Evaluate: _0^ 2 1 1+e^ sin x dx - Vidyadip

Question

Evaluate: $\int\limits_0^{2\pi}\frac{1}{1+e^{\text {sin x}}}\text{dx}$

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Answer

Let I = $\int\limits_0^{2\pi}\frac{1}{1+e^{\text {sin x}}}\text{dx}\ ......\text{(i)}$ Applying properties $\int\limits_0^{\alpha}f\text{(x)dx}=\int\limits_0^af\text{(a-x)dx}\text{ we get}$$\text{I}=\int\limits_{0}^{2\pi}\frac{dx}{1+e^{sin(2\pi-x)}}\int\limits_0^{2\pi}\frac{dx}{1+e^{-sin x}}=\int\limits_0^{2\pi}\frac{dx}{1+\frac{1}{e^{sinx}}}$
$\text{I}=\int\limits_{0}^{2\pi}\frac{e^{\text{sin x}}dx}{e^{\text{-sin x}}+1}\ .......\text{(ii)}$
Adding (i) and (ii) we get$\text{2I}=\int\limits_{0}^{2\pi}\frac{dx}{1+e^{\text{sin x}}}+\int\limits_{0}^{2\pi}\frac{e^{\text{sin x}}dx}{1+e^{\text{sin x}}}=\int\limits_{0}^{2\pi}\frac{1+e^{\text{sin x}}}{1+e^{\text{sin x}}}\text{dx}$
$=\int\limits_{0}^{2\pi}\text{dx} =\big[ \text{x}\big]^{2\pi}_{0}$
$\Rightarrow$ 2I = 2$\pi $
$\Rightarrow$ I = $\pi $.

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