First let 3y = x. Then y = (1/3)x and dy = (1/3)dx so:

Int[x^2/sqrt(9 - x^2)dx = Int[(3y)^2/sqrt(9 - (3y)^2)*3dy] = 9*Int[y^2/sqrt(1 - y^2)dy]

Now let y = sin(t). Then dy = cos(t) dt. Thus:

Int[x^2/sqrt(9 - x^2)dx = 9*Int[y^2/sqrt(1 - y^2)dy] = 9*Int[sin^2(t)/sqrt(1 - sin^2(t))*cos(t)dt]

= 9*Int[sin^2(t)/cos(t)*cos(t)dt] = 9*Int[sin^2(t)dt]

Now you take it from here.

-Dan