Step 1: Apply the product-to-sum identity $\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]$ to $\sin 2x \sin 3x$: $\sin 2x \sin 3x = \frac{1}{2} [\cos(2x - 3x) - \cos(2x + 3x)] = \frac{1}{2} [\cos(-x) - \cos 5x] = \frac{1}{2} [\cos x - \cos 5x]$.

Step 2: Multiply the result by $\sin x$: $\sin x \sin 2x \sin 3x = \frac{1}{2} \sin x (\cos x - \cos 5x) = \frac{1}{2} \sin x \cos x - \frac{1}{2} \sin x \cos 5x$.

Step 3: Apply the product-to-sum identity $\sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)]$ to both terms: $\frac{1}{2} (\frac{1}{2} [\sin 2x + \sin 0]) - \frac{1}{2} (\frac{1}{2} [\sin 6x + \sin(-4x)])$.

Step 4: Simplify the expression using $\sin 0 = 0$ and $\sin(-4x) = -\sin 4x$: $\frac{1}{4} \sin 2x - \frac{1}{4} (\sin 6x - \sin 4x) = \frac{1}{4} \sin 2x - \frac{1}{4} \sin 6x + \frac{1}{4} \sin 4x$.

Step 5: Integrate the simplified expression term by term: $\int (\frac{1}{4} \sin 2x - \frac{1}{4} \sin 6x + \frac{1}{4} \sin 4x) \, dx = \frac{1}{4} \left( -\frac{1}{2} \cos 2x \right) - \frac{1}{4} \left( -\frac{1}{6} \cos 6x \right) + \frac{1}{4} \left( -\frac{1}{4} \cos 4x \right) + C$.

Step 6: State the final answer: $-\frac{1}{8} \cos 2x - \frac{1}{16} \cos 4x + \frac{1}{24} \cos 6x + C$.