How do you find the antiderivative of #e^(2x)*sin(e^x)dx#?

1 Answer
Jan 17, 2017

Answer:

#inte^(2x)sin(e^x)dx=-e^xcos(e^x)+sin(e^x)+C#

Explanation:

#I=inte^(2x)sin(e^x)dx#

Let #t=e^x#. This implies that #dt=e^xdx#. Before substituting, note that #e^(2x)=e^x(e^x)#.

#I=inte^xsin(e^x)(e^xdx)=inttsin(t)dt#

To do this, use integration by parts. This takes the form #intudv=uv-intvdu#. Let:

#{(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 #t=e^x#:

#I=-e^xcos(e^x)+sin(e^x)+C#