intx^2/(sqrt(9-x^2)) dx= x^2/(sqrt(3^2-x^2))dx using our trig identities for integrals let us substitute 3sinphi = x and dx = 3cosphi dphi
therefore 3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9-(9sin^2phi))
3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9(1-(sin^2phi))
where 1-sin^2phi = cos^2phi
3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9(cos^2phi))
3int(9sin^2phi)(cancelcosphi)(dphi)/(3(cancelcosphi))
cancel3int(9sin^2phi)(dphi)/(cancel3)
int(9sin^2phi)(dphi) = 9int(sin^2phi)(dphi)
Use the identity from trig where cos(2phi) = 1-2sin^2phi rearranging to solve such that (cos(2phi) -1)/(-2)= sin^2phi or (1-cos(2phi))/(2)= sin^2phi
9int(1-cos(2phi))/(2)(dphi) = 9/2int(1-cos(2phi))(dphi) use subtraction rules of the integral
9/2int(dphi)-9/2intcos(2phi)dphi
9/2int(dphi) = (9/2)phi
-9/2intcos(2phi)dphi = -9/4sin(2phi)
Now piece together and substitute in
(9/2)phi-9/4sin(2phi)+C
phi = arcsin(x/3)
(9/2)phi-9/4sin(2phi)+C
(9/2)(arcsin(x/3)-1/2sin(2arcsin(x/3))+C