Help me with this integration please.

Question

I do not have the symbols in my mobile keyboard, so I will try to explain it here with the available ones. $$\int_{-π}^π \frac{\sin^2 x}{1+a^x} \,\mathrm{d}x. \quad a>0$$

I am still trying.

Answer 1

Hint:

Let $u=-x$.

\begin{align*} \int_{-π}^π \frac{\sin^2 x}{1+a^x}dx&=-\int_\pi^{-\pi}\frac{\sin^2(-u)}{1+a^{-u}}du\\ &=\int_{-\pi}^\pi\frac{a^u\sin^2u}{a^{u}+1}du\\ \end{align*}

So,

\begin{align*} \int_{-π}^π \frac{\sin^2 x}{1+a^x}dx&=\frac{1}{2}\int_{-π}^π\left[\frac{\sin^2 x}{1+a^x}+\frac{a^x\sin^2 x}{1+a^x}\right]dx\\ &=\frac{1}{2}\int_{-π}^π\sin^2xdx \end{align*}