Evaluate: \int_0^{2\pi} \frac{1}{1+e^{\sin x}} dx

Let I = \int_0^{2\pi} \frac{dx}{1+e^{\sin x}} -- (1)
We know that \int_0^a f(x)dx = \int_0^a f(a-x)dx
\int_0^{2\pi} \frac{dx}{1+e^{\sin(2\pi - x)}} = \int_0^{2\pi} \frac{dx}{1+e^{-\sin x}} \Rightarrow \int_0^{2\pi} \frac{e^{\sin x} dx}{e^{\sin x}+1} -- (2)
Adding 1 and 2, we get
2I = \int_0^{2\pi} \frac{dx}{1+e^{\sin x}} + \int_0^{2\pi} \frac{e^{\sin x} dx}{e^{\sin x}+1}
2I = \int_0^{2\pi} 1 dx \Rightarrow 2I = [x]_0^{2\pi} = 2\pi \therefore I = \pi