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

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}$

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