use a trig sub.
the Pythagorean identity cos^2 psi+ sin^2 psi = 1cos2ψ+sin2ψ=1 so let x = sin psi, dx = cos psi dpsix=sinψ,dx=cosψdψ
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
