# Could really use a hand with a complex exponential and integral problem

1. Oct 22, 2006

### Geronimo85

I have two homework problems that have been driving me nuts:

1.) evaluate the indefinite integral:

integral(dx(e^ax)cos^2(2bx))

where a and b are real positive constants. I just don't know where to start on it.

2.) Find all values of i^(2/3)

So far I have:

i^(2/3)
= e^(2/3*ln(i))
= e^(2/3*i*(Pi/2 + 2*n*Pi))
= e^(i*Pi/3)*e^(i*n*4Pi/3)

I know my three solutions should end up being (1+i*sqrt(3))/2, (1-i*sqrt(3))/2, -1. But I can't seem to get there. I'd really appreciate any help. Sorry if my shorthand is confusing.

2. Oct 22, 2006

### Hurkyl

Staff Emeritus
Can't you write it as nothing but complex exponentials?

Have you tried plugging in values for n?

Last edited: Oct 22, 2006
3. Oct 22, 2006

### Geronimo85

When I plugged in values for n, I drew out my

e^(i*Pi/3)* +/- (1/2 + i*sqrt(3)/2)

I don't see how I get any of my three answers out of that.

4. Oct 22, 2006

### Hurkyl

Staff Emeritus
(1) You know the value of e^(i*Pi/3).
(2) n ranges over all integers... not just 0 and 1.

5. Oct 22, 2006

### Geronimo85

I am assuming that it is just -1 as that is the only way it will end up as one of my answers, when n=0. I don't understand why though.

-Got that, I was having a brain fart and thinking that for n=2, theta was 10Pi/4

6. Oct 22, 2006

### Hurkyl

Staff Emeritus
But it's not -1... why do you think the n=0 solution has to be the one that's -1?

7. Oct 22, 2006

### Geronimo85

it should be 1/2 + i*sqrt(3)/2, but then how do I ever get -1 for a value?

8. Oct 22, 2006

### Geronimo85

because i'm an idiot... got it, thanks a bunch. Also, with the integral problem (I haven't done calc in a long long time), once I substitute

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

how do I solve?

9. Oct 22, 2006

### Hurkyl

Staff Emeritus
Well, what did you get when you simplified after substituting?

10. Oct 22, 2006

### Geronimo85

something really ugly that doesn't seem to cancel out:

integral(dx(e^ax)cos^2(2bx))

=(e^ax)*(2e^(-4*b^2*x^2)-2e^(4*b^2*x^2))

11. Oct 22, 2006

### Hurkyl

Staff Emeritus
Aha! There shouldn't be any x²'s in there. Remember your exponent laws: what is $(a^b)^c$?

12. Oct 22, 2006

### Geronimo85

but I have:

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

when I square [""] how do I not end up with x^2?

13. Oct 22, 2006

### Hurkyl

Staff Emeritus

First off, remember how to compute (a + b)².
Secondly, as I said in my previous post, remember your exponent laws: what is
$$\left( a^b \right)^c?$$

14. Oct 22, 2006

### Geronimo85

I know that (a + b)² = a² + 2ab + b² and (a^b)^c = a^b*c

so if:

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

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

why when I square inside the brackets don't I get

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

I'm just really not getting something here

15. Oct 23, 2006

$$\cos 2bx = \frac{e^{2ibx}+e^{-2ibx}}{2}$$
$$\cos 2bx = \frac{1}{2} \left[ (e^{2ibx}+e^{-2ibx})(e^{2ibx}+e^{-2ibx}) \right]$$
$$\cos^2 2bx = \frac{e^{4bix}}{4} + \frac{e^{-4bix}}{4} + \frac{1}{2}$$

right?

16. Oct 23, 2006

### quasar987

$$\cos^2 2bx = \frac{1}{4} \left[ (e^{2ibx}+e^{-2ibx})(e^{2ibx}+e^{-2ibx}) \right]$$

17. Oct 23, 2006

Woops. That line was all kinds of messed up.

18. Oct 23, 2006

### Geronimo85

yes, that makes sense, but how is that any easier to integrate when multiplied by e^ax?

19. Oct 23, 2006

What's $$(e^{ax})(e^{bix})$$ ?
$$(e^{ax})(e^{bix}) = e^{ax + bix} = e^{(a+bi)x}$$
$$\int e^{(a+bi)x} \, dx = \frac{1}{a+bi} e^{(a+bi)x}$$