Trig Integration: Solving \int sin^3(2x) dx Using Substitution Method

[ganondorf29]

Homework Statement

\int sin^3(2x) dx

Homework Equations

sin^2(x) + cos^2(x) = 1

The Attempt at a Solution

First I try to get the integral in the form of \intsin^3(u) and I do this by u-substitution.

\int sin^3(2x) dx
u = 2x
du = 2dx
dx = du/2

So the new integral looks as such:
\frac{1}{2}\int sin^3(u) du = \frac{1}{2}\int sin^2(u) * sin(x) du

= \frac{1}{2}\int (1-cos^2(u)) * sin(u) du

I do another substitution with 'w'

w = cos(u)
dw = -sin(u) du
du = -dw / sin(u)

\frac{1}{2}\int (1-w^2) * sin(u) -dw / sin(u)

\frac{-1}{2}[w - \frac{w^3}{3} + C]

Now I place u and w with their respective substitution

\frac{-1}{2}[cos(u) - \frac{\cos^3(u)}{3} + C]

\frac{-1}{2}[cos(2x) - \frac{cos^3(2x)}{3} + C]

I don't know where I'm going wrong. Can someone help tell me what/where I messed up

 

[Galadirith]

Hi ganondorf29,

I think actually your way over complicating this :D. I suppose much better than under complicating it.

So let's start from a good point you got to:

<br /> \frac{1}{2}\int sin^3(u) \ du = \frac{1}{2}\int sin^2(u)sin(u) \ du<br />

(careful you wrote x in the final sin in yours, obviously just error but may lose you a mark or two in an exam :-()

You then do:

<br /> \frac{1}{2}\int \left(1-cos^2(u)\right)sin(u) \ du<br />

now here's is where you made the mistake, well not necessary mistake but you didn't notice something important. Instead of jumping straight to a substitution how about:

<br /> \frac{1}{2}\left(\int sin(u) \ du \ - \ \int cos^2(u)sin(u)\ du \right)<br />

now look really carefully at this, you can integrate this directly now: consider what is the derivative of cos³(u) ? Can we use that with a bit of manipulation to solve this integral?

I hope that helps ganondorf29. I must say this too. It is very tempting to want to use substitution in every bit of integration you do, they seem to make like easier most of the time, I know I completely understand, when I started to learn about integral subs I used them where ever I could. But substitutions should only be used when nessesary, make sure that you have exhausted all other techniques and methods of trying to solve and integral. Sure sometimes you'll know a substitution is necessary straight away and that will come from time and experience, but subs can defiantly be overused :D

Have fun ganondorf29 ;D

 

[Bohrok]

Or,
∫(1-cos²u)sinu du

Let u = cosv

That might be a little easier.