let x = 3 sin(u) ==>sin(u)=x/3 and cos(u) =1/3v( 9-x^2), sin(2u) = (2/9)xv( 9-x^2) and u= sin^-1(x/3)
dx = 3 cos(u) du

now the integral becomes

?9 sin^2(u) ( 3 cos u) du/v( 9 - 9sin^2(u))

= 9? sin^2(u) cos(u) du/ cos(u)

= 9 ? sin^2(u) du

=9 ?(1/2)[ 1 - cos(2u) ] du

= 9/2 ?du - 9/2 ? cos(2u) du

= (9/2)u - (9/4)sin(2u) + C

substitute, sin(2u) = (2/9)xv( 9-x^2) and u= sin^-1(x/3)

= (9/2) sin^-1(x/3) - (1/2)xv( 9-x^2) + C

View the original article here