Int(1/sqrt(9-4x^2) dx) = Int(1/2 * 1/sqrt(9/4 - x^2) dx) (Where I have factored a sqrt(4) from the denominator.)

Note that 9/4 = (3/2)^2 is a perfect square. Now make the trig substitution:

x = 3/2 * sin(y) ==> y = asn(2x/3)

dx = 3/2 * cos(y) dy

Int(1/sqrt(9-4x^2) dx) = 1/2 * Int(1/sqrt(9/4 - 9/4sin^2(y)) 3/2*cos(y)dy)

= 1/2 * 3/2 * Int( 1/(3/2) * 1/sqrt(1 - sin^2(y)) cos(y)dy) (I have factored a sqrt(9/4) from the denominator.)

= 1/2 * 3/2 * 2/3 * Int(1/cos(y) * cos(y)dy)

= 1/2 * Int(dy)

= 1/2 * y + C

= 1/2 * asn(2x/3) + C.

-Dan