Question

The value of${\int }_0^{2\pi }\frac{1}{e^{\sin x}+1}dx$ is

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

Answer 1

The correct option is A

$\pi$

The explanation for the correct answer.

Solve for the value of the given integral.

$$\begin{gathered} I={\int }_0^{2\pi }\frac{1}{e^{\sin x}+1}dx \\ ={\int }_0^{2\pi }\frac{dx}{e^{\sin \left(2\mathrm{\pi }-\mathrm{x}\right)}+1} \\ ={\int }_0^{2\pi }\frac{dx}{e^{-\sin \left(\mathrm{x}\right)}+1} \\ ={\int }_0^{2\pi }\frac{e^{\sin \left(x\right)}dx}{1!+e^{\sin \left(\mathrm{x}\right)}} \\ 2I={\int }_0^{2\pi }\frac{e^{\sin \left(x\right)}dx}{1!+e^{\sin \left(\mathrm{x}\right)}} \\ ={\int }_0^{2\pi }dx \\ I=\mathrm{\pi } \end{gathered}$$

Hence, option(A) is correct.