Find arc length of a circle using a hyperbolic metric 
Given the hyperbolic metric $ds^2=\frac{dx^2+dy^2}{x^2}$ on the half plane $x > 0$, find the length of the arc of the circle $x^2+y^2=1$ from $(\cos\alpha,\sin\alpha)$ to $(\cos \beta, \sin\beta)$

I found that $ds^2=\displaystyle\frac{d\theta^2}{\cos^2\theta}$ but when I try to plug in $\pi/3, -\pi/3$, which should give me the arc length of $2\pi/3$, 
I get $4\pi/3=\sqrt{\displaystyle\frac{{(\pi/3-(-\pi/3))}^2}{cos^2{(\pi/3})}}$
I feel like I'm making a simple mistake but I cant place it

The circle $x^2 + y^2 = 1$ can be parametrised by $(\cos \theta, \sin \theta)$. If $x(\theta) = \cos \theta$ and $y(\theta) = \sin \theta$ then
$$ds^2 = \frac{dx^2+dy^2}{x^2} = \frac{(\sin^2\theta+\cos^2\theta) \, d\theta^2}{\cos^2\theta} = \sec^2\theta \, d\theta^2.$$
The arc-length that you are interested in is given by:
$$s = \int \sqrt{ds^2} = \int_{\alpha}^{\beta} |\sec \theta| \, d\theta \, . $$ 
