Question

The value of ${}_{-x}^{\pi }\frac{co{s}^{2}x}{1+a^x}dx,a>0,is$ [IIT Screening 2001; AIEEE 2005]

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

Answer 1

The correct option is C

$\frac{\pi }{2}$

$I={\int }_{-\pi }^{\pi }\frac{co{s}^{2}x}{1+a^x}dx={\int }_{\pi }^{-\pi }\frac{co{s}^{2}x}{1+a^{-x}}(-dx)(Putting-xforx)$
$={\int }_{-\pi }^{\pi }\frac{co{s}^{2}x}{1+a^{-x}}dx\Rightarrow I+I={\int }_{-\pi }^{\pi }co{s}^{2}x(\frac{1}{1+a^x}+\frac{1}{1+a^{-x}})dx$
$={\int }_{-\pi }^{\pi }co{s}^{2}xdx\Rightarrow 2I=2{\int }_0^{\pi }co{s}^{2}x.dx={\int }_0^{\pi }(1+cos2x)dx$
$\Rightarrow 2I=[x{]}_0^{\pi }+{\left[\frac{sin2x}{2}\right]}_0^{\pi }\Rightarrow 2I=\pi \Rightarrow I=\frac{\pi }{2}$