How do I evaluate e2xcosxdx by parts?

1 Answer
Feb 1, 2015

I´ll start by integrating e2x and leaving cos(x) as it is and then derive it leaving the integrated part as it is.

e2x2cos(x)e2x2(sin(x))dx=
=e2x2cos(x)+12e2x(sin(x))dx=

by parts again:
=e2x2cos(x)+12[e2x2(sin(x))e2x2cos(x)dx]=
=e2x2cos(x)+e2x4(sin(x))14e2xcos(x)dx

So your integral is:
e2xcos(x)dx=e2x2cos(x)+e2x4(sin(x))14e2xcos(x)dx

Now I can take to the left the integral: 14e2xcos(x)dx
Giving:

e2xcos(x)dx+14e2xcos(x)dx=e2x2cos(x)+e2x4(sin(x))

54e2xcos(x)dx=e2x2cos(x)+e2x4(sin(x))

and finally:
e2xcos(x)dx=45[e2x2cos(x)+e2x4(sin(x))]+c