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Integrate the function
$\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x=I($ say $)$
$\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x=I($ say $)$
Solution:
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Verified Answer
Put $\mathrm{x}=\cos \theta$, then $\mathrm{dx}=-\sin \theta \mathrm{d} \theta$ and
$\begin{aligned}
&\sqrt{\frac{1-x}{1+x}}=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\sqrt{\frac{2 \sin ^2 \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}}=\tan \frac{\theta}{2} \\
&=\int \tan ^{-1}\left(\tan \frac{\theta}{2}\right)(-\sin \theta) d \theta=-\int \frac{\theta}{2} \sin \theta d \theta \\
&I=-\frac{1}{2} \int \theta \sin \theta d \theta=\frac{1}{2} \theta \cos \theta-\frac{1}{2} \int \cos \theta d \theta \\
&=\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sin \theta+C \\
&=\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sqrt{1-\cos ^2 \theta}+C \\
&=\frac{1}{2}\left(x \cos -1 x-\sqrt{1-x^2}+C\right)
\end{aligned}$
$\begin{aligned}
&\sqrt{\frac{1-x}{1+x}}=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\sqrt{\frac{2 \sin ^2 \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}}=\tan \frac{\theta}{2} \\
&=\int \tan ^{-1}\left(\tan \frac{\theta}{2}\right)(-\sin \theta) d \theta=-\int \frac{\theta}{2} \sin \theta d \theta \\
&I=-\frac{1}{2} \int \theta \sin \theta d \theta=\frac{1}{2} \theta \cos \theta-\frac{1}{2} \int \cos \theta d \theta \\
&=\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sin \theta+C \\
&=\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sqrt{1-\cos ^2 \theta}+C \\
&=\frac{1}{2}\left(x \cos -1 x-\sqrt{1-x^2}+C\right)
\end{aligned}$
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