$I=\int\limits_0^{2 \pi} \frac{1}{1+e^{\sin x}} d x$ is equal to
Answer & explanation
Correct answer: option 1
Using property $\int\limits_0^{2 a} f(x) d x=\int\limits_0^a\{f(x)+f(2 a-x)\} d x$
$I=\int\limits_0^\pi\left\{\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{\sin (2 \pi-x)}}\right\} d x$
$\Rightarrow I=\int\limits_0^\pi\left\{\frac{1}{1+e^{\sin x}}+\frac{e^{\sin x}}{1+e^{\sin x}}\right\} d x=\int\limits_0^\pi 1 . d x=\pi$