Evaluate $\int\limits_{0}^{2\pi} \frac{dx}{1 + e^{\sin x}}.$

$\pi$

The correct answer is Option (2) → $\pi$

$I = \int\limits_{0}^{2\pi} \frac{dx}{1 + e^{\sin x}}$

$= \int\limits_{0}^{\pi} \left[ \frac{1}{1 + e^{\sin x}} + \frac{1}{1 + e^{\sin(2\pi - x)}} \right] dx$

$\left[ ∵\int\limits_{0}^{2a} f(x) dx = \int\limits_{0}^{a} \{f(x) + f(2a-x)\} dx \right]$

$= \int\limits_{0}^{\pi} \left[ \frac{1}{1 + e^{\sin x}} + \frac{1}{1 + e^{-\sin x}} \right] dx$

$= \int\limits_{0}^{\pi} \left[ \frac{1}{1 + e^{\sin x}} + \frac{e^{\sin x}}{1 + e^{\sin x}} \right] dx$

$= \int\limits_{0}^{\pi} \frac{1 + e^{\sin x}}{1 + e^{\sin x}} dx$

$= \int\limits_{0}^{\pi} 1 dx$

$= [x]_{0}^{\pi} = \pi$