Find the integral of the function $\sin ^{3}(2 x+1)$.

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