# Integral of e^x sin(lnx)

1. Sep 11, 2012

1. The problem statement, all variables and given/known data
Integrate
$\int e2xsin(ln(x)) dx$

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

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

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 $e^{2x}$ 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
$u = ln(x) \frac{du}{dx}= e^{x}$
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:

$z = sin u \frac{dz}{du}= cos u$

$\frac{dv}{dx} = \frac{du}{e^{x}} v= ln e^{x}$
The natural log of ex 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?

2. Sep 11, 2012

### 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?

3. Sep 11, 2012

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