The value of $$\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx} $$ is equal to:

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