The value of integral ∫_-(π)/2^(π)/2displaystyle1/(1+e^sin x) dx is

The correct answer is Option (2) → \( \displaystyle\frac{\pi}{2} \)

Let I be the integral. By the property \( \int_{-a}^{a}f(x)\,dx=\int_{-a}^{a}f(-x)\,dx \), and since \( \sin(-x)=-\sin x \):

    \( \displaystyle I=\int_{-\pi/2}^{\pi/2}\frac{1}{1+e^{-\sin x}}\,dx \)

    \( \displaystyle =\int_{-\pi/2}^{\pi/2}\frac{e^{\sin x}}{e^{\sin x}+1}\,dx \)

Adding the two forms of I:

    \( \displaystyle 2I=\int_{-\pi/2}^{\pi/2}\frac{1+e^{\sin x}}{1+e^{\sin x}}\,dx \)

    \( \displaystyle =\int_{-\pi/2}^{\pi/2}1\,dx \)

    \( \displaystyle =\pi \)

    \( \displaystyle I=\frac{\pi}{2} \)

Tip: For integrals of the form \( \frac{1}{1+e^{g(x)}} \) with g odd on [−a, a], the answer is always half the length of the interval, that is, a.