$\int\limits_0^\pi {\cfrac{{xdx}}{{1 + \sin x}}}$

Let $I = \int\limits_0^\pi {\cfrac{{xdx}}{{1 + \sin x}}}$

……(i)
$\Rightarrow$ $I = \int\limits_0^\pi {\cfrac{{\pi - x}}{{1 + \sin \left( {\pi - x} \right)}}dx}$

$= \int\limits_0^\pi {\cfrac{{\pi - x}}{{1 + \sin x}}dx}$

….(ii)
Adding (i) and (ii),

we get
$2I = \int\limits_0^\pi {\cfrac{{x + \pi - x}}{{1 + \sin x}}dx} = \pi \int\limits_0^\pi {\cfrac{1}{{1 + \sin x}}dx}$

$= \pi \int\limits_0^\pi {\cfrac{{1 - \sin x}}{{1 - {{\sin }^2}x}}} dx = \pi \int\limits_0^\pi {\cfrac{{1 - \sin \,x}}{{{{\cos }^2}x}}dx}$

$= \pi \int\limits_0^\pi {\left( {{{\sec }^2}x - \tan x\sec x} \right)dx = \pi \left[ {\tan x - \sec x} \right]_0^\pi }$

$= \pi \left[ {\left( {\tan \pi - \sec \pi } \right) - \left( {\tan 0 - \sec 0} \right)} \right] = \pi \left[ {\left( 0 \right) - \left( { - 1} \right) - \left( {0 - 1} \right)} \right] = 2\pi$

Hence, $I = \pi$