Evaluate: ∫₀^(2π) 1/(1 + e^(sin x)) dx

Let $I = \int_0^{2\pi}\frac{dx}{1+e^{\sin x}}$. By $\int_0^{a} f(x)dx = \int_0^{a} f(a-x)dx$ with $a = 2\pi$, $\sin(2\pi - x) = -\sin x$: $I = \int_0^{2\pi}\frac{dx}{1+e^{-\sin x}} = \int_0^{2\pi}\frac{e^{\sin x}}{1+e^{\sin x}}dx$ Adding: $2I = \int_0^{2\pi}\frac{1 + e^{\sin x}}{1+e^{\sin x}}dx = \int_0^{2\pi}dx = 2\pi$ Hence $I = \pi$.