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

The Correct Option is C

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): \(\frac {\pi}{2}\)