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

Question

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

Aicharya · Accepted Answer

Let $I=\int 	an ^{-1} \sqrt{\frac{1-x}{1+x}} d x$ Put $x=\cos 2 	heta$ $\begin{aligned} 	herefore \quad I & =\int 	an ^{-1} \sqrt{\frac{1-\cos 2 	heta}{1+\cos 2 	heta}} d x \ & =\int 	an ^{-1} \sqrt{	an ^2 	heta} d x \ & =\int 	heta d x \ & =\frac{1}{2} \int 1 \cdot \cos ^{-1} x d x \ & =\frac{1}{2}\left[\cos ^{-1} x \cdot x+\int \frac{x}{\sqrt{1-x^2}} d xight] \ & =\frac{1}{2}\left[x \cos ^{-1} x-\sqrt{1-x^2}ight]+c\end{aligned}$

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

Solution