1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Trig Integration

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int sin^3(2x) dx[/tex]

    2. Relevant equations
    [tex] sin^2(x) + cos^2(x) = 1 [/tex]

    3. The attempt at a solution

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

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

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

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

    I do another substitution with 'w'

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

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

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

    Now I place u and w with their respective substitution

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

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

    I don't know where I'm going wrong. Can someone help tell me what/where I messed up
  2. jcsd
  3. Oct 5, 2009 #2
    Hi ganondorf29,

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

    So lets start from a good point you got to:

    \frac{1}{2}\int sin^3(u) \ du = \frac{1}{2}\int sin^2(u)sin(u) \ du

    (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:

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

    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:

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

    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
  4. Oct 6, 2009 #3
    ∫(1-cos2u)sinu du

    Let u = cosv

    That might be a little easier.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook