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## anonymous one year ago integral (cos x)e^-x using euler's formula

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1. Loser66

What is the Euler's formula?

2. anonymous

cos=(e^ix+e^-ix)/2 and sin (e^ix-e^-ix)/2i

3. Loser66

Confirm: What is your problem? $\int cos(x) e^{-x}dx$ or $\int cos (x) e^{-ix} dx$ Which one?

4. anonymous

first one

5. 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 $$= \dfrac{1}{2}\dfrac{e^{x(i-1)}}{i-1}+C$$

6. 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.

7. anonymous

Thank you :)

8. IrishBoy123

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

9. 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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