Question

Evaluate ${\int }_0^{\pi }\frac{xdx}{1+{\cos }^{2}x}$

Answer 1

Solution: let $I={\int }_0^{\pi }\frac{xdx}{1+{\cos }^{2}x}={\int }_0^{\pi }\frac{(\pi -x)dx}{1+{\cos }^2(\pi -x)}={\int }_0^{\pi }\frac{xdx}{1+{\cos }^{2}x}-1$$\Rightarrow 2I={\int }_0^{\pi }\frac{xdx}{1+{\cos }^{2}x}=2\pi {\int }_0^{\pi /2}\frac{dx}{1+{\cos }^{2}x}=2\pi {\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{2+{\tan }^{2}x}$Let $\tan x=t$ so that for $x\to 0$, $t\to 0$ and for $x\to \pi /2$, $t\to \infty$.

Hence we can write.$I=\pi {\int }_0^{\pi }\frac{dt}{2+t^2}=\pi \frac{1}{\sqrt{2}}{\left[{\tan }^{-1}\frac{t}{\sqrt{2}}\right]}_0^{\pi }=\frac{{\pi }^2}{2\sqrt{2}}$