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Integration by Parts - Substitution

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate the following indefinite integral:
    ∫(sin(ln16x))/xdx


    2. Relevant equations



    3. The attempt at a solution
    let u = ln16x
    therefore du=16/16x=1/x

    ∫sinudu
    =-cosu
    =-cos(ln16x)

    Why is this showing as the wrong answer?
     
  2. jcsd
  3. Sep 10, 2012 #2
    you need to substitute with two variables, u & v.

    [itex] ln(16x) [/itex] has a function within itself, therefore substituting that whole part with u wont work.

    keep in mind: [tex] \int u\, du = uv - \int v\, du [/tex]
     
  4. Sep 10, 2012 #3
    Sorry, stupid question - if there's both u and v, how do I choose which one goes with which?
     
  5. Sep 10, 2012 #4

    ehild

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    Didn't you forget the arbitrary constant? The integral is -cos(ln16x))+C.

    ehild
     
  6. Sep 10, 2012 #5

    uart

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    Yep, I can't see anything wrong except that you forgot the constant.:smile:
     
  7. Sep 10, 2012 #6
    Substitute for 16x and then further substitute for log(16x). You should end up with Int([e^a][sin(a)])/16

    At 1st I did it just like u did, then I saw sikrut's answer... Reworked it & verified with Matlab. Turns out Matlab agrees with Sikrut.
     
  8. Sep 10, 2012 #7

    SammyS

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    This problem is adequately handled by substitution, exactly as you did it --- of course, include the constant of integration.

    No need for integration by parts. I don't see how that would even be helpful.
     
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