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Integrate cos(lnx)dx

  1. Jan 25, 2011 #1
    1. The problem statement, all variables and given/known data
    let u=lnx
    du=1/x*dx
    dv=cosdx
    v=-sin
    2. Relevant equations
    Now im confused as im getting nowhere with this substiution, i learned the LIPTE rule but its quite confusing, i have a function within a function


    3. The attempt at a solution
     
  2. jcsd
  3. Jan 25, 2011 #2
    for integration by parts to work, i would need two differentiable functions, but the cosdx is not differentiable would it need to be cosxdx for it to be differentiated?
     
  4. Jan 25, 2011 #3
    Is the integral:

    [tex]
    \int{\cos{(\ln{(x)})} \, dx}
    [/tex]

    If it is, make the substitution:

    [tex]
    t = \ln{(x)} \Rightarrow x = e^{t}
    [/tex]

    and substitute everywhere. The integral that you will get can be integrated by using integration by parts twice, or, if you know complex numbers, by representing the trigonometric function through the complex exponential.
     
  5. Jan 25, 2011 #4
    it is just cos(lnx)dx
     
  6. Jan 25, 2011 #5
    then proceed as I told you.
     
  7. Jan 25, 2011 #6
    ok im getting an integral of the form : coste^tdt. is this correct?
     
  8. Jan 25, 2011 #7
    yes. proceed by integration by parts or using complex exponentials.
     
  9. Jan 25, 2011 #8
    ok thank you for the help, quick question why do we have to let t=lnx?
     
  10. Jan 25, 2011 #9
    We have a composite function [itex]\cos{(\ln{x})}[/itex] of two elementary functions (trigonometric and logarithmic). As a combination, it does not have an immediate table integral. But, the method of substitution, which is nothing but inverting the chain rule for derivatives of composite functions, works exactly for such compound functions.
     
  11. Jan 25, 2011 #10
    thank you, lastly i end up with 1/2(e^tcost-e^tsint), do i sub back , so that x=e^t and t=lnx
    which leaves me with : 1/2x(cos(lnx)-sin(lnx)+c?
     
  12. Jan 25, 2011 #11
    I think the sign in front of the sine should be + and the result should be:

    [tex]
    \frac{1}{2} x \left[\cos{\left(\ln{(x)}\right)} + \sin{\left(\ln{(x)}\right)}\right]+ C
    [/tex]
     
  13. Jan 25, 2011 #12
    yes, made a slight mistake thank you very much for the help, i tried doing it another way by letting u=cos(lnx) and dv=dx and i arrived at the same answer
     
  14. Jan 25, 2011 #13
    yes, the x's will cancel.
     
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