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anonymous

  • one year ago

integral (cos x)e^-x using euler's formula

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  1. Loser66
    • one year ago
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    What is the Euler's formula?

  2. anonymous
    • one year ago
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    cos=(e^ix+e^-ix)/2 and sin (e^ix-e^-ix)/2i

  3. Loser66
    • one year ago
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    Confirm: What is your problem? \[\int cos(x) e^{-x}dx\] or \[\int cos (x) e^{-ix} dx\] Which one?

  4. anonymous
    • one year ago
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    first one

  5. Loser66
    • one year ago
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    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 \(= \dfrac{1}{2}\dfrac{e^{x(i-1)}}{i-1}+C\)

  6. Loser66
    • one year ago
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    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.

  7. anonymous
    • one year ago
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    Thank you :)

  8. IrishBoy123
    • one year ago
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    \[ (cos x)e^{-x} = \mathcal {Re} \ e^{ix} \ e^{-x}\] \[\ \implies \mathcal{Re} \ \int \ e^{(-1+i)x} \ dx \] \[= \mathcal{Re} \ \frac{1}{-1+i} \ e^{(-1+i)x}\] \[= \mathcal{Re} \ \frac{-1-i}{2} \ e^{-x} \ e^{ix}\] \[= \ \frac{ e^{-x}}{2} \mathcal{Re} \ (-1-i). (cos x + i sin x)\] \[= \ \frac{ e^{-x}}{2} \mathcal{Re} (-cosx + sin x + i[-cosx -sinx])\] \[=\frac{ e^{-x}}{2} (-cosx + sin x)\]

  9. IrishBoy123
    • one year ago
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    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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