Question

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

Answer 1

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

I=$\underset{0}{\overset{\pi }{\int }}\frac{4x\left(\sin x\right)}{1+{\cos }^{2}x}dx$ ...... 1

I =$\underset{0}{\overset{\pi }{\int }}\frac{4(\pi -x)\left(\sin x\right)}{1+{\cos }^{2}x}dx$ ...... 2

Adding 1 and 2

We get I=$\underset{0}{\overset{\pi }{\int }}\frac{2\pi \left(\sin x\right)}{1+{\cos }^{2}x}dx$ =$\frac{{\pi }^2}{2}$

$\int\limits_{ 0 }^\pi {\frac{{4x\left( { \sin x} \right)}}{{1 + {{\cos }^2}x}}} dx$ ...... 1ππ2x(1+sin⁡