Question

Prove that:
${\int }_0^{\pi }\frac{xdx}{1+\sin x}=\pi$

Answer 1

$I={\int }_0^{\pi }\left(\frac{x}{1+\sin x}\right)dx$

$I={\int }_0^{\pi }\left[\frac{x}{1+\sin x}\right]dx$ $\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\right]$

$2I={\int }_0^{\pi }\left(\frac{x}{1+\sin x}\right)dx$

$2I={\int }_0^{\pi }\left[\frac{x}{1+\sin x}\right]dx$ $\left[{\int }_a^{2b}f\left(x\right)dx=2{\int }_a^{b}f\left(x\right)dxiff\left(x\right)=7(2b-x)\right]$

$I={\int }_0^{\pi /2}\left(\frac{1}{1+\sin x}\right)dx$

$I={\int }_0^{\pi /2}\left(\frac{1-\sin x}{{\cos }^{2}x}\right)dx$

$I={\int }_0^{\pi /2}({\sec }^{2}x-\tan x\sec x)dx$

$I=\pi [\tan x-\sec x{]}_0^{\pi /2}$

$I=\pi {\left|\frac{\sin x-1}{\cos x}\right|}_0^{\pi /2}$

$I=\pi \left\{\underset{x\to \pi /2}{lim}\left(\frac{\sin x-1}{\cos x}\right)\left(\frac{0-1}{\cos 0}\right)\right\}$

$I=\pi \left\{\underset{x\to \pi /2}{lim}\left(\frac{\cos x}{-\sin x}\right)-(-1)\right\}$

$I=\pi \{0+1\}$

$I=\pi$.