$\int\limits_{-\pi}^{\pi}\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}dx$ is equal to

$\pi$

The correct answer is Option (3) → $\pi$

$ I=\int_{-\pi}^{\pi}\frac{e^{\sin x}}{\,e^{\sin x}+e^{-\sin x}\,}\,dx $

$ \frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}} =\frac{1}{1+e^{-2\sin x}} $

$ f(x)=\frac{1}{1+e^{-2\sin x}},\; f(-x)=\frac{1}{1+e^{2\sin x}} $

$ f(x)+f(-x)=1 $

$ I=\int_{-\pi}^{\pi}f(x)\,dx =\int_{-\pi}^{\pi}\frac{1}{2}(f(x)+f(-x))\,dx $

$ I=\frac{1}{2}\int_{-\pi}^{\pi}1\,dx $

$ I=\frac{1}{2}(2\pi)=\pi $

The value of the integral is $\pi$.