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Stuck on a step

  1. Aug 11, 2005 #1
    i'm trying to solve an equation, but i'm stuck on this step:
    (integral sign) e^2x*e^x(3sin2x+2cos2x)dx =?
     
  2. jcsd
  3. Aug 11, 2005 #2
    use the exponential form for the trig functions and multiply through. you'll get a few easy integrals of the form
    [tex]\int e^{(a+ib)x}dx[/tex]
     
  4. Aug 11, 2005 #3

    arildno

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    If you don't know about the complex exponential, here's a technique using integration by parts (I'll let you tailor it to your specific example):
    Suppose you are to find an anti-derivative (i.e, indefinite integral) of the function [tex]f(x)=e^{x}\sin(x)[/tex], that is, you are to find J, where J is given as:
    [tex]J=\int{e}^{x}\sin(x)dx(1)[/tex]
    The right-hand side can now be rewritten as:
    [tex]\int{e}^{x}\sin(x)dx=e^{x}\sin(x)-\int{e}^{x}\cos(x)dx=e^{x}\sin(x)-e^{x}\cos(x)-\int{e}^{x}\sin(x)dx=e^{x}\sin(x)-e^{x}\cos(x)-J(2)[/tex]
    where I have used integration by parts twice, along with (1).
    Thus, we have:
    [tex]J=e^{x}\sin(x)-e^{x}\cos(x)-J\to{J}=\frac{e^{x}\sin(x)-e^{x}\cos(x)}{2}[/tex]
    (I've not bothered with the constant of integration; this should also be included in the final expression).
     
  5. Aug 12, 2005 #4
    ok, thanks! i'll try that~
     
  6. Aug 12, 2005 #5
    come to think of it... is there a quicker way?
     
  7. Aug 12, 2005 #6

    arildno

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    Sure; we've got the "abra-kadabra" formula, but it is only taught to 50 year old professor with proven gentle disposition because of the formula's potential for abuse.
     
  8. Aug 13, 2005 #7
    haha... thanks! :)
     
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