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Reduction Formula and Integration by Parts

  1. Feb 16, 2008 #1
    I was sick and missed class:uhh:

    ∫ sin^2x dx = x/2 – sin2x/4 + C
    (I see that there is a trig identity in the answer as sin2x = 2sinxcosx)

    What I tried (copying the example in the book)
    u = sinx du = sinxcosx dx (why is du not cosx?)
    v = -cos x dv = sinx dx (why is this not just dx?)

    ∫ sin^2x dx = -cosxsinx + 1 ∫ sinx cos^2x dx (okay why did I add +1 and where did the funky ∫ come from) I see that cos^2x = 1 – sin^2x.

    So now (from the example in the book but why?)
    ∫ sin^2x dx = -cosxsinx + 1 ∫ sinx dx – 1 ∫ sin^2x dx

    2 ∫ sin^2x dx = -cosxsinx + 1 ∫ sinx dx (what happend to the -1?)

    ∫ sin^2x dx = -½cosxsinx + 1/2 ∫ sinx dx

    Which I thought should give

    -½cosxsinx – ½cosx + C

    Nothing like the above answer……. Can someone help?

    -Serena
     
  2. jcsd
  3. Feb 16, 2008 #2
    I don't want to read trough the whole thing, but I think it's easier to use the trig identity from the start:

    sin^2 x = 1/2 - 1/2 cos(2x)

    edit: this way you'll get to the answer easily too :)
     
    Last edited: Feb 16, 2008
  4. Feb 16, 2008 #3

    Vid

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    u = sin(x)
    du = cos)dx
    v = -cos(x)
    dv = sin(x)dx

    Int[sin^2(x)] = -sin(x)cos(x) + Int[cos^2(x)]
    Int[sin^2(x)] = -1/2sin(2x) + Int[1] - Int[sin^2x]
    2Int[sin^2(x)] = -1/2sin(2x) + x
    Int[sin^2(x)] = x/2 - 1/4sin(2x)
     
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