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Let $I=\int_{0}^{\pi} \frac{x d x}{1+\sin x}$                                  ...(1)

$\Rightarrow I=\int_{0}^{\pi} \frac{(\pi-x)}{1+\sin (\pi-x)} d x$                         $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$\Rightarrow I=\int_{0}^{\pi} \frac{(\pi-x)}{1+\sin x} d x$                    ...(2)

Adding (1) and (2), we obtain

$2 I=\int_{0}^{\pi} \frac{\pi}{1+\sin x} d x$

$\Rightarrow 2 I=\pi \int_{0}^{\pi} \frac{(1-\sin x)}{(1+\sin x)(1-\sin x)} d x$

$\Rightarrow 2 I=\pi \int_{0}^{\pi} \frac{1-\sin x}{\cos ^{2} x} d x$

$\Rightarrow 2 I=\pi \int_{0}^{\pi}\left\{\sec ^{2} x-\tan x \sec x\right\} d x$

$\Rightarrow 2 I=\pi[\tan x-\sec x]_{0}^{\pi}$

$\Rightarrow 2 I=\pi[2]$

$\Rightarrow I=\pi$