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)

$\frac{\pi}{2}$

Answer 2

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

So. the correct answer is (C): $\frac {\pi}{2}$