Evaluate the definite integral $\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x$.

(D) Let $I = \int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x$....$(1)$
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:
$I = \int_{0}^{\pi} \frac{(\pi-x) \tan(\pi-x)}{\sec(\pi-x)+\tan(\pi-x)} d x$
Since $\tan(\pi-x) = -\tan x$ and $\sec(\pi-x) = -\sec x$,we have:
$I = \int_{0}^{\pi} \frac{-(\pi-x) \tan x}{-(\sec x+\tan x)} d x = \int_{0}^{\pi} \frac{(\pi-x) \tan x}{\sec x+\tan x} d x$....$(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{0}^{\pi} \frac{\pi \tan x}{\sec x+\tan x} d x = \pi \int_{0}^{\pi} \frac{\sin x}{1+\sin x} d x$
$2I = \pi \int_{0}^{\pi} \frac{1+\sin x - 1}{1+\sin x} d x = \pi \int_{0}^{\pi} (1 - \frac{1}{1+\sin x}) d x$
$2I = \pi [x]_{0}^{\pi} - \pi \int_{0}^{\pi} \frac{1-\sin x}{\cos^2 x} d x$
$2I = \pi^2 - \pi \int_{0}^{\pi} (\sec^2 x - \sec x \tan x) d x$
$2I = \pi^2 - \pi [\tan x - \sec x]_{0}^{\pi}$
$2I = \pi^2 - \pi [(\tan \pi - \sec \pi) - (\tan 0 - \sec 0)]$
$2I = \pi^2 - \pi [(0 - (-1)) - (0 - 1)] = \pi^2 - \pi [1 + 1] = \pi^2 - 2\pi$
$I = \frac{\pi}{2}(\pi - 2)$