How do I evaluate #inte^(2x)cosx dx# by parts?

Question

1962 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-02-01T00:31:13

I´ll start by integrating #e^(2x)# and leaving #cos(x)# as it is and then derive it leaving the integrated part as it is.

#e^(2x)/2*cos(x)-inte^(2x)/2*(-sin(x))dx=#
#=e^(2x)/2*cos(x)+1/2inte^(2x)*(sin(x))dx=#

by parts again:
#=e^(2x)/2*cos(x)+1/2[e^(2x)/2*(sin(x))-inte^(2x)/2*cos(x)dx]=#
#=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))-1/4inte^(2x)*cos(x)dx#

So your integral is:
#inte^(2x)cos(x)dx=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))-1/4inte^(2x)*cos(x)dx#

Now I can take to the left the integral: #-1/4inte^(2x)*cos(x)dx#
Giving:

#inte^(2x)cos(x)dx+1/4inte^(2x)*cos(x)dx=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))#

#5/4inte^(2x)*cos(x)dx=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))#

and finally:
#inte^(2x)*cos(x)dx=4/5[e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))]+c#