Using substitution with $u=\sin{x}$, The integral of $\sin{x}\cos{x}$ is equal to $\frac{1}{2}\sin^2{x}$. This is the solution I require.
However, why is it that when I use the double angle formula $\sin{2x} = 2\sin{x} \cos{x}$, and integrate $\frac{1}{2}\sin{2x}$, I get a completely different answer of $-\frac{1}{4}\cos{2x}$ even though $\sin{2x} = 2\sin{x} \cos{x}$.
It's actually not a completely different answer. By the double-angle formulae for cosine and the pythagorean identity, you have $$ \cos{2x} = \cos^2{x}-\sin^2{x} = 1-2\sin^2{x}, $$ so $-\frac{1}{4}\cos{2x}$ differs from $\frac{1}{2}\sin^2{x}$ by a constant. Since an indefinite integral is only determined up to a constant anyway, this is fine.