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Homework Help: Integral of e^x sin(lnx)

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\int e2xsin(ln(x)) dx[/itex]

    2. Relevant equations
    Well, I'm not exactly sure which rule to apply here, but I'm going to assume integration by parts:

    [itex]\int u \frac{dv}{dx} = uv - \int v \frac{du}{dx}[/itex]

    3. The attempt at a solution
    I'm a little thrown off because since the sine and e are recursive. But, should I start by making [itex]e^{2x}[/itex] equal one variable? like a? so then I'll have

    [itex]\int a sin(ln x) dx[/itex] then proceed to say that [itex]a \int sin(ln x) [/itex]
    then I'll let
    [itex]u = ln(x)
    \frac{du}{dx}= e^{x}[/itex]
    I figured that doing a u substitution may be easier for this.
    [itex]\frac{du}{dx}= e^{x}[/itex]
    [itex]\frac{dx}{du}= \frac{1}{e^{x}}[/itex]
    [itex]dx= \frac{du}{e^{x}}[/itex]
    [itex]\int sinu \frac{du}{e^{x}}[/itex]
    Then integration by parts (I might make u = z to make things easier):
    [itex]\int z \frac{dv}{dx} = zv - \int v \frac{dz}{dx}[/itex]

    [itex]z = sin u
    \frac{dz}{du}= cos u[/itex]

    [itex]\frac{dv}{dx} = \frac{du}{e^{x}}
    v= ln e^{x}[/itex]
    The natural log of ex is simply x
    [itex] v=x [/itex]

    [itex]\int sin u \frac{du}{e^{x}} = sin u x - \int x cos u[/itex]

    At this point I don't know how to continue, because now I have u and x, and when I sub in ln x as u, I'll end up getting cos lnx which is pretty much where I started from (only difference was I used sine).

    Could someone give me a hint?
  2. jcsd
  3. Sep 11, 2012 #2
    You have made a number of mistakes. First, you cannot pull ## e^{2x} ## out of the integral: it depends on the variable of integration. Second, given ## u = \ln x ##, ## \frac {du}{dx} \ne e^x ##.

    Finally, I suspect this integral cannot be expressed in elementary functions. Are you really supposed to find an indefinite integral?
  4. Sep 11, 2012 #3
    Oh...crap...silly error

    [itex]u = lnx \frac{du}{dx}= \frac{1}{x}[/itex]

    Yes, we are looking for the indefinite integral.
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