Question

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

Answer 1

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