Question

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

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

Answer 1

Answer: (C)

$\frac{\pi }{2}$

$I={\int }_{-\pi }^{\pi }\frac{co{s}^{2}x}{1+a^x}dx$ __________ (1)

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

$\Rightarrow I={\int }_{-\pi }^{\pi }\frac{co{s}^2(-x)}{1+a^{-x}}dx={\int }_{-\pi }^{\pi }\frac{co{s}^{2}x}{(1+a^x)}{a}^{x}dx$ _______ (2)

(1)+ (2)

$2I={\int }_{-\pi }^{\pi }(1+a^x)\frac{co{s}^{2}x}{(1+a^x)}dx$

$\Rightarrow I=\frac{1}{2}{\int }_{-\pi }^{\pi }co{s}^{2}xdx$

$=\frac{1}{4}{\int }_{-\pi }^{\pi }2co{s}^{2}xdx=\frac{1}{4}{\int }_{-\pi }^{\pi }(1+cosx)dx$

$I={\int }_{-\pi }^{\pi }\frac{1}{4}dx+{\int }_{-\pi }^{\pi }\frac{1}{4}cos2xdx$

$I=\frac{1}{4}[\pi -(-\pi \left)\right]+\frac{1}{8}sin2x{]}_{-\pi }^{\pi }=\frac{2\pi }{9}=\frac{\pi }{2}$

$\therefore {\int }_{-\pi }^{\pi }\frac{co{s}^{2}x}{1+a^x}dx=\frac{\pi }{2}$