Find the integral of the function $\sin x \sin 2x \sin 3x$.

We use the trigonometric identity $\sin A \sin B = \frac{1}{2} \{\cos(A-B) - \cos(A+B)\}$.
First,consider $\sin 2x \sin 3x = \frac{1}{2} \{\cos(2x-3x) - \cos(2x+3x)\} = \frac{1}{2} \{\cos(-x) - \cos(5x)\} = \frac{1}{2} \{\cos x - \cos 5x\}$.
Now,the integral becomes $\int \sin x \cdot \frac{1}{2} \{\cos x - \cos 5x\} \, dx = \frac{1}{2} \int (\sin x \cos x - \sin x \cos 5x) \, dx$.
Using $\sin x \cos x = \frac{\sin 2x}{2}$ and $\sin A \cos B = \frac{1}{2} \{\sin(A+B) + \sin(A-B)\}$:
$= \frac{1}{2} \int \frac{\sin 2x}{2} \, dx - \frac{1}{2} \int \frac{1}{2} \{\sin(x+5x) + \sin(x-5x)\} \, dx$
$= \frac{1}{4} \int \sin 2x \, dx - \frac{1}{4} \int (\sin 6x - \sin 4x) \, dx$
$= \frac{1}{4} \left( \frac{-\cos 2x}{2} \right) - \frac{1}{4} \left( \frac{-\cos 6x}{6} + \frac{\cos 4x}{4} \right) + C$
$= -\frac{\cos 2x}{8} + \frac{\cos 6x}{24} - \frac{\cos 4x}{16} + C$.