By using the properties of definite integrals,evaluate the integral $\int_{0}^{\pi} \frac{x \, dx}{1+\sin x}$.

(D) Let $I = \int_{0}^{\pi} \frac{x \, dx}{1+\sin x} \quad \dots (1)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get:
$I = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{1+\sin(\pi-x)} = \int_{0}^{\pi} \frac{\pi-x}{1+\sin x} \, dx \quad \dots (2)$
Adding $(1)$ and $(2)$:
$2I = \int_{0}^{\pi} \frac{x + \pi - x}{1+\sin x} \, dx = \pi \int_{0}^{\pi} \frac{1}{1+\sin x} \, dx$
Multiply numerator and denominator by $(1-\sin x)$:
$2I = \pi \int_{0}^{\pi} \frac{1-\sin x}{1-\sin^2 x} \, dx = \pi \int_{0}^{\pi} \frac{1-\sin x}{\cos^2 x} \, dx$
$2I = \pi \int_{0}^{\pi} (\sec^2 x - \sec x \tan x) \, dx$
Integrating the terms:
$2I = \pi [\tan x - \sec x]_{0}^{\pi}$
$2I = \pi [(\tan \pi - \sec \pi) - (\tan 0 - \sec 0)]$
$2I = \pi [(0 - (-1)) - (0 - 1)] = \pi [1 + 1] = 2\pi$
Therefore,$I = \pi$.