Question

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

• A. $4\mathrm{\pi }$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{8}$
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{4}$

Explanation for the correct option:

Step 1: Rewrite the given integral.

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

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

We know that, ${\int }_{-a}^{a}f\left(x\right)dx={\int }_0^a\left[f\left(x\right)+f(-x)\right]dx$.

So,

$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{{\sin }^{2}x}{1+2^x}+\frac{{\sin }^2(-x)}{1+2^{-x}}\right]dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{{\sin }^{2}x}{1+2^x}+\frac{{\sin }^2\left(x\right)}{1+{\displaystyle \frac{1}{2^x}}}\right]dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{{\sin }^{2}x}{1+2^x}+\frac{2^x{\sin }^2\left(x\right)}{2^x+1}\right]dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^{2}x\left[\frac{1+2^x}{1+2^x}\right]dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^{2}xdx...1 \end{gathered}$$

Step 2: Find the value of the given integral.

We know that,$\sin \left(\frac{\mathrm{\pi }}{2}-x\right)=\cos x$.

Therefore, equation $1$ becomes.

$I={\int }_0^{\frac{\mathrm{\pi }}{2}}{\cos }^{2}xdx...2$ $\left[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\right]$

Add equation $1$ and equation $2$.

$$\begin{gathered} 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}1dx \\ \Rightarrow I=\frac{{\left[x\right]}_0^{{\displaystyle \frac{\mathrm{\pi }}{2}}}}{2} \\ \Rightarrow I=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Therefore, the value of the given integral is $\frac{\mathrm{\pi }}{4}$.

Hence, option (B) is the correct answer.