Integration with Eulers formula.

  • Thread starter cragar
  • Start date
  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
132
Hey cragar.

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

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