# Integral problem

1. Oct 22, 2006

### Geronimo85

I'm supposed to integrate the following expression, and supposedly there is a very simple way to do so. Maple comes up with something rediculous, so I'd appreciate any input. Sorry about the short hand, don't know how to make everything pretty on here:

Integral[(e^ax)cos^2(2bx)dx] where a and b are positive constants

So far all i've got is:

(e^ax)cos^2(2bx)= (e^ax)*[(e^(i*2*b*x) - e^(-i*2*b*x))/2]^2

because: cosx = (e^ix - e^-ix)/2

squaring inside the brackets gets me:

(e^ax)* [((e^(i*2*b*x)-e^(-i*2*b*x)/2)^2]

I'm just really not getting something here

2. Oct 23, 2006

### Max Eilerson

$$e^{ax}.e^{-bx} = e^{(a-b)x}$$

You should be able to work from that, just factor the exponents. The final integral won't look pretty.

3. Oct 23, 2006

### HallsofIvy

Staff Emeritus
How many times are you going to post this same question?

4. Oct 23, 2006

### Meir Achuz

Use cos^2(2bx)=[1+cos(4bx)]/2.

5. Oct 23, 2006

### Geronimo85

sorry about reposting it, I just feel like i'm going in circles