Let I = \(\int\limits_0^{2\pi} \frac1{1 + e^{\sin x}}dx\) ....(i)
Applying property \(\int \limits_0^a f(x) dx= \int \limits_0 ^a f(a -x) dx,\) we get
\(I =\int \limits_0^{2\pi } \frac {dx}{1 +e^{\sin (2\pi -x)}}\)
\(= \int \limits_0^{2\pi} \frac{dx}{1 + e^{-\sin x}} \)
\(= \int\limits_{0}^{2\pi} \frac{dx}{1 + \frac 1{e^{\sin x}}}\)
\(I = \int \limits_0^{2\pi} \frac{e^{\sin x}dx}{e^{\sin x + 1}}\) .....(ii)
Adding (i) and (ii), we get
\(2I = \int \limits_0^{ 2{\pi}} \frac{dx}{1 + e^{\sin x}} + \int \limits_0^{2\pi} \frac{e^{\sin x}}{1 + e^{\sin x}}\)
\(= \int \limits_0^{2\pi} \frac{1 + e^{\sin x}}{1 + e^{sin x}} dx\)
\(= \int\limits_0^{2x} dx \)
\(= [x]_0^{2\pi}\)
⇒ \(2I = 2\pi\)
⇒ \(I = \pi\)
