Question

$\int {\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int {\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx \\ =2\int \underset{\mathrm{II}}{1}.\underset{\mathrm{I}}{{\tan }^{-1}}xdx\left[\because \cos {}^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x\right] \\ =2\left[{\tan }^{-1}x\int 1dx-\int \left\{\frac{d}{dx}\left({\tan }^{-1}x\right)\int 1dx\right\}dx\right] \\ =2\left[{\tan }^{-1}x.x-\int \frac{1}{1+x^2}\times xdx\right] \\ =2x{\tan }^{-1}x-\int \frac{2x}{1+x^2}dx \\ \mathrm{Putting}1+x^2=t \\ \Rightarrow 2xdx=dt \\ \therefore I=2x{\tan }^{-1}x-\int \frac{dt}{t} \\ =2x{\tan }^{-1}x-\ln \left|t\right|+C \\ =2x{\tan }^{-1}x-\ln \left|1+x^2\right|+C\left[\because t=1+x^2\right] \end{gathered}$$