Let $I=\int\limits_{0}^{\pi} \frac{x\tan x}{\sec x+\tan x} dx$ ...(i) Also, $I=\int\limits_{0}^{\pi} \frac{(\pi-x)\tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)} dx$ $=\int\limits_{0}^{\pi} \frac{(\pi-x)(-\tan x)}{-\sec x)+(-\tan x)} dx$ $=\int\limits_{0}^{\pi} \frac{(\pi-x)\tan x}{\sec x+\tan x} dx$ ...(ii) Adding (i) and (ii), we get$2I=\int\limits_{0}^{\pi} \frac{(x+\pi-x)\tan x}{\sec x+\tan x} dx$ $=\pi\int\limits_{0}^{\pi} \frac{\tan x}{\sec x+\tan x} dx$ $=\pi\int\limits_{0}^{\pi} \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} dx$ $=\pi\int\limits_{0}^{\pi} \frac{\sin x}{1+\sin x} dx$ $=\pi\int\limits_{0}^{\pi} \frac{(1+\sin x)-1}{1+\sin x} dx$ $=\pi\int\limits_{0}^{\pi} \left(1-\frac{1}{1+\sin x}\right) dx$$=\pi\int\limits_{0}^{\pi} 1 dx - \pi\int\limits_{0}^{\pi} \frac{1-\sin x}{1-\sin^2 x} dx$$=\pi[x]^{\pi}_0 - \pi\int\limits_{0}^{\pi} \frac{1-\sin x}{\cos^2 x} dx$$=\pi(\pi-0) - \pi\int\limits_{0}^{\pi} (\sec^2 x - \sec x \tan x) dx$$=\pi^2 - \pi[\tan x - \sec x]^{\pi}_0$$=\pi^2 - \pi[(\tan \pi - \tan 0) - (\sec \pi - \sec 0)]$$=\pi^2 - \pi(0-0) + \pi(-1-1)$$=\pi^2 - 2\pi = \pi(\pi-2)$$\therefore I = \frac{\pi}{2} (\pi-2)$