\(\int_{0}^{\pi}\frac{e^{\cos x}}{e^{\cos x+e^{-\cos x}}}dx\)

\(\pi\) 0 \(\frac{\pi}{2}\) \(\frac{\pi}{4}\)

\(\frac{\pi}{2}\)

$I=\int\limits_{0}^{\pi}\frac{e^{\cos x}}{e^{\cos x+e^{-\cos x}}}dx$  ...(1) $I=\int_{0}^{\pi}\frac{e^{-\cos x}}{e^{-\cos x+e^{\cos x}}}dx$  ...(2)  [as $\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{b}f(a+b-x)dx$] so $2I=\int_{0}^{\pi}1dx⇒I=\frac{\pi}{2}$