Inetgral

\hspace{-16}$Let $\bf{\mathbb{I} = \int_{0}^{\pi}\frac{1}{1+\cos^2 x}dx = 2\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cos^2x}dx}$\\\\\\ Using The Formula.......\\\\\\ $\bf{\Rightarrow \int_{0}^{2a}f(x)dx = 2\int_{0}^{a}f(x)dx}\;,\; $ If $\bf{f(2a-x)=f(x)}$\\\\\\ Now Didide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\bf{\cos^2 x\;,}$ we get\\\\\\ $\bf{\Rightarrow \mathbb{I}=2\int_{0}^{\frac{\pi}{2}}\frac{\sec^2 x}{1+\tan^2x+1}dx = 2\int_{0}^{\frac{\pi}{2}}\frac{\sec^2 x}{\tan^2x+\left(\sqrt{2}\right)^2}dx}$\\\\\\ Now Let $\bf{\tan x= t}$ and $\bf{\sec^2x dx = dt}$ and Changing Limit, we get\\\\\\ $\bf{\rightarrow \mathb{I} = 2\int_{0}^{\infty}\frac{1}{t^2+\left(\sqrt{2}\right)^2}dt = \frac{2}{\sqrt{2}}\left\{\tan^{-1}(\infty)-\tan^{-1}(0)\right\}}$\\\\\\ So $\bf{\mathbb{I} = \int_{0}^{\pi}\frac{1}{1+\cos^2 x}dx = \frac{\pi}{\sqrt{2}}}$

you may also try this way just simplify cos^x in the form of cos2x then just apply tanx formula of cos2x and after taking tanx=Z you will get an expression which will be easier to solve