How do you integrate #e^2xcosx dx#?

1 Answer
Feb 7, 2015

Starting from:
#inte^2xcos(x)dx=# I would take the constant #e^2# out of the integral:
#e^2intxcos(x)dx=# then I would use integration by parts where you have:
#intf(x)*g(x)dx=F(x)g(x)-intF(x)g'(x)dx#
Where:
#F(x)=intf(x)dx#
#g'(x)# is the derivative of #g(x)#

In your case choosing #g(x)=x# you get:

#e^2[x*sin(x)-intsin(x)*1dx]=#
#e^2[xsin(x)+cos(x)]+c#

hope it helps