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

[miniradman]

Homework Statement

Integrate
\int e^2xsin(ln(x)) dx

Homework Equations

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

\int u \frac{dv}{dx} = uv - \int v \frac{du}{dx}

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 e^{2x} equal one variable? like a? so then I'll have

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

z = sin u <br /> \frac{dz}{du}= cos u

\frac{dv}{dx} = \frac{du}{e^{x}}<br /> v= ln e^{x}
The natural log of e^x is simply x
v=x

\int sin u \frac{du}{e^{x}} = sin u x - \int x cos u

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

u = lnx \frac{du}{dx}= \frac{1}{x}

Yes, we are looking for the indefinite integral.