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!

Integral of e^x sin(lnx)

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Integrate
    [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]
    where:

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




Similar Discussions: Integral of e^x sin(lnx)
  1. Integrate sin(lnx)dx (Replies: 5)

Loading...