anonymous
  • anonymous
integral (cos x)e^-x using euler's formula
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Loser66
  • Loser66
What is the Euler's formula?
anonymous
  • anonymous
cos=(e^ix+e^-ix)/2 and sin (e^ix-e^-ix)/2i
Loser66
  • Loser66
Confirm: What is your problem? \[\int cos(x) e^{-x}dx\] or \[\int cos (x) e^{-ix} dx\] Which one?

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More answers

anonymous
  • anonymous
first one
Loser66
  • Loser66
I am doubt myself when I take it easy!! ha!! \[\int cos(x) e^{-x}dx =\int \dfrac{e^{ix}+e^{-ix}}{2}e^{-x}dx\] =\[1/2\int e^{x(i-1)}dx +1/2\int e^{-x(i+1)}dx\] For the firs one, it is equal \(= \dfrac{1}{2}\dfrac{e^{x(i-1)}}{i-1}+C\)
Loser66
  • Loser66
Do the same with the second one and simplify if it is needed. But I am not confident on it since it is quite easy like that.
anonymous
  • anonymous
Thank you :)
IrishBoy123
  • IrishBoy123
\[ (cos x)e^{-x} = \mathcal {Re} \ e^{ix} \ e^{-x}\] \[\ \implies \mathcal{Re} \ \int \ e^{(-1+i)x} \ dx \] \[= \mathcal{Re} \ \frac{1}{-1+i} \ e^{(-1+i)x}\] \[= \mathcal{Re} \ \frac{-1-i}{2} \ e^{-x} \ e^{ix}\] \[= \ \frac{ e^{-x}}{2} \mathcal{Re} \ (-1-i). (cos x + i sin x)\] \[= \ \frac{ e^{-x}}{2} \mathcal{Re} (-cosx + sin x + i[-cosx -sinx])\] \[=\frac{ e^{-x}}{2} (-cosx + sin x)\]
IrishBoy123
  • IrishBoy123
oooops! that is a complete solution which is a bad, bad thing - mea culpa, my bad - but you presumably know he answer anyway and are looking into some methods to do it and this is @Loser66's approach with a slant.

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