How do you integrate #int e^(2x)cosx# by integration by parts method?

Question

13934 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-04T20:44:44

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

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

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

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

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

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

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

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

#= e^(2x) sin x - 2 ( -e^(2x) cos x + 2 I + C)#

#= e^(2x) sin x + 2 e^(2x) cos x - 4 I + C#

#5 I = e^(2x) sin x + 2 e^(2x) cos x + C#

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