Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int {\tan }^{-1}\sqrt{\frac{1-x}{1+x}}dx \\ \mathrm{Putting}x=\cos \mathit{}\theta \\ \Rightarrow dx=-\sin \theta \mathit{}d\theta \\ \&\theta ={\cos }^{-1}x \\ \therefore I=\int {\tan }^{-1}\sqrt{\frac{1-\cos \mathit{}\theta }{1+\cos \theta }}\left(-\sin \theta \right)d\theta \\ =\int {\tan }^{-1}\sqrt{\frac{2{\sin }^2{\displaystyle \frac{\theta }{2}}}{2{\cos }^2{\displaystyle \frac{\theta }{2}}}}\left(-\sin \theta \right)d\theta \\ =\int {\tan }^{-1}\left(\tan \frac{\theta }{2}\right)\left(-\sin \theta \right)d\theta \\ =-\frac{1}{2}\int \underset{I}{\theta }\underset{\mathrm{II}}{\sin \theta }d\theta \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=-\frac{1}{2}\left[\theta \int \sin \theta d\theta -\int \left\{\left(\frac{d}{d\theta }\theta \right)\int \sin \theta d\theta \right\}d\theta \right] \\ =-\frac{1}{2}\left[\theta \left(-\cos \theta \right)-\int 1.\left(-\cos \mathit{}\theta \right)d\theta \right] \\ =-\frac{1}{2}\left[-\theta \cos \theta +\sin \mathit{}\theta \right]+C \\ =-\frac{1}{2}\left[-\theta .\cos \mathit{}\theta +\sqrt{1-{\cos }^2\theta }\right]+C \\ =-\frac{1}{2}\left[-{\cos }^{-1}x.x+\sqrt{1-x^2}\right]+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\mathit{\because }\mathit{}\theta ={\cos }^{-1}\mathit{}x\right] \\ =\frac{x{\cos }^{-1}x}{2}-\frac{\sqrt{1-x^2}}{2}+C \end{gathered}$$