Question
Evaluate the following integrals:
$\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}$

Answer

Let $\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin(2\pi-\text{x})}}{\text{e}^{\sin(2\pi-\text{x})}+\text{e}^{-\sin(2\pi-\text{x})}}\text{ dx}$ $\Bigg(\int\limits^\text{a}_0\text{f(x)}\text{dx}=\int\limits^\text{a}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg)$ 
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{-\sin\text{x}}}{\text{e}^{-\sin\text{x}}+\text{e}^{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{2\pi}_0\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{2\pi}_0$
$\Rightarrow2\text{I}=2\pi-0$
$\Rightarrow\text{I}=\pi$

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