Question

For $a>0$ the value of ${\int }_{-\pi }^{\pi }\frac{{\sin }^{2}x}{1+a^x}dx$

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

Answer 1

The correct option is A $\frac{\pi }{2}$

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

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

Adding (1) and (2)

$2I={\int }_{-\pi }^{\pi }{\sin }^{2}dx$

$2I=2{\int }_0^{\pi }{\sin }^{2}xdx$

$=2{\int }_0^{\pi }\left[\frac{1-\cos 2x}{2}\right]dx$

$={\int }_0^{\pi }(1-\cos 2x)dx$

$={[x-\frac{\sin 2x}{2}]}_0^{\pi }$

$=\left[\pi \right]$

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