A Calculus Problem from Aly Ahmed 200716

$\begin{aligned} I & = \int_0^\pi \frac {2x\sin x}{3+\cos (2x)} dx & \small \blue{\text{By identity }\int_a^b f(x)\ dx = \int_a^b f(a+b-x)\ dx} \\ & = \frac 12 \int_0^\pi \left(\frac {2x\sin x}{3+ \cos(2x)} + \frac {2(\pi-x)\sin (\pi-x)}{3+ \cos (2(\pi -x))} \right) dx \\ & = \frac 12 \int_0^\pi \left(\frac {2x\sin x}{3+ \cos(2x)} + \frac {(2\pi-2x)\sin x}{3+ \cos (2x)} \right) dx \\ & = \int_0^\pi \frac {\pi\sin x}{3+ \cos(2x)} dx \\ & = \int_0^\pi \frac {\pi\sin x}{2+ 2\cos^2 x} dx & \small \blue{\text{Let }u = \cos x \implies du = - \sin x \ dx} \\ & = \frac \pi 2 \int_{-1}^1 \frac {dx}{1+u^2} dx \\ & = \frac {\pi \tan^{-1} u}2 \bigg|_{-1}^1n= \frac {\pi^2}4 \approx \boxed{2.47} \end{aligned}$