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Trigonometric integral

  1. Sep 25, 2005 #1
    [tex]\int sin^6 x dx[/tex]
    Progress so far:
    [tex]\int (sin^2 x)^3 dx[/tex]
    [tex]\frac{1}{8} \int (1-cos2x)^3 dx [/tex]
    [tex]\frac 1 8 \int (1 - 3cos2x + 3cos^22x - cos^32x) dx[/tex]

    Any help is appreciated.

    I can see using a half angle identity for cos^2(2x), but what do I do with the cos^3(2x)?

    Last edited: Sep 25, 2005
  2. jcsd
  3. Sep 25, 2005 #2
    Try looking up some sine integral reduction formulas on google. They take care of integrals involving powers of sine pretty nicely.
  4. Sep 25, 2005 #3
    I've found one in my text. Would I simply continue using the reduction formula until I get to sin^0(x)?

  5. Sep 25, 2005 #4
    Yes thats pretty much how we did it.
  6. Sep 25, 2005 #5
    Thanks whozum.

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