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$\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$ is equal to :
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
Let
$I=\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$
Put $x=\cos \theta \Rightarrow d x=-\sin \theta d \theta$
$\therefore \quad I=\int_{\pi / 2}^0 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)(-\sin \theta) d \theta$
$=\int_0^{\pi / 2} \sin \left(2 \tan ^{-1} \tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \sin \theta d \theta$
$=\int_0^{\pi / 2} \sin (\pi-\theta) \sin \theta d \theta=\int_0^{\pi / 2} \sin ^2 \theta d \theta$
$=\int_0^{\pi / 2}\left[\frac{1-\cos 2 \theta}{2}\right] d \theta=\frac{1}{2}\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\pi / 2}$
$=\frac{1}{2}\left[\frac{\pi}{2}-0\right\rceil=\frac{\pi}{4}$
$I=\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$
Put $x=\cos \theta \Rightarrow d x=-\sin \theta d \theta$
$\therefore \quad I=\int_{\pi / 2}^0 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)(-\sin \theta) d \theta$
$=\int_0^{\pi / 2} \sin \left(2 \tan ^{-1} \tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \sin \theta d \theta$
$=\int_0^{\pi / 2} \sin (\pi-\theta) \sin \theta d \theta=\int_0^{\pi / 2} \sin ^2 \theta d \theta$
$=\int_0^{\pi / 2}\left[\frac{1-\cos 2 \theta}{2}\right] d \theta=\frac{1}{2}\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\pi / 2}$
$=\frac{1}{2}\left[\frac{\pi}{2}-0\right\rceil=\frac{\pi}{4}$
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