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.
Best Answer
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
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 😅