$\int_0^\pi {\frac{x}{{1 + \sin x}}}$

Let $I = \int_0^\pi {\frac{x}{{1 + \sin x}}} dx$ …….(i)

and $I = \int_0^\pi {\frac{{\pi - x}}{{1 + \sin (\pi - x)}}} dx = \int_0^\pi {\frac{{\pi - x}}{{1 + \sin x}}} dx$ ………(ii)

On adding Equations (i) and (ii), we get
$2I = \pi \int_0^\pi {\frac{1}{{1 + \sin x}}} dx$

$= \pi \int_0^\pi {\frac{{(1 - \sin x)dx}}{{(1 + \sin x)(1 - \sin x)}}}$

$= \pi \int_0^\pi {\frac{{(1 - \sin x)dx}}{{{{\cos }^2}x}}}$

$= \pi \int_0^\pi {\left( {{{\sec }^2}x - \tan x \cdot \sec x} \right)} dx$

$= \pi \int_0^\pi {{{\sec }^2}} xdx - \pi \int_0^\pi {\sec } xx \cdot \tan xdx$

$= \pi [\tan x]_0^\pi - \pi [\sec x]_0^\pi$
$= \pi [\tan x - \sec x]_0^\pi$

$= \pi [\tan \pi - \sec \pi - \tan 0 - \sec 0]$

$\Rightarrow$ $2I = \pi [0 + 1 - 0 + 1]$

$2I = 2\pi$

therefore,$I = \pi$