Question

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

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

Answer 1

The correct option is C $\frac{\pi }{4}$

Let $I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}x}{1+2^x}dx$.......(1)

or, $I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^2(-\frac{\pi }{2}+\frac{\pi }{2}-x)}{1+2^{(-\frac{\pi }{2}+\frac{\pi }{2}-x)}}dx$ ....... $\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\right]$

$I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}x}{1+2^{-x}}dx$

$I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{2^x{\sin }^{2}x}{1+2^x}dx$ ....... $\left(2\right)$

Now adding $\left(1\right)$ and $\left(2\right)$ we get,

$2I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2}xdx$

$2I=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2}xdx$ ...... [ $\because {\sin }^{2}x$ is an even function]

$2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}(1-\cos 2x)dx$

$2I=\frac{\pi }{2}$

$\Rightarrow I=\frac{\pi }{4}$