Let $I=\int_{-\pi}^{\pi}\dfrac{sin^2x}{1+a^x}dx,a>0$.Find the value of $I$ I have taken $a=e$ but using partial integration, getting no luck.Please help.I avoided purposefully 6 pages of needless scribble as that may deviate the answerer.

We know that $\displaystyle\int_{-a}^af(x)\,dx = \int_0^af(x)\,dx+\int_0^af(-x)dx$

so,$I = \displaystyle\int_{-\pi}^\pi\dfrac{\sin^2(x)}{1+a^x}dx $

$I = \displaystyle\int_0^\pi\dfrac{\sin^2(x)}{1+a^x}+\dfrac{\sin^2(x)}{1+a^{-x}}\,dx$

$I = \displaystyle\int_0^\pi\dfrac{\sin^2(x)}{1+a^x}+\dfrac{a^x\cdot\sin^2(x)}{1+a^{x}}\,dx$

$I = \displaystyle\int_0^\pi\dfrac{\sin^2(x)(1+a^x)}{1+a^x}$

$I =\displaystyle\int_0^\pi\sin^2(x)\,dx$

I assume you can continue from here. ask if you need help

HINT: let $\sin^2(x) = \dfrac{1-\cos(2x)}{2}$