Find the integral of the function $\sin ^{3} x \cos ^{3} x$.

Let $I = \int \sin ^{3} x \cos ^{3} x \, dx$.
We can rewrite the integral as:
$I = \int \cos ^{3} x \cdot \sin ^{2} x \cdot \sin x \, dx$.
Using the identity $\sin ^{2} x = 1 - \cos ^{2} x$:
$I = \int \cos ^{3} x (1 - \cos ^{2} x) \sin x \, dx$.
Let $\cos x = t$. Then,$-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.
Substituting these into the integral:
$I = -\int t^{3} (1 - t^{2}) \, dt$.
$I = -\int (t^{3} - t^{5}) \, dt$.
Integrating with respect to $t$:
$I = -\left( \frac{t^{4}}{4} - \frac{t^{6}}{6} \right) + C$.
$I = \frac{t^{6}}{6} - \frac{t^{4}}{4} + C$.
Substituting back $t = \cos x$:
$I = \frac{\cos ^{6} x}{6} - \frac{\cos ^{4} x}{4} + C$,where $C$ is an arbitrary constant.