$\int_{-π}^{π}\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}dx$ equals:
Answer & explanation
Correct answer: option 2
$I=\int\limits_{-π}^{π}\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}dx$
$\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{b}f(a+b-x)dx$
$⇒I=\int\limits_{-π}^{π}\frac{e^{-\sin x}}{e^{\sin x}+e^{-\sin x}}dx$
$⇒2I=\int\limits_{-π}^{π}dx⇒I=π$