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Question:

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

Solution:

$I=\tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$

Let $x=\cos \theta \Rightarrow d x=-\sin \theta d \theta$

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

$=-\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \sin \theta d \theta$

$=-\int \tan ^{-1} \tan \frac{\theta}{2} \cdot \sin \theta d \theta$

$=-\frac{1}{2} \int \theta \cdot \sin \theta d \theta$

$=-\frac{1}{2}\left[\theta \cdot(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta\right]$

$=-\frac{1}{2}[-\theta \cos \theta+\sin \theta]$

$=+\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sin \theta$

$=\frac{1}{2} \cos ^{-1} x \cdot x-\frac{1}{2} \sqrt{1-x^{2}}+\mathrm{C}$

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

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

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