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

Answer: (B)

Let $I={\int }_{-\pi }^{\pi }\frac{{\cos }^{2}x}{1+a^x}dx,a>0$ ...(i)
$I={\int }_{-\pi }^{\pi }\frac{{\cos }^{2}x}{1+a^{-x}}dx$ (put $x=-x$ ) ...(ii)
On adding Eqs. (i) and (ii), we get
$2I={\int }_{-\pi }^{\pi }\frac{(1+a^x)\cos x}{(1+a^x)}dx$
$\Rightarrow$ $2I={\int }_{-\pi }^{\pi }{\cos }^{2}xdx$
$\Rightarrow$ $2I={\int }_{-\pi }^{\pi }\left(\frac{\cos 2x+1}{2}\right)dx$
$\Rightarrow$ $2I=\frac{1}{2}{\left[\frac{\sin 2x}{2}+x\right]}_{-\pi }^{\pi }$
$\Rightarrow$ $2I=\frac{1}{2}(\pi +\pi )$
$\Rightarrow$ $I=\frac{\pi }{2}$