∫_0^π(e^cos x)/(e^cos x+e^-cos x) dx =

The correct answer is Option (4) → \(\displaystyle\frac{\pi}{2}\)

Let

    \( \displaystyle I = \int_{0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}\,dx \qquad (1) \)

By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx\), replace x by \(\pi - x\). Since \(\cos(\pi - x) = -\cos x\):

    \( \displaystyle I = \int_{0}^{\pi}\frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}\,dx \qquad (2) \)

Adding (1) and (2), the numerators add up to the common denominator:

    \( \displaystyle 2I = \int_{0}^{\pi} 1\,dx = \pi \)

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

Tip: An integral of the form \(\int_{a}^{b}\frac{f(x)}{f(x)+f(a+b-x)}\,dx\) always equals \(\frac{b-a}{2}\), half the length of the interval. Here it is \(\frac{1}{2}\times\pi = \frac{\pi}{2}\).