# Integration using eulers formula and complex numbers

## Main Question or Discussion Point

does any one know how to integrate e^(-2x)(cos(3x)dx
using eulers formula e^(ix)= i(sinx)+cosx

does any one know how to integrate e^(-2x)(cos(3x)dx
using eulers formula e^(ix)= i(sinx)+cosx
Use eular's identity to express the above equation as an exponential function.

HallsofIvy
Homework Helper
From $e^{ix}= cos x+i sinx$, changing x to -x and remembering that cosine is an "even" function and sine is an "odd" function, $e^{-ix}= cos x- i sin s$ so adding, $e^{ix}+ e^{-ix}= 2 cos x$ so $cos x= (e^{ix}+ e^{-ix})/2$.

$$cos(3x)= \frac{e^{3ix}+ e^{-3ix}}{2}$$

$$e^{2x}cos(3x)= \frac{e^{(2+3i)x}+ e^{(2-3i)}}{2}$$

ok i get it now thanks

k i think i get it now but how would we integrate e^(2x)*(sin(-2x))dx

HallsofIvy
$$e^{2x}sin(-2x)= -e^{2x}sin(2x)[tex] [tex]sin(2x)= \frac{e^{2x}- e^{-2x}}{2i}$$