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Integral using Euler's formula.

  1. Mar 28, 2012 #1
    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?
     
  2. jcsd
  3. Mar 28, 2012 #2
    e^i3x = sin3x+isin2x , so the imaginary part is different from i(sinxsin2x)
     
  4. Mar 28, 2012 #3
    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.
     
  5. Mar 28, 2012 #4
    sry typos , its sin3x
     
  6. Mar 28, 2012 #5
    how come one part is not cos(3x)
     
  7. Mar 28, 2012 #6
    another typos , sry =='
     
  8. Mar 29, 2012 #7
    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)
     
  9. Mar 29, 2012 #8
    but then ur integral cant become e^i3x now , can it
     
  10. Mar 29, 2012 #9
    ok, im not sure exactly what you mean, How do you recommend I approach the problem.
     
  11. Mar 29, 2012 #10
    sina*sinb = -0.5[cos(a+b)+cos(a-b)] , then u have 2 solvable integrals
     
  12. Mar 30, 2012 #11
    oh i see thanks for your answer.
     
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