Question

If $I={\int }_{-\pi }^{\pi }\frac{{\sin }^{2}x}{1+a^x}dx$ for $a>0,$

then $I$ is equal to?

• A. $\pi$
• B. $\pi /2$
• C. $a\pi$
• D. $a\pi /2$

Answer 1

The correct option is B $\pi /2$

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

$I={\int }_{-\pi }^{\pi }\frac{a^x{\sin }^{2}x}{1+a^x}dx...................\left(2\right)$
Adding (1) and (2), we get
$2I={\int }_{-\pi }^{\pi }{\sin }^{2}xdx$
$=2{\int }_0^{\pi }{\sin }^{2}xdx$
$={\int }_0^{\pi }(1-\cos 2x)dx$
$={(x-\frac{\sin 2x}{2})}_0^{\pi }=\pi$

$\Rightarrow I=\pi /2$