Integration from Purcell

$\begin{aligned} I & = \int_0^{2 \pi} \frac {x|\sin x|}{1+\cos^2 x} dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^{2 \pi} \left( \frac {x|\sin x|}{1+\cos^2 x} + \frac {(2\pi - x)|\sin x|}{1+\cos^2 x} \right) dx \\ & = \pi \int_0^{2 \pi} \frac {|\sin x|}{1+\cos^2 x} dx & \small \color{#3D99F6} \text{Since the integrand is even} \\ & = 2\pi \int_0^\pi \frac {\sin x}{1+\cos^2 x} dx & \small \color{#3D99F6} \text{The integrand is symmetrical about }\frac \pi 2. \\ & = 4\pi \int_0^\frac \pi2 \frac {\sin x}{1+\cos^2 x} dx & \small \color{#3D99F6} \text{Let }u = \cos x \implies du = - \sin x \ dx. \\ & = 4\pi \int_0^1 \frac 1{1+u^2} du \\ & = 4\pi \tan^{-1} u \bigg|_0^1 \\ & = \boxed{\pi^2} \end{aligned}$