$I = \int\limits_{ - \pi }^\pi {\frac{{{{\sin }^2}x}}{{1 + {e^x}}}dx} = ?$

Let $x=-t$, so $dx=-dt$

$I = \int\limits_\pi ^{ - \pi } { - \frac{{{{\sin }^2}t}}{{1 + {e^{ - t}}}}dt} = \int\limits_{ - \pi }^\pi {\frac{{{e^t}{{\sin }^2}t}}{{1 + {e^t}}}dt} = \int\limits_{ - \pi }^\pi {\frac{{{e^x}{{\sin }^2}x}}{{1 + {e^x}}}dx}$

$I + I = \int\limits_{ - \pi }^\pi {\frac{{{{\sin }^2}x}}{{1 + {e^x}}}dx} + \int\limits_{ - \pi }^\pi {\frac{{{e^x}{{\sin }^2}x}}{{1 + {e^x}}}dx}$

$\Rightarrow 2I = \int\limits_{ - \pi }^\pi {\frac{{(1 + {e^x}){{\sin }^2}x}}{{1 + {e^x}}}dx} = \int\limits_{ - \pi }^\pi {{{\sin }^2}xdx}$

$\Rightarrow 2I = \int\limits_0^\pi {2{{\sin }^2}xdx} = \int\limits_0^\pi {(1 - \cos 2x)dx}$

$\Rightarrow 2I = \left. {\left( {x - \frac{{\sin 2x}}{2}} \right)} \right|_0^\pi = \pi$

$\Rightarrow I = \frac{\pi }{2}$