$\int \sin x \cos x \ dx \neq \frac {\sin^2x}{2}$ if we use the identity $\sin A \cos B = \frac12[\sin(A-B) + \sin(A+B)]?$ How is it that we can get two different answers for an integral depending on whether we apply an identity or not? 
Typically, $$\int \sin x \cos x \ dx = \frac {\sin^2x}{2}+C~.$$
However if we apply the trigonometric identity $$\sin A \cos B = \frac12[\sin(A-B) + \sin(A+B)]~,$$ then the integral becomes $$\frac12 \int (\sin(0) + \sin(2x)) dx =\frac12 \int \sin(2x) dx = -\frac14 \cos(2x) $$
So we end up with a different answer. Have I made a mistake here, or is this just a property of integration/trigonometry?

You forgot the arbitrary constant in your second method.
The non-constant part of the first result may be transformed as follows: $$\frac{\sin^2x}{2}=\frac{1-\cos 2x}{4}=-\frac14\cos 2x+C,$$ since $C$ is arbitrary.
