Question

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

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

Answer 1

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

Let, $f\left(x\right)={\int }_{-\pi }^{\pi }\frac{{\cos }^{2}x}{1+a^x}dx(a>0)$ ..(1)

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

$\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)$

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

Adding (1) and (2), we get
$2f\left(x\right)={\int }_{-\pi }^{\pi }{\cos }^{2}xdx=2{\int }_0^{\pi }{\cos }^{2}xdx$
$=2f\left(x\right)=4\times \frac{1}{2}\times \frac{\pi }{2}$
$\Rightarrow f\left(x\right)=\frac{\pi }{2}$
Ans: A