Find the integral of the function $\sin 3x \cos 4x$.

We use the trigonometric identity: $\sin A \cos B = \frac{1}{2} \{\sin(A+B) + \sin(A-B)\}$.
Applying this to the integral:
$\int \sin 3x \cos 4x \, dx = \int \frac{1}{2} \{\sin(3x+4x) + \sin(3x-4x)\} \, dx$
$= \frac{1}{2} \int \{\sin 7x + \sin(-x)\} \, dx$
Since $\sin(-x) = -\sin x$,we have:
$= \frac{1}{2} \int (\sin 7x - \sin x) \, dx$
$= \frac{1}{2} \int \sin 7x \, dx - \frac{1}{2} \int \sin x \, dx$
$= \frac{1}{2} \left( \frac{-\cos 7x}{7} \right) - \frac{1}{2} (-\cos x) + C$
$= -\frac{\cos 7x}{14} + \frac{\cos x}{2} + C$,where $C$ is an arbitrary constant.