∫_-π^π(e^sin x)/(e^sin x+e^-sin x) dx is equal to

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

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

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

Adding the two forms of I, the numerators add up to the common denominator:

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

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

    \(I = \pi\)

Tip: A result worth remembering: for any g with \(g(x) + g(-x) \ne 0\), \(\displaystyle\int_{-a}^{a} \frac{g(x)}{g(x) + g(-x)}\,dx = a\), half the length of the interval.