If $I=\int\limits_{-\pi}^\pi \frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}} d x$, then I equals
Answer & explanation
Correct answer: option 3
Using property $\int\limits_{-a}^a f(x) d x=\int\limits_0^a\{f(x)+f(-x)\} d x$
$I=\int\limits_0^\pi\left\{\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}+\frac{e^{-\sin x}}{e^{-\sin x}+e^{\sin x}}\right\} d x=\int\limits_0^\pi 1 . d x=\pi$