Integral using Euler's formula.

  • Thread starter cragar
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
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e^i3x = sin3x+isin2x , so the imaginary part is different from i(sinxsin2x)
 
  • #3
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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.
 
  • #4
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sry typos , its sin3x
 
  • #5
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how come one part is not cos(3x)
 
  • #6
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another typos , sry =='
 
  • #7
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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)
 
  • #8
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but then ur integral cant become e^i3x now , can it
 
  • #9
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ok, im not sure exactly what you mean, How do you recommend I approach the problem.
 
  • #10
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sina*sinb = -0.5[cos(a+b)+cos(a-b)] , then u have 2 solvable integrals
 
  • #11
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oh i see thanks for your answer.
 

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