\(\int_{0}^{\pi}\frac{e^{\cos x}}{e^{\cos x+e^{-\cos x}}}dx\)
Answer & explanation
Correct answer: option 3
$I=\int\limits_{0}^{\pi}\frac{e^{\cos x}}{e^{\cos x+e^{-\cos x}}}dx$ ...(1)
$I=\int_{0}^{\pi}\frac{e^{-\cos x}}{e^{-\cos x+e^{\cos x}}}dx$ ...(2) [as $\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{b}f(a+b-x)dx$]
so $2I=\int_{0}^{\pi}1dx⇒I=\frac{\pi}{2}$