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Integrate 48sin^4(3x)

  1. Nov 20, 2009 #1
    I think i've done it at least once every day for the past week.
    I week getting it wrong on the homework site.
    Worst off all, it only alows 5 entries. I've used 4. Please help.
     
  2. jcsd
  3. Nov 20, 2009 #2
    Just curious, are you taking the anti-derivative, or a definite integral?

    If it's a definite integral, it looks like a textbook case for Wallis' formula.
     
  4. Nov 20, 2009 #3
    Use this formula [tex] \int \sin^n x \, dx = - \frac{\sin^{n-1} {x} \cos {x}}{n} + \frac{n-1}{n} \int \sin^{n-2}{x} \, dx[/tex]
     
  5. Nov 21, 2009 #4

    HallsofIvy

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    noblegas's formula is probably simplest but the general method for even powers of sine or cosine is to use the double angle formula: cos(2x)= 1- 2sin2(x)= 2cos^2(x)- 1. From that [itex]sin^2(x)= (1/2)(1- cos(2x))[/itex] and [itex]cos^2(x)= (1/2)(1+ cos(2x))[/itex].

    For a fourth power, use that twice: [itex]sin^4(x)= ((1/2)(1- cos(2x)))^2[/itex][itex]= (1/4)(1- cos(2x)- cos^2(2x))[/itex]. 1/4 and (1/4) cos(2x) can be integrated directly. To integrate (1/4)cos2(2x) do it again: [itex](1/4)cos^2(2x)= (1/4)(1+ cos(4x))[/itex].
     
  6. Nov 21, 2009 #5

    CompuChip

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    Also, if you want to check your answer before entering it on the homework site, you can always post your concerns here.
    We're not going to give you the answer, but if you show what you did we can always take a look.
     
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