$$\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {\left( {1 + {{\cos }^2}x} \right)\left( {{e^{\cos x}} + {e^{ - \cos x}}} \right)}}dx} $$

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)

$$2I = \int\limits_{ - 1}^1 {{{dt} \over {1 + {t^2}}}} $$

$$2I = \left. {{{\tan }^{ - t}}} \right|_{ - 1}^1$$

$$2I = {\pi \over 4} - \left( { - {\pi \over 4}} \right)$$

$$2I = {\pi \over 2}$$

$$I = {\pi \over 4}$$