Question

Evaluate each of the following integrals:

${\int }_0^{2\mathrm{\pi }}\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}dx$

Answer 1

Let I = ${\int }_0^{2\mathrm{\pi }}\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}dx$ .....(1)

Then,

$$\begin{gathered} I={\int }_0^{2\mathrm{\pi }}\frac{e^{\sin \left(2\mathrm{\pi }-x\right)}}{e^{\sin \left(2\mathrm{\pi }-x\right)}+e^{-\sin \left(2\mathrm{\pi }-x\right)}}dx\left({\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right) \\ ={\int }_0^{2\mathrm{\pi }}\frac{e^{-\sin x}}{e^{-\sin x}+e^{\sin x}}dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_0^{2\mathrm{\pi }}\frac{e^{\sin x}+e^{-\sin x}}{e^{\sin x}+e^{-\sin x}}dx \\ \Rightarrow 2I={\int }_0^{2\mathrm{\pi }}dx \\ \Rightarrow 2I={\overline{)x}}_0^{2\mathrm{\pi }} \\ \Rightarrow 2I=2\mathrm{\pi }-0 \\ \Rightarrow I=\mathrm{\pi } \end{gathered}$$