Question

Solve: $I={\int }_o^{\pi }\frac{xdx}{1+\sin x}$

Answer 1

$I={\int }_0^n\frac{xdx}{1+{\sin }^{2}x}$

$={\int }_0^n\frac{x}{{\cos }^{2}x/(1-\sin x)}$

$=\int \frac{x}{{\cos }^{2}x}-\frac{x\sin x}{{\cos }^{2}s}dx$

$I_1=2.=\int \frac{x}{1+\cos 2x}$

$2=\int \frac{1}{2}x\tan x-\int \frac{1}{2}\tan dx$

$x\tan x-(-\ln |\cos x|)$

$I_2=\int \frac{x\sin x}{{\cos }^{2}x}dx$

$=x\sin x\tan x-\int (\sin x+x\cos x)\tan xdx$

$\int \tan x\sin xdx+\int x\tan x\cos dx$

$\Rightarrow (\ln |\tan x+\sec x|-\sin x)-(x\cos x+\sin x)$

$x\tan x+\ln |\cos x|-(x\sin x\tan x+x\cos x-\ln |\tan x+\sec x|)$

$\Rightarrow \tan x+\ln |\cos x|-x\sin x\tan x-x\cos x+\ln (\tan x+\sec x)+C$

$\Rightarrow =\underset{x+0}{lim}\to 0$

$=\underset{x+n}{lim}\Rightarrow \pi$

$\therefore \pi -0,I=\pi$