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Integral using Euler's formula.

by cragar
Tags: euler, formula, integral
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cragar
#1
Mar28-12, 07:34 AM
P: 2,466
If I have [itex] \int e^{2x}sin(x)sin(2x) [/itex]
And then I use Eulers formula to substitute in for the sine terms.
So I have [itex] \int e^{2x}e^{ix}e^{2ix} [/itex]
then I combine everything so i get
[itex] e^{(2+3i)x} [/itex]
so then we integrate the exponential, so we divide by 2+3i
and then i multiply by the complex conjugate. now since sine is the imaginary part of his
formula I took the imaginary part when I back substituted for e^(3i)
but I didn't get the correct answer doing this, so am i not using Eulers formula correctly?
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NeroKid
#2
Mar28-12, 08:32 AM
P: 44
e^i3x = sin3x+isin2x , so the imaginary part is different from i(sinxsin2x)
cragar
#3
Mar28-12, 04:11 PM
P: 2,466
why does e^i3x = sin3x+isin2x , i guess im not seeing it off hand I probably should look at it more and try to manipulate it more.

NeroKid
#4
Mar28-12, 09:29 PM
P: 44
Integral using Euler's formula.

sry typos , its sin3x
cragar
#5
Mar28-12, 09:41 PM
P: 2,466
how come one part is not cos(3x)
NeroKid
#6
Mar28-12, 10:16 PM
P: 44
another typos , sry =='
cragar
#7
Mar29-12, 04:28 AM
P: 2,466
But we could get it in the form of
[itex] sin(x)e^{2ix}=isin(2x)sin(x)+cos(2x)sin(x) [/itex]
Do we need to get an expression where we have just exponentials on the left hand side
and then isin(x)sin(2x)+cos(2x)cos(x)
NeroKid
#8
Mar29-12, 06:32 AM
P: 44
but then ur integral cant become e^i3x now , can it
cragar
#9
Mar29-12, 06:01 PM
P: 2,466
ok, im not sure exactly what you mean, How do you recommend I approach the problem.
NeroKid
#10
Mar29-12, 06:45 PM
P: 44
sina*sinb = -0.5[cos(a+b)+cos(a-b)] , then u have 2 solvable integrals
cragar
#11
Mar30-12, 05:25 AM
P: 2,466
oh i see thanks for your answer.


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