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### Answers (1)

Answer:$\frac{1}{2}\left[x \cos ^{-1} x-\sqrt{1-x^{2}}\right]+C$

Hint: to solve this question we will presume x as $\cos \theta$

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

Solution:

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

$\operatorname{let} x=\cos 2 \theta$

$\frac{d x}{d \theta}=-2 \sin 2 \theta$

$d x=-2 \sin 2 \theta d \theta$

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

$\cos 2 \theta=2 \cos ^{2} \theta-1$

$\cos 2 \theta=1-2 \sin ^{2} \theta$

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

$I=\int \tan ^{-1} \sqrt{\tan ^{2} \theta} \times-2 \sin 2 \theta d \theta$

$I=-2 \int \theta \sin 2 \theta d \theta$

$I=-2\left[\theta \int \sin 2 \theta d \theta-\int \frac{d \theta}{d \theta} \int \sin 2 \theta d \theta\right] d x . \text { .applying Byparts }$

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

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

$I=-2\left[-\theta \frac{\cos 20}{2}+\int \frac{\cos 20}{2}\right] d \theta$

$I=-2\left[-\frac{\cos 2 \theta}{2}+\frac{\sin 2 \theta}{4}\right]+C$

$I=-2\left[\left(\frac{-1}{2} \cos ^{-1} x\right) \frac{x}{2}+\frac{\sin \left(\cos ^{-1} x\right)}{4}+C\right]$

$I=-2\left[\frac{-x}{4} \cos ^{-1} x+\frac{\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)}{4}\right]+C$

$I=-\frac{2}{4}\left[-x \cos ^{-1} x+\sqrt{1-x^{2}}\right]+C$

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

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