Evaluate: `int(sin^(-1)x)^2\ dx`

Explanation

Putting x=sin t and dx=cos t dt, we get
`int(sin^(-1)x)^(2)dx=intt^(2)costdt`
`=t^(2)*(sint)-int2t(sint)dt" "` [integrating by parts]
`=t^(2)sint-2[t(-cost)-int1*(-cost)dt]" "` [integrating t (sin t) by parts]
`=t^(2)sint+2tcost-2sint+C`
`=x(sin^(-1)x)^(2)+2(sin^(-1)x)sqrt(1-x^(2))-2x+C`.