How do you find the integral of e^(-x)sin(2x)dx using integration by parts? Here's my work: u = sin(2x) dv to be e^(-x)dx du = 2cos(2x)dx v =...

You need
to avoid the perpetual cycle, hence, you should use the following notation `int ``e^(-x)sin 2x
dx`  = I and you need to use parts to solve the integral such that:

`u =
e^(-x) => du = -e^(-x)dx`

`dv = sin 2x dx => v = -(cos 2x)/2`

Using the following formula yields:

`int udv = uv - int
vdu`

`int e^(-x)sin 2x dx = -(e^(-x)*cos 2x)/2 - (1/2)int e^(-x)cos 2x dx
`

You should solve the integral `int e^(-x)cos 2x dx`  using parts such
that:

`u =...