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}$