Practicing Success
$I=\int\limits_0^{2 \pi} \frac{1}{1+e^{\sin x}} d x$ is equal to |
$\pi$ $2~\pi$ $\frac{\pi}{2}$ none of these |
$\pi$ |
Using property $\int\limits_0^{2 a} f(x) d x=\int\limits_0^a\{f(x)+f(2 a-x)\} d x$ $I=\int\limits_0^\pi\left\{\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{\sin (2 \pi-x)}}\right\} d x$ $\Rightarrow I=\int\limits_0^\pi\left\{\frac{1}{1+e^{\sin x}}+\frac{e^{\sin x}}{1+e^{\sin x}}\right\} d x=\int\limits_0^\pi 1 . d x=\pi$ |