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Trig substitution

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int{cos^4 6x sin^3 6x dx}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I've gotten this far but now I'm stuck:

    [tex]\int{cos^4 6x sin 6x sin^2 6x dx} = \int{cos^4 6x}*(\frac{1-cos 12x}{2})sin 6x dx[/tex]
     
    Last edited: Sep 22, 2008
  2. jcsd
  3. Sep 22, 2008 #2
    From you having the integrand as this: cos^4 6x sin^3 6x dx

    Use a substitution u = cos^4 6x
     
  4. Sep 22, 2008 #3
    I don't understand exactly. Are you saying use substitution after using the power-reducing formula? Or do I not use power-reducing at all here? I briefly saw him explaining it to one person, and he mentioned something about reducing it from sin^3 to sin^2... I could have misunderstood though.
     
  5. Sep 22, 2008 #4
    Sorry, that substitution was wrong. Tomorrow when I'm more awake (it's 3:40 AM here) I'll have another look at it.
     
  6. Sep 22, 2008 #5

    danago

    User Avatar
    Gold Member

    Use the fact that [tex]sin^3 (6x) =sin^2 (6x) . sin(6x)[/tex]. Then perhaps make use of the pythagorean identity, and you will be 1 simple substitution away from a solution :smile:
     
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