Question

$\underset{0}{\overset{1}{\int }}{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx$

Answer 1

$$\begin{gathered} {\int }_0^1{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx \\ \mathrm{Let},x=\tan \theta ,dx=se{c}^2\theta d\theta \\ \mathrm{When},x\to 0;\theta \to 0 \\ \mathrm{and}x\to 1;\theta \to \frac{\mathrm{\pi }}{4} \\ \mathrm{Therefore},\mathrm{the}\mathrm{integral}\mathrm{becomes} \\ {\int }_0^{\frac{\mathrm{\pi }}{4}}{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)se{c}^2\theta d\theta \\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}{\cos }^{-1}\left(\cos 2\theta \right)se{c}^2\theta d\theta \\ =2{\int }_0^{\frac{\mathrm{\pi }}{4}}\theta se{c}^2\theta d\theta \\ =2{\left[\theta t\mathrm{an}\theta \right]}_0^{\frac{\mathrm{\pi }}{4}}-2{\int }_0^{\frac{\mathrm{\pi }}{4}}\tan \theta d\theta \\ =2{\left[\theta \tan \theta \right]}_0^{\frac{\mathrm{\pi }}{4}}+2{\left[\log \left(\cos \theta \right)\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ =2\left(\frac{\mathrm{\pi }}{4}-0\right)+2\left[\log \frac{1}{\sqrt{2}}-0\right] \\ =\frac{\mathrm{\pi }}{2}-\log 2 \end{gathered}$$