Let x = 2sinθ Then, dx = 2cosθdθ
So, ∫ 9/√(4-x2)dx = 9∫(2cosθ)/√(4-4sin2θ)dθ
= 9∫(2cosθ)/(2√(1-sin2θ))dθ
= 9∫(2cosθ)/(2cosθ)dθ
= 9∫dθ
= 9θ + C [Since sinθ = x/2, θ = Sin-1(x/2)]
= 9 sin-1(x/2) + C