Integral dx/√(9-x²)=?

Let x = 3 sin(u)
Then, dx = 3 cos(u) du and sin(u) = x/3

∫ dx / √(9-x²)
= ∫ 3 cos(u) du / √[9 - 9 cos²(u)]
= ∫ cos(u) du / √[1 - cos²(u)]
= ∫ cos(u) du / sin(u)
= ∫ cot(u) du
= ln|sin(u)| + C
= ln|x/3| + C

I = ∫ dx / [ √ [3² - x² ]
I = (1/3) sin^(-1) [ x/2 ] + C

This is easier if you remember that

d/dx (sin⁻¹ x) = 1/√(1 - x²)

...which makes the indefinite integral:

∫ dx/√(1 - x²) = 1/√(1 - x²) + C

Now, back to your integral:
∫ dx/(9 - x²) = (1/3) ∫ dx / [1 - (x/3)²]

Now let x/3 = u, (1/3)dx = du to get:

∫ du/√(1 - u²) = sin⁻¹ u + C

...and back substitute to get sin⁻¹ (x/3) + C

Set x=3sint , t€[-π/2,π/2], t=arcsin(x/3)
x=3sint , dx=3costdt
I = integral 3costdt/√(9-9sin^2t)
=intg[ (3cost)dt/3cost
=intg[dt]
= t+c
=arcsin(x/3)+ c