Question

The value of ${\int }_{-\pi }^{\pi }\frac{{\cos }^{2}x}{1+a^x}$ dx, $a>0$, is

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

Answer 1

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

$I={\int }_{-\pi }^{\pi }\frac{{\cos }^{2}x}{1+a^x}dx$ ....(i)
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{-\pi }^{\pi }\frac{{\cos }^{2}x}{1+a^{-x}}dx$ ..(ii)
Adding (i) and (ii)
$I=\frac{1}{2}{\int }_{-\pi }^{\pi }{\cos }^{2}x$ dx= $\frac{\pi }{2}$.