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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 $\theta_0$.

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  • $\begingroup$ 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. $\endgroup$ Mar 17, 2016 at 18:13
  • $\begingroup$ Please show your own efforts $\endgroup$
    – Yuriy S
    Mar 17, 2016 at 18:16
  • $\begingroup$ 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. $\endgroup$ Mar 17, 2016 at 18:17
  • $\begingroup$ I edited in my best efforts. There really isn't much to show though. $\endgroup$ Mar 17, 2016 at 18:23

1 Answer 1

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Integral is equal to

$$-\int_{\theta_0}^{\pi} \frac{d(\cos{\theta})}{\sqrt{(\cos{\theta_0}-\cos{\theta})(1+\cos{\theta})}} = \int_{-1}^{\cos{\theta_0}} \frac{du}{\sqrt{(\cos{\theta_0} - u)(1+u) }} = \int_0^{1+\cos{\theta_0}}\frac{dv}{\sqrt{v (1+\cos{\theta_0}-v)}} = 2 \int_0^{s_0} \frac{dv}{\sqrt{s_0^2-v^2}}$$

where $s_0^2 = 1+\cos{\theta_0}$. I hope you can take it from here.

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  • $\begingroup$ Let $v=s_0\sin\phi$ in that last integral and you really do get a constant $\endgroup$ Mar 17, 2016 at 18:31
  • $\begingroup$ Yeah, I just noticed that! Thanks for the advice to both of you! $\endgroup$ Mar 17, 2016 at 18:32

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