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Quick question on integration by parts

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm following an example in the textbook that states:

    http://img24.imageshack.us/img24/1672/33686252.jpg [Broken]

    I was just wondering what happened to the 2 out the front, I would have been more inclined to think this would be the next step:

    http://img34.imageshack.us/img34/4854/a1aa.jpg [Broken]

    Any explanations?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 28, 2009 #2

    lanedance

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    when you perform the integration by parts you get the original integral on both sides of the equation, which leads to the cancelling of the factor of 2 on the right hand side
     
  4. Sep 28, 2009 #3
    I'm not following 100%.
    The integral on the RHS is not the original integral on the LHS?

    This it the formula I used (and I am familar with):
    05fdc1c25e10bae02245a3334d109965.png
    (of course with respect to t here not x)

    fyi: I let f = Gsintcost and g' = Gdot

    everything works out fine, its just the 2 that has me puzzled
     
  5. Sep 28, 2009 #4

    lanedance

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    ok so in some bastardised short hand

    du = G'
    u = G
    v = Gsc
    dv = G'sc + G(c2 - s2)

    [tex] \int du.v = uv| - \int u.dv [/tex]

    [tex] \int G'.Gsc = G.Gsc| - \int G.(G'sc + G(c^2-s^2)) [/tex]

    [tex] \int G'.Gsc = G^2 sc| - \int G.G'sc -\int G^2(c^2-s^2) [/tex]

    [tex] 2 \int G'.Gsc = G^2 sc| - \int G^2(c^2-s^2) [/tex]

    [tex] 2 \int G'.Gsc = G^2 sc| + \int G^2(s^2-c^2) [/tex]
     
    Last edited: Sep 28, 2009
  6. Sep 28, 2009 #5
    I think dv should be dv = G'sc + G(c^2 - s^2)
     
    Last edited: Sep 28, 2009
  7. Sep 28, 2009 #6

    lanedance

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    yeah i think you're right, but hopefully the bit about the orginal intergal aappearing on the RHS is celar
     
  8. Sep 28, 2009 #7
    bugger, I think dv = G'sc + G(c^2 - s^2), but then if I change that, then I don't get the correct answer!

    I understand what you mean by the original integral showing up on the RHS (cos now I can take it over to the lhs and get 2* original integral).


    edit: found the mistake. When I broke up the integral I forgot to take the minus inside both.

    got it all sorted now.

    Cheers!
     
  9. Sep 28, 2009 #8

    lanedance

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    no worries - i updated the working above as well
     
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