<s>[INST] Integration of $\int\sin^{3}2x\, dx.$ using IBP I am attempting to solve this integral:
$$\int\sin^{3}2x\, dx.$$
Using an identity, I have manipulated the integral into this:
$$\int\frac{1}{2}\left(1-\cos4x\right)\sin2x dx.$$
From here, using IBP, I let $u$ = $1-cos4x$ and $v'$ = sin2x
However, I obtained an answer of $\cos2x+\frac{1}{2}\cos2x\cos4x\ -\ \frac{1}{6}\cos6x$, which is nowhere near the intended answer in the solutions of $\frac{1}{6}\cos^{3}2x-\frac{1}{2}\cos2x$.
Any help on why my solution/method is incorrect will be appreciated.
 [/INST] Integration by parts is interesting as below:
$$
\begin{aligned}
&I=\int \sin ^3 2 x d x=-\frac{1}{2} \int \sin ^2 2 x d(\cos 2 x)\\
&=-\frac{1}{2} \sin ^2 2 x \cos 2 x+\frac{1}{2} \int 4 \sin 2 x \cos ^2 2 xdx\\
&=-\frac{1}{2} \sin ^2 2 x \cos 2 x+2 \int \sin 2 x\left(1-\sin ^2 2 x\right) d x\\
&=-\frac{1}{2} \sin ^2 2 x \cos 2 x+2 \int \sin 2 x d x-2 I\\
I&=-\frac{\cos 2 x}{6}\left(\sin ^2 2 x+2\right)+C \\(\textrm{  OR}&  =\frac16 \cos^3 2x-\frac12\cos 2x+C)
\end{aligned}
$$
</s>