Evaluate: ∫₀^π (x tan x)/(sec x + tan x) dx

$\frac{\tan x}{\sec x + \tan x} = \frac{\sin x}{1 + \sin x}$. Let $I = \int_0^\pi \frac{x\sin x}{1+\sin x}dx$. Using $\int_0^a f(x)dx = \int_0^a f(a-x)dx$: $I = \int_0^\pi \frac{(\pi - x)\sin x}{1+\sin x}dx$. Adding: $2I = \pi\int_0^\pi \frac{\sin x}{1+\sin x}dx = \pi\int_0^\pi\left(1 - \frac{1}{1+\sin x}\right)dx$. $\int_0^\pi \frac{dx}{1+\sin x} = \int_0^\pi \frac{1-\sin x}{\cos^2 x}dx = \left[\tan x - \sec x\right]_0^\pi = 1 + 1 = 2$. So $2I = \pi(\pi - 2)$. Hence $I = \frac{\pi(\pi - 2)}{2} = \frac{\pi^2}{2} - \pi$.