The value of the integral $\int\limits_0^\pi \frac{1}{e^{\cos x}+1} d x$, is

$\frac{\pi}{2}$

Let $I=\int\limits_0^\pi \frac{1}{e^{\cos x}+1} d x$            ........(i)

$I =\int\limits_0^\pi \frac{1}{e^{\cos (\pi-x)}+1} d x$

$\Rightarrow I =\int\limits_0^\pi \frac{1}{e^{-\cos x}+1} d x$         [Using $\int\limits_0^a f(x) dx = \int\limits_0^a f(a-x) dx$]

$\Rightarrow I=\int\limits_0^\pi \frac{e^{\cos x}}{e^{\cos x}+1} d x$         ....(ii)

Adding (i) and (ii), we get

$2 I=\int\limits_0^\pi 1 d x=\pi \Rightarrow I=\frac{\pi}{2}$