How do you integrate #int e^-x*cos2xdx# using integration by parts?

Question

1930 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-12T22:21:48

# = -1/5 e^-x*cos2x + 2 e^-x sin 2x + C#

#I = int e^-x*cos2x \ dx#

using IBP

#I = int (-e^-x)'*cos2x \ dx#

#I = -e^-x*cos2x - int \-e^-x (cos 2x)' \ dx#

#= -e^-x*cos2x - 2int \e^-x sin 2x \ dx#

#= -e^-x*cos2x - 2J#

#J = int \e^-x sin 2x \ dx#

# = int (-\e^-x)' sin2x \ dx#

# = -e^-x sin 2x - int -\e^-x (sin 2x)' \ dx#

# = -e^-x sin 2x + 2 int \e^-x cos 2x \ dx#

# = -e^-x sin x + 2I#

#I = -e^-x cos2x - 2(- \e^-x sin 2x + 2I)#

# = -e^-x cos2x + 2 e^-x sin 2x -4I + C#

#5I = -e^-x cos2x + 2 e^-x sin 2x + C#

#I = -1/5 e^-x cos2x + 2 e^-x sin 2x + C#