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Integrate sin(lnx)dx

  1. Feb 9, 2006 #1
    we are doing 'integration by parts' right now. but we never learned anything like this. (having a function inside a function).
    would u=lnx du=1/x .. v= ??? dv = sin... ????
    please help.
     
  2. jcsd
  3. Feb 9, 2006 #2
    Let u = sin(ln x)
    and dv = dx

    It works out very nicely.
     
  4. Feb 9, 2006 #3
    hummm.

    so...
    u=sin(lnt) du=cos(lnt)(1/t)dt
    v=t dv=dt

    intergration of sin(lnt)dt =
    tsin(lnt)- ∫tcos(lnt)(1/t)dt (and then do integration by parts again?)

    could you just tell me if i took the derivative of u correctly? thank you so much by the way~ :)
     
  5. Feb 9, 2006 #4
    Yea looks right and don't forget to cancel the t and 1/t inside the integral, then do parts again the exact same way and you should get it.
     
  6. Feb 9, 2006 #5
    :biggrin: THANK YOU SO MUCH! :) you are so kind. hehehe
     
  7. Feb 9, 2006 #6
    No problem glad I could help, and I'll take that to mean that you got the answer lol...
     
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