On an upside-down cycloid the slider takes the same time … to…

On an upside-down cycloid the slider takes the same time $T$ to reach bottom _wherever it starts_. Starting at $\theta =\alpha$, write $1-\cos \theta =2{\sin }^2\theta /2$ and $1-\cos \alpha =2{\sin }^2\alpha /2$ to show that
$$T={\int }_{\alpha }^{\pi }\frac{\sqrt{2a^2(1-\cos \theta )}d\theta }{\sqrt{2ag(\cos \alpha -\cos \theta )}}=\pi \sqrt{\frac{a}{g}}.$$

Its source prints no answer for this problem, so there's nothing to check against.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §12.2 Plane Motion: Projectiles and Cycloids, #39