Integration of $\int\sin^{3}2x\, dx.$ using IBP

Question

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.

Answer 1

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 :)

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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.

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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

Answer 2

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?

Answer 3

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}

Answer 4

Integration by parts is interesting as below: $$ \begin{aligned} &I=\int \sin ^3 2 x d x=-\frac{1}{2} \int \sin ^2 2 x d(\cos 2 x)\\ &=-\frac{1}{2} \sin ^2 2 x \cos 2 x+\frac{1}{2} \int 4 \sin 2 x \cos ^2 2 xdx\\ &=-\frac{1}{2} \sin ^2 2 x \cos 2 x+2 \int \sin 2 x\left(1-\sin ^2 2 x\right) d x\\ &=-\frac{1}{2} \sin ^2 2 x \cos 2 x+2 \int \sin 2 x d x-2 I\\ I&=-\frac{\cos 2 x}{6}\left(\sin ^2 2 x+2\right)+C \\(\textrm{ OR}& =\frac16 \cos^3 2x-\frac12\cos 2x+C) \end{aligned} $$