int tan ^{-1}left(sqrt{frac{1-x}{1+x}}right) | vedclass

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(A) Let $I = \int 	an ^{-1} \sqrt{\frac{1-x}{1+x}} d x$. Substitute $x = \cos 2	heta$,so $dx = -2\sin 2	heta d	heta$. Then $\sqrt{\frac{1-x}{1+x}}

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$\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^2}\right)+c$

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(A) Let $I = \int 	an ^{-1} \sqrt{\frac{1-x}{1+x}} d x$. Substitute $x = \cos 2	heta$,so $dx = -2\sin 2	heta d	heta$. Then $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 2	heta}{1+\cos 2	heta}} = \sqrt{\frac{2\sin^2 	heta}{2\cos^2 	heta}} = 	an 	heta$. Thus,$I = \int 	heta (-2\sin 2	heta) d	heta = -2 \int 	heta \sin 2	heta d	heta$. Using integration by parts: $I = -2 \left[ 	heta \left( -\frac{\cos 2	heta}{2} ight) - \int 1 \cdot \left( -\frac{\cos 2	heta}{2} ight) d	heta ight] = 	heta \cos 2	heta - \int \cos 2	heta d	heta = 	heta \cos 2	heta - \frac{\sin 2	heta}{2} + c$. Since $x = \cos 2	heta$,then $	heta = \frac{1}{2} \cos^{-1} x$ and $\sin 2	heta = \sqrt{1-x^2}$. Substituting these back: $I = \frac{1}{2} x \cos^{-1} x - \frac{1}{2} \sqrt{1-x^2} + c = \frac{1}{2} \left( x \cos^{-1} x - \sqrt{1-x^2} ight) + c$.

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

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