Question

The value of the integral ${\int }_{-\pi }^{\pi }\frac{co{s}^{2}x}{1+a^x}dx$, where a > 0, is

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

Answer 1

Answer: (A)

$\pi$

We have, $I={\int }_{-\pi }^{\pi }\frac{{\cos }^{2}x}{a^x+1}$ ...... $\left(i\right)$

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

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

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

Adding (i) and (ii), we get

$2I={\int }_{-\pi }^{\pi }{\cos }^{2}x\because \cos 2x=2{\cos }^{2}x-1{\cos }^{2}x=\frac{\cos 2x+1}{2}\therefore 2I={\int }_{-\pi }^{\pi }\frac{\cos 2x}{2}+{\int }_{-\pi }^{\pi }\frac{1}{2}dx={\left|\frac{sin2x}{4}\right|}_{-\pi }^{\pi }+{\left|\frac{x}{2}\right|}_{-\pi }^{\pi }=0+\frac{\pi -(-\pi )}{2}\therefore I=\pi$