# How to integrate $\int_{\theta_0}^{\pi} \sqrt{\frac{(1-\cos \theta)}{(\cos \theta_0 - \cos \theta)}}d\theta$?

I was wondering if you could solve this integral for me. It is part of a much larger problem, and I can't seem to solve it at all. I know that the answer should be some constant. Thanks in advance!

$\int_{\theta_0}^{\pi} \sqrt{\frac{(1-\cos \theta)}{(\cos \theta_0 - \cos \theta)}}d\theta$

I have tried substituting $u = \cos \theta$. However, this did now work since one of the limits of integration then depends on $\theta_0$.

• It might help to include the context in order to provide details. For instance, my guess would be that this is related to the oscillation of a classical pendulum. – Semiclassical Mar 17 '16 at 18:13
• Please show your own efforts – Yuriy S Mar 17 '16 at 18:16
• It is related to a ball falling along a cycloid-shaped ramp. The point is to prove that no matter where the ball is placed, the time to reach the lowest point of the cycloid is the same. – Birdman2246 Mar 17 '16 at 18:17
• I edited in my best efforts. There really isn't much to show though. – Birdman2246 Mar 17 '16 at 18:23

$$-\int_{\theta_0}^{\pi} \frac{d(\cos{\theta})}{\sqrt{(\cos{\theta_0}-\cos{\theta})(1+\cos{\theta})}} = \int_{-1}^{\cos{\theta_0}} \frac{du}{\sqrt{(\cos{\theta_0} - u)(1+u) }} = \int_0^{1+\cos{\theta_0}}\frac{dv}{\sqrt{v (1+\cos{\theta_0}-v)}} = 2 \int_0^{s_0} \frac{dv}{\sqrt{s_0^2-v^2}}$$
where $s_0^2 = 1+\cos{\theta_0}$. I hope you can take it from here.
• Let $v=s_0\sin\phi$ in that last integral and you really do get a constant – user5713492 Mar 17 '16 at 18:31