Evaluate the following integrals:

Question:

Evaluate $\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$

Solution:

Put $x=\cos 2 t ; d x=-2 \sin 2 t$

$=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} \mathrm{~d} x=\int \tan ^{-1} \sqrt{\frac{1-\cos 2 t}{1+\cos 2 t}}(-2 \sin 2 t) d t$

$=\int \tan ^{-1} \sqrt{\frac{1-\cos 2 t}{1+\cos 2 t}}(-2 \sin 2 t) d t$

$=-2 \int \tan ^{-1} t \tan t \sin 2 t d t$

$=-2 \int t \sin 2 t d t$

$=-2\left[-\frac{t \cos 2 t}{2}+\frac{1}{2} \int \cos 2 t d t\right]$

$=t \cos 2 t-\frac{\sin 2 t}{2}+c$

$=\frac{x \cos ^{-1} x}{2}-\frac{\sqrt{1-x^{2}}}{2}+c$

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