Question

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

Answer 1

`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`