Nasty Integral

Call the integral $\large\mathscr{R}$ . $\underbrace{\overbrace{\displaystyle\int_{-\pi}^{\pi} \dfrac{2x}{1+ \cos^2x }\mathrm{d}x}^{=0}}_{\text{Odd ~Function}} + \underbrace{ \displaystyle\int_{-\pi}^{\pi} \dfrac{2x\sin x}{1+ \cos^2x} \mathrm{d}x }_{\text{Even Function}}$

$\large =4 \displaystyle\int_{0}^{\pi} \dfrac{x\sin x}{1+ \cos^2x} \mathrm{d}x$

Use $\small{\displaystyle\int_a^b f(x)\mathrm{d}x=\displaystyle\int_a^b f(a+b-x)\mathrm{d}x}$ .

$\implies \mathscr{R}= 4\pi\displaystyle\int_{0}^{\pi} \dfrac{\pi\sin x}{1+ \cos^2x} \mathrm{d}x-\underbrace{4 \displaystyle\int_{0}^{\pi} \dfrac{x\sin x}{1+ \cos^2x} \mathrm{d}x}_{\mathscr R}$

$\large\implies \mathscr{R}=2\pi\displaystyle\int_{0}^{\pi} \dfrac{\sin x}{1+ \cos^2x} \mathrm{d}x$

$\large =\left[2\pi\tan^{-1}\left(\cos x\right)\right]_{\pi}^0$

$\Large =\pi^2\approx\boxed{\color{#007fff}{9.86}}$