$I = \int\limits_0^\pi {\frac{{2(\pi – x)\sin (\pi – x)}}{{3 + \cos 2(\pi – x)}}dx} $

$ = \int\limits_0^\pi {\frac{{2(\pi – x)\sin x}}{{3 + \cos 2x}}dx} $

$2I = \int\limits_0^\pi {\frac{{2x\sin x}}{{3 + \cos 2x}}dx} + \int\limits_0^\pi {\frac{{2(\pi – x)\sin x}}{{3 + \cos 2x}}dx} = \int\limits_0^\pi {\frac{{2\pi \sin x}}{{3 + \cos 2x}}dx} $

$ \Rightarrow I = \pi \int\limits_0^\pi {\frac{{\sin x}}{{3 + \cos 2x}}dx} $

$ = \pi \int\limits_0^\pi {\frac{{\sin x}}{{3 + 2{{\cos }^2}x – 1}}dx} $

$ = \frac{\pi }{2}\int\limits_0^\pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}dx} $

Let, $\cos x = t$

$ \Rightarrow – \sin xdx = dt$

$I = \frac{\pi }{2}\int\limits_1^{ – 1} {\frac{{ – dt}}{{1 + {t^2}}}} $

$ = \frac{\pi }{2}\int\limits_{ – 1}^1 {\frac{{dt}}{{1 + {t^2}}}} $

$ = \frac{{\pi \times 2}}{2}\int\limits_0^1 {\frac{{dt}}{{1 + {t^2}}}} $

$ = \pi \left. {{{\tan }^{ – 1}}t} \right|_0^1$

$ = \frac{{{\pi ^2}}}{4}$