Evaluate the following : ^ -1 1- x 1+ x d x - Vidyadip

Question

Evaluate the following : $\int \tan ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}} \cdot d x$

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Answer

$
\begin{aligned}
I & =\int \tan ^{-1} \sqrt{\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}} \cdot d x \\
& =\int \tan ^{-1} \sqrt{\frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}} \cdot d x \\
& =\int \tan ^{-1} \sqrt{\tan ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} \cdot d x \\
& =\int \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right] \cdot d x \\
& =\int\left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot d x \\
& =\frac{\pi}{4} x-\frac{1}{2} \cdot \frac{x^2}{2}+c \\
& =\frac{\pi}{4} x-\frac{x^2}{4}+c
\end{aligned}
$

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