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

Solution:

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

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