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Intergration by parts

  1. Oct 9, 2006 #1
    kay I am having difficulties with this..
    Knowing the gerneral formuala of

    |uv'=uv- |vu'

    i using a nonehomework question i was trying to make sure i had it down pat was having problems..

    | x cos 5x dx

    but for some reason i don't get the right answer when it done...
    If I have u=x, du=1
    v'=cos 5x dx

    v= 1/5 sin5x??? or did i crew up some where I been having trouble here ..

    so i can do the substion and all but this stuff is screwing with help plz.
  2. jcsd
  3. Oct 9, 2006 #2
    [tex] \int x\cos 5x = \frac{x}{5}\sin 5x - \frac{1}{5}\int \sin 5x dx [/tex].

    [tex] \int x\cos 5x = \frac{x}{5}\sin 5x +\frac{1}{25}\cos 5x [/tex]

    So [tex] \int udv = uv-\int vdu [/tex].

    [tex] \int \sin 5x = -\frac{\cos 5x}{5} [/tex].
    Last edited: Oct 9, 2006
  4. Oct 9, 2006 #3
    kay my major problem is that 1/5 where does it come from.

    how does cos 5x dx = 1/5 sin 5x..
  5. Oct 9, 2006 #4
    when you take the derivative of 1/5 sin 5x, you get cos 5x by doing the chain rule...you have to take the derivative of the argument because it is more complex than just an x
  6. Oct 9, 2006 #5
    t6hanks Maps Think i undersand it now... hopefully the homework will be easier now
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