1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Intergration by parts
  1. Intergration by parts (Replies: 2)

  2. Intergral ? (Replies: 4)