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

In summary: What's (e^{ax})(e^{bix}) ?Remember how nice exponentials play with multiplication:(e^{ax})(e^{bix}) = e^{ax + bix} = e^{(a+bi)x}
  • #1
Geronimo85
20
0
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.
 
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  • #2
integral(dx(e^ax)cos^2(2bx))
Can't you write it as nothing but complex exponentials?

e^(i*Pi/3)*e^(i*n*4Pi/3)
Have you tried plugging in values for n?
 
Last edited:
  • #3
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
(1) You know the value of e^(i*Pi/3).
(2) n ranges over all integers... not just 0 and 1.
 
  • #5
(1) You know the value of e^(i*Pi/3).

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.

(2) n ranges over all integers... not just 0 and 1.

-Got that, I was having a brain fart and thinking that for n=2, theta was 10Pi/4
 
  • #6
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.
But it's not -1... why do you think the n=0 solution has to be the one that's -1?
 
  • #7
it should be 1/2 + i*sqrt(3)/2, but then how do I ever get -1 for a value?
 
  • #8
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
Well, what did you get when you simplified after substituting?
 
  • #10
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
Aha! There shouldn't be any x²'s in there. Remember your exponent laws: what is [itex](a^b)^c[/itex]?
 
  • #12
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
Remember your basic arithmetic.

First off, remember how to compute (a + b)².
Secondly, as I said in my previous post, remember your exponent laws: what is
[tex]\left( a^b \right)^c?[/tex]
 
  • #14
Hurkyl said:
Remember your basic arithmetic.

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

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
Geronimo85 said:
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

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

right?
 
  • #16
Provided your second line reads instead,

[tex] \cos^2 2bx = \frac{1}{4} \left[ (e^{2ibx}+e^{-2ibx})(e^{2ibx}+e^{-2ibx}) \right] [/tex]
 
  • #17
quasar987 said:
Provided your second line reads instead,

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

Woops. That line was all kinds of messed up.
 
  • #18
yes, that makes sense, but how is that any easier to integrate when multiplied by e^ax?
 
  • #19
Geronimo85 said:
yes, that makes sense, but how is that any easier to integrate when multiplied by e^ax?

What's [tex] (e^{ax})(e^{bix}) [/tex] ?

Remember how nice exponentials play with multiplication:

[tex] (e^{ax})(e^{bix}) = e^{ax + bix} = e^{(a+bi)x} [/tex]


So,
[tex] \int e^{(a+bi)x} \, dx = \frac{1}{a+bi} e^{(a+bi)x} [/tex]
 

FAQ: Could really use a hand with a complex exponential and integral problem

1. How do I simplify a complex exponential and integral problem?

To simplify a complex exponential and integral problem, you can use the properties of exponential and logarithmic functions, as well as integration techniques such as substitution or integration by parts.

2. What is the purpose of using complex numbers in exponential and integral problems?

Complex numbers allow us to represent and manipulate quantities that involve both real and imaginary numbers. In exponential and integral problems, they can help us to solve equations and describe physical phenomena more accurately.

3. Can you provide an example of a complex exponential and integral problem?

One example is solving the integral of e^(2x+3) dx. We can use the property that e^(a+b) = e^a * e^b to rewrite the integral as e^(2x) * e^3 dx. Then, we can use the power rule for integrals to solve it as (1/2)e^(2x) + Ce^3.

4. How can I check if my solution to a complex exponential and integral problem is correct?

You can check your solution by differentiating it and seeing if you get back to the original function. For example, if you integrated e^(2x+3) and got (1/2)e^(2x) + Ce^3, you can differentiate it to get (1/2)(2e^(2x)) = e^(2x), which is the original function.

5. Are there any tips for solving complex exponential and integral problems?

Some tips include recognizing patterns and using properties, being familiar with integration techniques, and practicing regularly. It can also be helpful to break down the problem into smaller parts and approach it step by step.

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