Evaluate: ∫₀^π (xtan 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: $2I = \pi\int_0^\pi\frac{\sin x}{1+\sin x}dx = \pi\int_0^\pi\left(1-\frac{1}{1+\sin x}\right)dx$ Since $1+\sin x = 2\sin^2\left(\frac\pi4+\frac x2\right)$: $\int_0^\pi\frac{dx}{1+\sin x} = \int_0^\pi\frac12\operatorname{cosec}^2\left(\frac\pi4+\frac x2\right)dx = \left[-\cot\left(\frac\pi4+\frac x2\right)\right]_0^\pi = 1+1 = 2$. $2I = \pi(\pi-2)$ Hence $I = \frac{\pi(\pi-2)}{2}$.