How do you integrate #int e^xcos(2x)# by parts?

Question

10004 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-14T19:24:10

#int e^x cos(2x) dx = 1/5(e^xcos2x+2e^xsin2x) + C#

As #d(e^x) = e^xdx# we can integrate by parts as:

#int e^x cos(2x) dx = int cos(2x) d(e^x) = e^xcos2x + 2int e^xsin(2x)dx#

We integrate the resulting integral by parts again:

#int e^xsin(2x)dx = int sin(2x)d(e^x) = e^xsin2x -2int e^xcos(2x)dx#

So if we name:

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

we get the following equation:

#I = e^xcos2x+2e^xsin2x-4I#

and solving for #I#:

#I= 1/5(e^xcos2x+2e^xsin2x)#

so that:

#int e^x cos(2x) dx = 1/5(e^xcos2x+2e^xsin2x) + C#