Question

Evaluate:${\int }_0^{\pi }\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}dx$

Answer 1

Let $I={\int }_0^{\pi }\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}dx$ ......$\left(1\right)$

Replace $x\to \pi -x$

$I={\int }_0^{\pi }\frac{e^{\cos (\pi -x)}}{e^{\cos (\pi -x)}+e^{-\cos (\pi -x)}}d(\pi -x)$

$={\int }_0^{\pi }\frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}dx$ ......$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

$2I={\int }_0^{\pi }\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}dx+{\int }_0^{\pi }\frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}dx$

$2I={\int }_0^{\pi }\frac{e^{\cos x}+e^{-\cos x}}{e^{\cos x}+e^{-\cos x}}dx$

$2I=\int dx$

$2I={\left[x\right]}_0^{\pi }=\pi$

$\therefore I=\frac{\pi }{2}$