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∫ xSin(x)Sin(2x) dx

  1. Nov 7, 2003 #1
    A few weeks ago, I saw a post regarding the area under the function xsinxsin2x dx, (x = pi*x/a). After cogitations, I have found the answer:

    ∫x*Sin(x)*Sin(2x) dx = - xSin(2x)Cos(x) + ∫Cos(x) dx
    = Sin(x) - xSin(2x)Cos(x) + C

    By assuming u = xsin2x, du = 2xcos2x + sin2x, dv = sinx, v = -cosx. And using the Parts formula.
     
  2. jcsd
  3. Nov 7, 2003 #2

    Hurkyl

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    The IBP formula is

    ∫ u dv = uv - ∫ v du

    not ∫ v dx


    BTW, the key to this is the trig identity
    2 sin x cos x = sin 2x
     
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