Evaluate the following integrals:

Question:

Evaluate the following integrals:

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

Solution:

Let $\mathrm{I}=\int \tan ^{-1} \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}} \mathrm{dx}$

$\mathrm{x}=\cos \theta ; \mathrm{dx}=-\sin \theta \mathrm{d} \theta$

$I=\int \tan ^{-1}\left(\tan \frac{\theta}{2}\right)-\sin \theta d \theta$

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

Using integration by parts,

$=-\frac{1}{2}\left[\theta \int \sin \theta d \theta-\int \frac{d}{d \theta} \theta \int \sin \theta d \theta\right]$

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

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

$I=\frac{1}{2}\left[-x \cos ^{-1} x+\sqrt{1-x^{2}}\right]+c$

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