Question
Evaluate: $\int\limits_0^{2\pi}\frac{1}{1+e^{\text {sin x}}}\text{dx}$
$\text{I}=\int\limits_{0}^{2\pi}\frac{dx}{1+e^{sin(2\pi-x)}}\int\limits_0^{2\pi}\frac{dx}{1+e^{-sin x}}=\int\limits_0^{2\pi}\frac{dx}{1+\frac{1}{e^{sinx}}}$
$\text{I}=\int\limits_{0}^{2\pi}\frac{e^{\text{sin x}}dx}{e^{\text{-sin x}}+1}\ .......\text{(ii)}$
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 $.
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$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$
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