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Q. The value of $\int\limits_0^\pi \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x$ is

Integrals

Solution:

$I=\int\limits_0^\pi \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x=\int\limits_0^\pi \frac{e^{\cos (\pi-x)}}{e^{\cos (\pi-x)}+e^{-\cos (\pi-x)}} d x$
$=\int\limits_0^\pi \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} d x$
$\therefore 2 I=\int\limits_0^\pi \frac{e^{\cos x}+e^{-\cos x}}{e^{\cos x}+e^{-\cos x}} d x=\int\limits_0^\pi d x=\pi$
$I=\frac{\pi}{2}$