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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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    is it possible i am misunderstanding the purpose of the question? when someone asks for a taylor expansion, can i do a bunch of trig substitutions and then do a different type of expansion and feel like everything is okey dokey?2012-09-07
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    Is it possible that by "up to $O(\theta^4)$" your instructor meant $\cos(\theta) = 1 - \theta^2/2 + O(\theta^4)$, i.e., just the first two terms? That would certainly make the problem easier (the integral is then of the form $\int \frac{d\theta}{\sqrt{a - b\theta^2}}$). Also, there's no need to expand $\cos(\theta_0)$ in a Taylor series - it's just a constant.2012-09-07
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    hmm maybe your right2012-09-07
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    in genereal if someone says up to O(x^4) does that mean including x^4 or just all the ones before it?2012-09-07
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    I take that to mean "for small x, the error is bounded by $Cx^4$" (i.e. just the terms before $x^4$), but I think this is ambiguous enough that you should ask your instructor for clarification - especially since you mentioned something about "not just the standard formula, but with a correction term".2012-09-07
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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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    i'm sorry could you explain what you did a little more... i'm trying to figure out the expansion... sorry to pester you2012-09-07
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    @Tim, the integrand is a nice (technically: analytic) function of $\theta$ in a neighborhood of $\theta=0$, so it has an expansion in powers of $\theta$, a Maclaurin series. You know how to find the Maclaurin series for a function? I think that's all Eric has done.2012-09-07
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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