- #1

miniradman

- 195

- 0

## Homework Statement

Integrate

[itex]\int e

^{2x}sin(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?