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Integration with Eulers formula.

  1. Oct 18, 2012 #1
    I want to integrate [itex] \frac{e^x}{cos(x)} [/itex] with eulers formula.
    I start by writing [itex] \frac{e^x}{e^{ix}} [/itex]
    then I integrate that as usual.
    So after I integrate I get [itex] \frac{e^{(1-i)x}}{1-i} [/itex]
    Normally I would multiply and divide by the complex conjugate and then back substitute in
    e^(ix) and the take the real part.

    can I just back substitute in isin(x)+cos(x) on the bottom and then multiply it by (1-i)
    and then take the real part. That seems to easy.
    Does anyone have suggestions. This is not a homework problem.
     
  2. jcsd
  3. Oct 18, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey cragar.

    You should be able to integrate the expression and then take the real part as your answer.
     
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