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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:

    [tex]
    \frac{1}{2}\int sin^3(u) \ du = \frac{1}{2}\int sin^2(u)sin(u) \ du
    [/tex]

    (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:

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

    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:

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

    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
    Or,
    ∫(1-cos2u)sinu du

    Let u = cosv

    That might be a little easier.
     
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