Question

Solve: ${\int }_{-\pi }^{\pi }\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx=$

Answer 1

Let $I={\int }_{-\pi }^{\pi }\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$

$={\int }_{-\pi }^{\pi }\frac{2x}{1+{\cos }^{2}x}dx+{\int }_{-\pi }^{\pi }\frac{2x\sin x}{1+{\cos }^{2}x}dx$

$=I_1+I_2$

Consider $f\left(x\right)=\frac{2x}{1+{\cos }^{2}x}$

Now, $f(-x)=\frac{2(-x)}{1+{\cos }^2(-x)}=\frac{-2x}{1+{\cos }^{2}x}=-f\left(x\right)$

$\therefore I_1={\int }_{-\pi }^{\pi }\frac{2x}{1+{\cos }^{2}x}dx=0$ since ${\int }_{-a}^{a}f\left(x\right)dx=0$ if $f(-x)$=-f\left(x\right)$

Consider $g\left(x\right)=\frac{2x\sin x}{1+{\cos }^{2}x}$

$g(-x)=\frac{2(-x)\sin (-x)}{1+{\cos }^2(-x)}=\frac{2x\sin x}{1+{\cos }^{2}x}=g\left(x\right)$

$\therefore I_2={\int }_{-\pi }^{\pi }\frac{2x\sin xdx}{1+{\cos }^{2}x}$

$=2\times 2{\int }_0^{\pi }\frac{x\sin xdx}{1+{\cos }^{2}x}$

$=4{\int }_0^{\pi }\frac{x\sin xdx}{1+{\cos }^{2}x}$

Then $I_2=4{\int }_0^{\pi }\frac{(\pi -x)\sin (\pi -x)dx}{1+{\cos }^2(\pi -x)}$

$=4{\int }_0^{\pi }\frac{(\pi -x)\sin xdx}{1+{\cos }^{2}x}$ ......$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$,we get

$2I_2=4{\int }_0^{\pi }\frac{\pi \sin xdx}{1+{\cos }^{2}x}$

$\Rightarrow 2I_2=4\pi {\int }_0^{\pi }\frac{\sin xdx}{1+{\cos }^{2}x}$

Put $\cos x=z\Rightarrow -\sin xdx=dz$

When $x\to 0,z\to 0$

When $x\to \pi ,z\to -1$

$\therefore 2I_2=-4\pi {\int }_1^{-1}\frac{dz}{1+z^2}$

$=-4\pi \times {\left[{\tan }^{-1}z\right]}_1^{-1}$

$=-4\pi [{\tan }^{-1}(-1)-{\tan }^{-1}1]$

$=-4\pi (\frac{-\pi }{4}-\frac{-\pi }{4})$

$=2{\pi }^2$

$\Rightarrow I_2={\pi }^2$

$\therefore I=I_1+I_2=0+{\pi }^2={\pi }^2$