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#
