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.
 A: You could use the sine reduction formula, but let’s try it with your method.
First, $2x\mapsto u$ to get
$$u=2x,\quad\frac{du}{dx}=2,\quad dx=\frac{1}{2}du$$
$$\frac12\int\sin^3(u)du$$
Now, since $$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
We have
$$\frac12\int\frac{\sin(u)-\sin(u)\cos(2u)}{2}du$$
Now, separate this into two integrals. The first one is trivial. For the second one, instead of parts, recall from product to sum formulas that $$\sin(A)\cos(B)=\frac{\sin(A+B)+\sin(A-B)}{2}$$
You should be able to take it from here :)

Rest of solution

\begin{align*}
 \frac12\int\frac{\sin(u)-\sin(u)\cos(2u)}{2}du&=\frac14\int\sin(u)du-\frac14\int\sin(u)\cos(2u)du\\
&=-\frac14\cos(u)-\frac18\int \sin(3u)-\sin(u) du\\
&=-\frac14\cos(2x)-\frac18\cos(2x)+\frac1{24}\cos(6x)\\
&=\frac1{24}\cos(6x)-\frac38\cos(2x)
 \end{align*}

As to convert this answer, I guess I'm somehow rusty on my trig manipulation skills because I was not able to but whatever. You'd probably have to use a bunch of product to sum or something but mathematica confirms they are equal so here. 


Okay, so regarding parts, you technically can use it. I highly do not recommend doing so. A handy thing to keep in mind that whenever you have an integral that has sine and cosine multiplied with each other, you want to use product to sum, not parts.
I tried parts just then, and basically after 1 iteration, we have an integral with just sine and cosine. You can use the classic trick where we integrate by parts twice and subtract the integral (like in $\int e^x\sin(x)dx$) but this is like 3 integration by parts. My attempt took up half a page and I definetely will not be latexing it 
A: Integrate by parts as follows
\begin{align}
\int\sin^{3}2x\, dx
=&\ \frac16\int e^{-\frac32\tan^2 2x}d(e^{\frac32\tan^2 2x} \cos^3 2x )\\
=& \ \frac16 \cos^3 2x+ \int \sin 2x\ dx
=\frac16 \cos^3 2x-\frac12\cos 2x
\end{align}
A: HINT
As an alternative, I would recommend you the following identity
\begin{align*}
\sin(3\alpha) = 3\sin(\alpha) - 4\sin^{3}(\alpha)
\end{align*}
Based on it, the proposed integral reduces to
\begin{align*}
\int\sin^{3}(2x)\mathrm{d}x = \int\left[\frac{3\sin(2x) - \sin(6x)}{4}\right]\mathrm{d}x
\end{align*}
Can you take it from here?
A: Integration by parts is interesting as below:
$$
\begin{aligned}
&I=\int \sin ^3 2 x d x=-\frac{1}{2} \int \sin ^2 2 x d(\cos 2 x)\\
&=-\frac{1}{2} \sin ^2 2 x \cos 2 x+\frac{1}{2} \int 4 \sin 2 x \cos ^2 2 xdx\\
&=-\frac{1}{2} \sin ^2 2 x \cos 2 x+2 \int \sin 2 x\left(1-\sin ^2 2 x\right) d x\\
&=-\frac{1}{2} \sin ^2 2 x \cos 2 x+2 \int \sin 2 x d x-2 I\\
I&=-\frac{\cos 2 x}{6}\left(\sin ^2 2 x+2\right)+C \\(\textrm{  OR}&  =\frac16 \cos^3 2x-\frac12\cos 2x+C)
\end{aligned}
$$
