Choose the branch for $(1-\zeta^2)^{1/2}$ that makes it holomorphic in the upper half-plane and positive when $-1<\zeta<1$ From Stein/Shakarchi's Complex Analysis page 232: 

...We consider for $z\in \mathbb{H}$, $$f(z)=\int_0^z
 \frac{d\zeta}{(1-\zeta^2)^{1/2}},$$ where the integral is taken from
  $0$ to $z$ along any path in the closed upper half-plane. We choose
  the branch for $(1-\zeta^2)^{1/2}$ that makes it holomorphic in the
  upper half-plane and positive when $-1<\zeta<1$. As a result,
  $$(1-\zeta^2)^{-1/2}=i(\zeta^2-1)^{-1/2}\quad \text{when }\zeta>1.$$

I'm a little confused on the branch they're choosing. Do they mean that the we cut along $\mathbb{R}\setminus [-1,1]$? But then $(\zeta^2-1)^{-1/2}$ is undefined for $\zeta>1$...

Another (probably not as good) way to look at it is to think
$$
\sqrt{1-\zeta^2} = i \sqrt{\zeta -1} \sqrt{\zeta + 1} 
$$
Choose the (possibly different) branches for each of the square roots on the right side to give you what you need. It might look at first like you will have problems on $z \in [-1,1]$, but the discontinuities cancel each other.
For example, use the principle branch of $\sqrt{w}$ on $\zeta - 1$, but for $\zeta + 1$ use the branch of $\sqrt{w}$ where $-2\pi < \arg(w) < 0$.
EDIT:
Use $-2 \pi \le \arg(\zeta+1) < 0$ and $-\pi < \arg(\zeta -1) \le \pi$.
Then $-\pi \le \arg(\sqrt{\zeta+1}) < 0$ and $-\pi/2 < \arg(\sqrt{\zeta -1}) \le \pi/2$.  
So on $-1 < z < 1$ you get
$$
\arg(i \sqrt{\zeta -1} \sqrt{\zeta + 1} ) = \pi/2 + (-\pi) + \pi/2 = 0
$$
On $1 < z$ you get
$$
\arg(i \sqrt{\zeta -1} \sqrt{\zeta + 1} ) = \pi/2 + (-\pi) + 0 = -\pi/2
$$
so
$$
\arg(\frac{1}{i \sqrt{\zeta -1} \sqrt{\zeta + 1}} ) = \pi/2 = \arg(\frac{i}{\sqrt{t^2-1}})
$$
where $t = |\zeta| > 1$ and $\sqrt{t^2-1}$ uses the regular square root on $\mathbb{R}^+$.
