`int(dx)/(sqrt(2x-x^2)`

This is not in the form of either of our new formulas, but we can do some juggling to get it into a useful form.

First, we recognise that

`2x-x^2=-(x^2-2x)`

We now add `1` at the front, then compensate for it inside the bracket:

`=1-(x^2-2x+1)`

`=1-(x-1)^2`

With `a = 1`; `u = x - 1`, and `du = dx`, our integral becomes:

`int(dx)/(sqrt(2x-x^2))=int(du)/(sqrt(1-u^2))`

`=arcsin\ u+K`

`=arcsin(x-1)+K`

Get the Daily Math Tweet!

IntMath on Twitter