Definite integral evaluation of $\frac{\sin^2 x}{2^x + 1}$ 
Compute
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} \ dx.$$

So looking at the limits I first checked whether the function is even or odd by substituting $x = -x$ but this gave me:
$$\frac{\sin^2 x}{2^x + 1} \cdot 2^x$$
so the function is clearly neither odd or even and I don't know what to infer from this. I can't think of a good substitution either.
Any help would be appreciated!

This video provides a formula that is very useful in your case, check the video for a proof: https://www.youtube.com/watch?v=xiIsPEqyTqU
$$\int_{-a}^{a} \frac{E(x)}{t^{O(x)} + 1} \ dx = \int_{0}^{a} E(x) \ dx$$
where $E(x)$ and $t$ are even functions, and $O(x)$ is an odd function. In your case: $\sin^2(x)=E(x); x = O(x)$, and so your integral simplifies to:
$$\int_{0}^{\pi/2} \sin^2(x) \ dx = \pi/4$$
