Question

The value of ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+e^{\sin x}}dx$ is:

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\mathrm{\pi }$
• D. $\frac{3\mathrm{\pi }}{2}$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{2}$

Explanation for the correct option:

Step 1: Rewrite the given integral.

Given, ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+e^{\sin x}}dx$

Assume that, $I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+e^{\sin x}}dx...1$.

We know that, ${\int }_a^{b}f\left(x\right)dx={\int }_a^b\left[f(a+b-x)\right]dx$.

So,

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

Step 2: Find the value of the given integral.

Add equation $1$ and equation $2$.

$$\begin{gathered} 2I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left[\frac{1+e^{\sin x}}{e^{\sin x}+1}\right]dx \\ \Rightarrow I=\frac{{\left[x\right]}_{-{\displaystyle \frac{\mathrm{\pi }}{2}}}^{{\displaystyle \frac{\mathrm{\pi }}{2}}}}{2} \\ \Rightarrow I=\frac{\mathrm{\pi }}{2} \end{gathered}$$

Therefore, the value of the given integral is $\frac{\mathrm{\pi }}{2}$.

Hence, option (A) is the correct answer.