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!

You did the right thing. Using your substitution, you have
$$
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} \ dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} 2^x \ dx.
$$
Adding the two equivalent formulations you have
$$
2 I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} (2^x +1) \ dx
= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sin^2 x} \ dx = \int_0^{\frac{\pi}{2}} (1-\cos(2x))\ dx=\frac{\pi}{2}$$
which gives $I = \frac{\pi}{4}$ as the answer.
