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