Integrating e^x sin(lnx) using Integration by Parts

[miniradman]

Homework Statement

Integrate
[itex]\int e^2xsin(ln(x)) dx[/itex]

Homework 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]

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 e^x 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?

 

[voko]

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?

 

[miniradman]

Oh...crap...silly error

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

Yes, we are looking for the indefinite integral.