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I have to deal with this integral in order to compute the period of a pendulum

$ \int^{\theta_{0}}_{0}\frac{d\theta}{\sqrt{\cos\theta_{0}-\cos\theta}} $

I was asked by my instructor to solve this with a taylor expansion for cos up to $O(\theta^4)$ I plugged in

$ \cos\theta_{0} = 1 - \frac{\theta_{0}^2}{2!} + \frac{\theta_{0}^4}{4!} $ $ \cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} $

$ \int^{\theta_{0}}_{0}\frac{d\theta}{\sqrt{\frac{\theta^2}{2}-\frac{\theta_0^2}{2}-\frac{\theta^4}{4!}+\frac{\theta_0^4}{4!} }} $

but the following integral eluded simplication ( I spent alot of time here). Later, I was able to solve this problem by using the substitution $\cos\theta = 1-2\sin^2\frac{\theta}{2}$ the integral is then solvable by series in terms of a binomial expansion in terms of $k^2x^2$ where $\sin x = \frac{\sin\frac{\theta}{2}}{\sin\frac{\theta_0}{2}}$

However, my task was not to do the expansion of a binomial but rather to solve the integral with an expansion for cos. Thus i am still lost as to how to proceed.

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    Please use$\sqrt{stuff+more stuff}$to get $\sqrt{stuff+more stuff}$ instead of$\surd{stuff+more stuff}$to get $\surd{stuff+more stuff}$ as it is clear what terms are under the square root sign.2012-09-07

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I'm pretty sure the integrand should be $ f\left(\theta\right) = \frac{1}{\sqrt{\cos \theta - \cos \theta_0}}, $ in which case I would just integrate its Taylor series about $\theta = 0$: $ \begin{eqnarray} f\left(\theta\right) &=& \sum_{n=0}^{\infty} \frac{1}{n!}\frac{d^n f}{d\theta^n}\Bigg|_{\theta = 0} \ \theta^n \\ &=& \frac{1}{\sqrt{1-\cos \theta_0}}\left[1 + \frac{1}{4\left(1-\cos \theta_0\right)} \theta^2 + \frac{2 \cos \theta_0 + 7}{96\left(1-\cos \theta_0\right)^2}\theta^4 + O\left(\theta^6\right)\right], \end{eqnarray} $ which is easy to integrate. Note that I just used this for the last part.

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    That's correct, I (Wolfram, technically) just wrote out the Taylor series (Maclaurin, since it's about $\theta = 0$) of the integrand.2012-09-08