Question

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

Answer 1

$I=\int {\tan }^{-1}\sqrt{\frac{1-x}{1+x}}dx$
Let $x=\cos \theta \Rightarrow dx=-\sin \theta d\theta$
$\therefore I=\int {\tan }^{-1}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}(-\sin \theta d\theta )$
$=-\int {\tan }^{-1}\sqrt{\frac{2{\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}}\sin \theta d\theta$
$=-\int {\tan }^{-1}\tan \frac{\theta }{2}\cdot \sin \theta d\theta$
$=-\frac{1}{2}\int \theta \cdot \sin \theta d\theta$
$=-\frac{1}{2}[\theta \cdot (-\cos \theta )-\int 1\cdot (-\cos \theta \left)d\theta \right]$
$=-\frac{1}{2}[-\theta \cos \theta +\sin \theta ]$
$=\frac{1}{2}\theta \cos \theta -\frac{1}{2}\sin \theta$
$=\frac{1}{2}{\cos }^{-1}x\cdot x-\frac{1}{2}\sqrt{1-x^2}+C$
$=\frac{x}{2}{\cos }^{-1}x-\frac{1}{2}\sqrt{1-x^2}+C$
$=\frac{1}{2}(x{\cos }^{-1}x-\sqrt{1-x^2})+C$