In the exercise, we have to prove that the time to get to the lowest part of this curve is $\pi \sqrt{\frac{a}{g}}$ no matter the initial point (which I am considering as $\theta = \alpha$).
I have done the following
$y = a(\theta - sin(\theta))$
$x = a(1 - cos(\theta))$
$t = \int \frac{ds}{v}$
$v = \sqrt{2g(x - x_0)}$
$x_0 = a(1 - cos(\alpha))$
$v = \sqrt{2ga(cos(\alpha) - cos(\theta)}$
$t = \int_{\theta = \alpha}^{\theta = \pi} \sqrt{\dfrac{a}{g}} \dfrac{\sqrt{1 - cos(\theta)}}{\sqrt{cos(\alpha) - cos(\theta)}} d\theta$
Currently, I am in trouble to evaluate this integral. But I guess I am not solving the entire exercise the right way. Any hint?