To do this, I think you will need to realise that:
int 1/(sqrt(1-t^2)) dt = sin^(-1) t + C implies d/dt ( sin^(-1) t) = 1/(sqrt(1-t^2))
So
int sin^(-1) sqrt x \ dx
= int (x)^prime sin^(-1) sqrtx \ dx
And by IBP
=x sin^(-1) sqrtx - int x ( sin^(-1) sqrt x)^prime \ dx
=x sin^(-1) sqrtx - int x 1/sqrt(1-x) (sqrtx)' \ dx
=x sin^(-1) sqrtx - 1/2color(red)( int sqrtx /sqrt(1-x) \ dx) qquad triangle
For the red bit, if we let x = sin^2 y, dx = 2 sin y cos y dy, we have:
int sqrtx /sqrt(1-x) \ dx = int siny /sqrt(1- sin^2 y) \ 2 sin y cos y \ dy
= int 2 sin^2y \ dy
From the double angle formula:
= int 1 - cos 2y \ dy
= y - 1/2 sin 2y + C
Reversing out of the sub:
= sin sqrt x - 1/2 sin (2 sin^(-1) sqrtx) + C
Again a double angula formula for middle term:
= sin sqrt x - sin ( sin^(-1) sqrtx) cos ( sin^(-1) sqrtx) + C
= sin sqrt x -sqrtx sqrt(1-x) + C
If we park this back in triangle:
=x sin^(-1) sqrtx - 1/2(sin sqrt x - sqrt(x(1-x)) + C)
=(x - 1/2) sin^(-1) sqrtx - sqrt(x(1-x)) + C
