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Integration using Eulers formual

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data
    e^(-2x)*sin(3x)dx


    2. Relevant equations
    using e^(ix)=cosx+isinx


    3. The attempt at a solution
    so i get e^(-2+i3)(x)

    so then after i integrate i get (e^(-2+i3)(x))/(-2+i3) then i multiply by the conjegate to get the i out of the bottom then i multiply it by cos(3x)+isin(3x) then pick out the real part is this correct.
     
  2. jcsd
  3. Mar 15, 2009 #2

    Mark44

    Staff: Mentor

    So how do you get from e-2xsin(3x) to e(-2 + 3i)x?
     
  4. Mar 15, 2009 #3
    cause i think e^(3ix)=cos(3x)+isin(3x) i dont know i know it works with
    e^(-2x)*(cos(3x)
     
  5. Mar 15, 2009 #4

    rock.freak667

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    Homework Helper

    then you'd want to extract the imaginary component to get the required integral, right?
     
  6. Mar 15, 2009 #5
    yes but the sin(3x) is throwing me
    so in the formula e^(ix)=cos(x)+isin(x) in that formula the cosx is the real part
    but in our case we need the sinX to be the real part so do we need to manipluate the formula to make it work am i no the right track.
     
  7. Mar 15, 2009 #6

    rock.freak667

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    [tex]\int e^{2+3i}x dx = \int e^{2x}e^{3xi}dx=\int (e^{2x}cos3x+ie^{2x}sin3x)dx=A+Bi[/tex]

    A+Bi is what you realized when you integrated exp(2+3i)


    [tex]\int (e^{2x}cos3x+ie^{2x}sin3x)dx = \int e^{2x}cos3x dx + i \int e^{2x}sin3x dx =A+iB[/tex]


    So if you equate the real and imaginary components...

    the imaginary part B= [itex]\int e^{2x}sin3x dx[/itex]
     
  8. Mar 15, 2009 #7

    Mark44

    Staff: Mentor

    I believe you're on the wrong track. Another approach is to integrate by parts twice. After the first application the part you integrate will be something like e^(-2x)cos(3x). Apply integration by parts one more time to get an integral similar to the one you started with. You'll should have an equation in which you can solve algebraically for the integral you want.
     
  9. Mar 15, 2009 #8
    the whole reason for doing this is to avoid integration by parts which i know how to do it that way. i want to learn how to do it this way .
     
  10. Mar 15, 2009 #9

    Mark44

    Staff: Mentor

    OK, then rock.freak's and your direction is the way to go...
     
  11. Mar 15, 2009 #10
    ok so would i then divide everything by i in the formula to make sinx the real part then use that so i would have e^(ix)/i=cosx/i + sinx or is that wrong
     
  12. Mar 15, 2009 #11

    rock.freak667

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    Homework Helper

    I still don't get why you want to make sin on the left side in the integral for.When you get the right side in the form A+Bi. B is supposed to be the required integral.
     
  13. Mar 15, 2009 #12
    k i think i get it now.
     
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