Question

The value of $\int\limits_{0}^{2\pi} \frac{dx}{e^{sinx} +1}$ is

• A. $\pi$
• B. $0$
• C. $3\pi$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (A)

Let $I =\int\limits_{0}^{2\pi} \frac{dx}{e^{sinx} +1} \quad...\left(i\right) $

$\Rightarrow I =\int\limits_{0}^{2\pi } \frac{dx}{e^{sin\left(2\pi -x\right)} +1} \,\,\,\, $(by property)

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

$ \Rightarrow I =\int\limits_{0}^{2\pi } \frac{e^{sin x}}{e^{sinx} +1} \quad...\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$, we get

$ 2I =\int\limits_{0}^{2\pi } 1\cdot dx = 2\pi \quad\therefore I = \pi $