Question

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

Answer 1

What is the Euler's formula?

Answer 2

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

Answer 3

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

Answer 4

first one

Answer 5

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

Answer 6

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.

Answer 7

Thank you :)

Answer 8

\[ (cos x)e^{-x} = \mathcal {Re} \ e^{ix} \ e^{-x}\]
\[\ \implies \mathcal{Re} \ \int \ e^{(-1+i)x} \ dx \]

\[= \mathcal{Re} \ \frac{1}{-1+i} \ e^{(-1+i)x}\]

\[= \mathcal{Re} \ \frac{-1-i}{2} \ e^{-x} \ e^{ix}\]

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

\[=\frac{ e^{-x}}{2} (-cosx + sin x)\]

Answer 9

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.