can anyone tell me how to start to take the integral of dx/sqrt[9-x^2]?

Answer 1

trig substitution.
Let x = 3sin(u)
or you could use x = 3cos(u)
It doesn't matter.

Answer 2

you can use direct integral if you know the formula.
otherwise you can use substitution method.
let x = 3 sin(y)
dx = 3 cos(y)dy
now Integral of dx/squrt[9-x^2] will be equal to integral of 3cos(y) dy/sqrt[9 -9sin^2y]
=arcsin(x/3) + c
where c is integral constant

Answer 3

let 3u = x, then 3du = dx

Once you have made the substitutions factor out the 3^2 out of the square root and you will find this form listed in the table of integrals which integrates to the inverse Sin function.

Answer 4

Make the substitution:

x = 3sin θ
dx = (3cos θ)dθ

Then
√(9 - x²) = √(9 - (3sin θ)²) = √(9 - 9sin² θ) = √(9cos² θ) = 3cos θ

So
∫{dx/√(9 - x²)} = ∫{(3cos θ)/(3cos θ)}dθ = ∫dθ

You can do the rest.