How do you find the antiderivative of #e^( -x)*cos2x#?

#int dx \ e^(-x) cos 2x#

#= mathbb(Re) ( int dx \ e^(-x) e^(2 i x) )#

#= mathbb(Re) ( 1/(- 1 + 2 i) e^((- 1 + 2 i) x) )#

#=e^(-x) mathbb(Re) ( (- 1 - 2 i)/(5) (cos 2x + i sin 2x) )#

#=1/5e^(-x) ( - cos 2x + 2 sin 2x) + C#

#=-1/5 e^(-x) ( cos 2x - 2 sin 2x) + C#

Let:

# I = int \ e^(-x) \ cos(2x) \ dx #

We can use integration by parts:

Let # { (u,=cos2x, => (du)/dx=-2sinx), ((dv)/dx,=e^(-x), => v=-e^(-x) ) :}#

Then plugging into the IBP formula:

# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #

gives us

# int \ (cos2x)(e^(-x)) \ dx = (cos2x)(-e^(-x)) - int \ (-e^(-x))(-2sin2x) \ dx #
# :. I = -e^(-x) cos2x - 2 \ int \ e^(-x) \ sin2x \ dx # .... [A]

At first it appears as if we have made no progress, as now the second integral is similar to #I#, having exchanged #sin2x# for #sin2x#, but if we apply IBP a second time then the progress will become clear:

Let # { (u,=sin2x, => (du)/dx=2cos2x), ((dv)/dx,=e^(-x), => v=-e^(-x) ) :}#

Then plugging into the IBP formula, gives us:

# int \ (sin2x)(e^(-x)) \ dx = (sin2x)(-e^(-x)) - int \ (-e^(-x))(2cos2x) \ dx #
# :. int \ e^(-x) \ sin2x \ dx = -e^xsin2x + 2I #

Inserting this result into [A] we get:

# I = -e^(-x) cos2x - 2 (-e^xsin2x + 2I) + A#

Now we can solve fior #I#

# I = -e^(-x) cos2x + 2e^xsin2x - 4I + A#

# :. 5I = e^x{2sin2x - cos2x} + A#

# :. I = e^x/5{2sin2x - cos2x} + A#