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Integration using eulers formula and complex numbers

  1. Mar 11, 2009 #1
    does any one know how to integrate e^(-2x)(cos(3x)dx
    using eulers formula e^(ix)= i(sinx)+cosx
     
  2. jcsd
  3. Mar 11, 2009 #2
    Use eular's identity to express the above equation as an exponential function.
     
  4. Mar 11, 2009 #3

    HallsofIvy

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    From [itex]e^{ix}= cos x+i sinx[/itex], changing x to -x and remembering that cosine is an "even" function and sine is an "odd" function, [itex]e^{-ix}= cos x- i sin s[/itex] so adding, [itex]e^{ix}+ e^{-ix}= 2 cos x[/itex] so [itex]cos x= (e^{ix}+ e^{-ix})/2[/itex].

    [tex]cos(3x)= \frac{e^{3ix}+ e^{-3ix}}{2}[/tex]

    [tex]e^{2x}cos(3x)= \frac{e^{(2+3i)x}+ e^{(2-3i)}}{2}[/tex]
     
  5. Mar 11, 2009 #4
    ok i get it now thanks
     
  6. Mar 14, 2009 #5
    k i think i get it now but how would we integrate e^(2x)*(sin(-2x))dx
     
  7. Mar 15, 2009 #6

    HallsofIvy

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    [tex]e^{2x}sin(-2x)= -e^{2x}sin(2x)[tex]
    [tex]sin(2x)= \frac{e^{2x}- e^{-2x}}{2i}[/tex]
     
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