Question

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

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

Answer 1

Answer: (A)

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

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

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

$\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}\frac{e^{sinx}}{e^{sinx}+1}...\left(ii\right)$

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

$2I=\underset{0}{\overset{2\pi }{\int }}1\cdot dx=2\pi \therefore I=\pi$