Indefinate integral of sin-1 (x? i know you have to use integral by parts: u = sin-1 (x v = x

Yes

you let u = sin-1 x and dv = 1 (so v = x)

and use the rule

INT udv = uv - INT vdu

= x sin^-1 x - INT x sin-1 x

u = sin-1 x so x = sin u

dx/du = cos u, du/dx = 1 / cos u = 1 / sqrt (1 - sin^2) = 1 / sqrt (1-x) ^2

so substituting we have

INT sin-1 (x) = x sin-1x - INT (x / sqrt (1 - x^2)

= x sun-1 x - ( - sqrt (1 - x^2)

= x sin-1 x + sqrt (1 - x^2) + C