Prove

Let $I=\int \sin ^{3}(2 x+1)$

$\begin{aligned} \Rightarrow \int \sin ^{3}(2 x+1) d x &=\int \sin ^{2}(2 x+1) \cdot \sin (2 x+1) d x \\ &=\int\left(1-\cos ^{2}(2 x+1)\right) \sin (2 x+1) d x \end{aligned}$

Let $\cos (2 x+1)=t$

$\Rightarrow-2 \sin (2 x+1) d x=d t$

$\Rightarrow \sin (2 x+1) d x=\frac{-d t}{2}$

$\Rightarrow I=\frac{-1}{2} \int\left(1-t^{2}\right) d t$

$=\frac{-1}{2}\left\{t-\frac{t^{3}}{3}\right\}$

$=\frac{-1}{2}\left\{\cos (2 x+1)-\frac{\cos ^{3}(2 x+1)}{3}\right\}$

$=\frac{-\cos (2 x+1)}{2}+\frac{\cos ^{3}(2 x+1)}{6}+C$