How do you calculate this integral

I have the following integral:
int[(-y/(1-y^2)^(1/2))dy]

I'm just not sure what to use for the substitution if it should be u=y^2 or u=(1-y^2). Please show all work and explain how you did it and I will give 10 points to the best answer. The answer should be (1-y^2)^(1/2) + c. Thanks

Let u = 1 - y^2. It will make things easier

u = 1 - y^2
du = -2y * dy

-y * dy / sqrt(1 - y^2) =>
(1/2) * du / sqrt(u)

Now integrate:

(1/2) * 2 * u^(1/2) + C =>
sqrt(u) + C =>
sqrt(1 - y^2) + C

........... ∫ [ (-y) / √(1-y²) ] dy

= (1/2) ∫ [ 1 / √(1-y²) ] · ( -2y ) dy

= (1/2) ∫ ( 1 / √u ) du, ..................... u = 1-y², du = (-2u) du

= (1/2)· ( 2√u ) + C

= √(1-y²) + C ........................................… Ans.
____________________________

Let u = 1-y^2
du/dy = -2y
-ydy = du/2
integral becomes
(1/2) u^(-1/2) du
= u^(1/2) + C
= (1-y^2)^(1/2) + C