use a trig sub.
the Pythagorean identity #cos^2 psi+ sin^2 psi = 1# so let #x = sin psi, dx = cos psi dpsi#
thus #int dx qquad sqrt(1-x^2)# becomes #int d psi qquad cos psi sqrt(1 - sin^2 psi)#
#= int d psi qquad cos^2 psi#
from there double angle it from #cos 2A = 2 cos^2 A - 1#
#= int d psi qquad (cos 2psi + 1)/2 #
#= 1/2 int d psi qquad (cos 2psi + 1)#
#= 1/2 ( 1/2 sin 2psi + psi) + C#
#= color{blue}{(sin 2psi)}/4 + psi/2 + C qquad star#
again double angling the blue term from #sin 2A = 2 sin A cos A#
we have
#sin 2 psi = 2 sin psi cos psi#
#=2 x sqrt(1-x^2)#
putting that into #star# we get
#= (2 x sqrt(1-x^2))/4 + (arcsin x)/2 + C #
#= 1/2 x sqrt(1-x^2) + 1/2 arcsin x + C #
