Question

How do you integrate `int tsin(2t)` by integration by parts method?

Answer 1

`I = -t cos(2t)/2 + 1/4 sin(2t)`

Explanation:

Set `sin(2t)dt = dv`
then `v = -cos(2t)/2`
`u = t`
`du = dt`

Using `int udv = [uv] - int v du`
now you get
`I = -t cos(2t)/2 + 1/2 int cos(2t) dt`
`I = -t cos(2t)/2 + 1/4 sin(2t)`