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### Answers (1)

Answer: $\frac{1}{2}\left ( x\cos ^{-1}x-\sqrt{1-x^{2}} \right )+c$

Hint: Let $x=\cos \Theta$

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

Solution: Let $x=\cos \Theta$

$dx=-\sin \Theta \: d\: \Theta$

$\therefore \int \tan ^{-1}\sqrt{\frac{1-x}{1+x}}dx=\int \tan ^{-1}\sqrt{\frac{1-\cos \Theta }{1+\cos \Theta }}\left ( -\sin \Theta \: d\Theta \right )$

\begin{aligned} &=-\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} \sin \theta d \theta \\ &=-\frac{1}{2} \int \theta \sin \theta d \theta \\ &=-\frac{1}{2}\left[\theta(-\cos \theta)-\int 1(-\cos \theta) d \theta\right] \end{aligned}

Applying by parts

$=\frac{1}{2}\left [ -\Theta \cos \Theta +\sin \Theta \right ]$

\begin{aligned} &=\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sin \theta \\ &=\frac{1}{2} \cos ^{-1} x x-\frac{1}{2} \sqrt{1-x^{2}}+c \\ &=\frac{x}{2} \cos ^{-1} x-\frac{1}{2} \sqrt{1-x^{2}}+c \\ &=\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^{2}}\right)+c \end{aligned}

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