How do you find the antiderivative of #e^(2x)*sin(e^x)dx#?
1 Answer
Jan 17, 2017
Explanation:
#I=inte^(2x)sin(e^x)dx#
Let
#I=inte^xsin(e^x)(e^xdx)=inttsin(t)dt#
To do this, use integration by parts. This takes the form
#{(u=t" "=>" "du=dt),(dv=sin(t)dt" "=>" "v=-cos(t)):}#
Then:
#I=uv-intvdu=-tcos(t)-int(-cos(t))dt#
#I=-tcos(t)+intcos(t)dt=-tcos(t)+sin(t)#
Returning to the original variable using
#I=-e^xcos(e^x)+sin(e^x)+C#