Official solution
\[\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {\left( {1 + {{\cos }^2}x} \right)\left( {{e^{\cos x}} + {e^{ - \cos x}}} \right)}}dx} \]
Let \[\cos x = t\]
\[\sin xdx = dt\]
\[ = \int\limits_1^{ - 1} {{{ - {e^t}dt} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}} \]
\[I = \int\limits_{ - 1}^1 {{{{e^t}} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}dt} \] ...... (i)
\[I = \int\limits_{ - 1}^1 {{{{e^{ - t}}} \over {\left( {1 + {t^2}} \right)\left( {{e^{ - t}} + {e^t}} \right)}}dt} \] .... (ii)
Adding (i) and (ii)
\[2I = \int\limits_{ - 1}^1 {{{dt} \over {1 + {t^2}}}} \]
\[2I = \left. {{{\tan }^{ - t}}} \right|_{ - 1}^1\]
\[2I = {\pi \over 4} - \left( { - {\pi \over 4}} \right)\]
\[2I = {\pi \over 2}\]
\[I = {\pi \over 4}\]