 	heta) \sin 	heta d 	heta = \int_0^{\pi/2} \sin^2 	heta d 	heta$. Using $\sin^2 	heta = \frac{1-\cos 2	heta}{2}$,we get $I = \int_0^{\pi/2} \frac{1-\cos 2	heta}{2} d 	heta$. $I = \frac{1}{2} [	heta - \frac{\sin 2	heta}{2}]_0^{\pi/2} = \frac{1}{2} [(\frac{\pi}{2} - 0) - (0 - 0)] = \frac{\pi}{4}$.

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

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