Question

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

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

Answer 1

Answer: (C)

$\frac{\pi }{2}$

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