INTEGRALS — Maths STD 12 Science — Question

$\text{I}=\int\limits_{0}^{2\pi}\frac{dx}{1+e^{sin(2\pi-x)}}\int\limits_0^{2\pi}\frac{dx}{1+e^{-sin x}}=\int\limits_0^{2\pi}\frac{dx}{1+\frac{1}{e^{sinx}}}$

$\text{I}=\int\limits_{0}^{2\pi}\frac{e^{\text{sin x}}dx}{e^{\text{-sin x}}+1}\ .......\text{(ii)}$

$\text{2I}=\int\limits_{0}^{2\pi}\frac{dx}{1+e^{\text{sin x}}}+\int\limits_{0}^{2\pi}\frac{e^{\text{sin x}}dx}{1+e^{\text{sin x}}}=\int\limits_{0}^{2\pi}\frac{1+e^{\text{sin x}}}{1+e^{\text{sin x}}}\text{dx}$

$=\int\limits_{0}^{2\pi}\text{dx} =\big[ \text{x}\big]^{2\pi}_{0}$

$\Rightarrow$ 2I = 2$\pi $

$\Rightarrow$ I = $\pi $.