2016/024)integrate $\int_{x = 0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} dx$

as we have $\int_{x = a}^{b} f(a) = \int_{x = a}^{b} f(b)$
$\int_{x = 0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} dx$
$=\int_{x = 0}^{\pi}\frac{e^{\cos(\pi-x) }}{e^{\cos (\pi-x) }+e^{-\cos (\pi- x) }} dx$
$=\int_{x = 0}^{\pi}\frac{e^{^-\cos x}}{e^{- \cos (x) }+e^{\cos  x}} dx$
$=\int_{x = 0}^{\pi}\frac{e^{^-\cos x}}{e^{\cos (x) }+e^{-\cos  x}} dx$
hence
$\int_{x = 0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} dx= \int_{x = 0}^{\pi}\frac{e^{^-\cos x}}{e^{\cos (x) }+e^{-\cos  x}} dx$
$=\frac{1}{2}\int_{x = 0}^{\pi}(\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} + \frac{e^{^-\cos x}}{e^{\cos (x) }+e^{-\cos  x}})dx$
$=\frac{1}{2}\int_{x = 0}^{\pi} 1 dx = \frac{\pi}{2}$