Mathematical pendulum - Arbitrary-amplitude period

[dingo_d]

Homework Statement

Hi! I have a problem with mathematical pendulum. I had to prove that the right expression for the period of the mathematical pendulum is:
$T=\sqrt{\frac{8l}{g}}\int_0^{\theta_0}\frac{d\theta}{\sqrt{\cos \theta- \cos \theta_0}}$
I did that via energy, but now I have to transform that integral with:
$\cos\theta=1-2\sin^2(\theta/2)$ and with this substitution
$\sin x =\sin(\theta/2)/\sin(\theta_0/2)$ so that I could expand that integral into Taylor series.

The attempt at a solution

I've tried with substitution and I get:
$2\sqrt{\frac{l}{g}}\int_0^{\theta}\frac{d\theta}{\sqrt{\sin^2(\theta_0/2)-\sin^2(\theta/2)}}$
Now I have seen what the expansion is on wikipedia, but there is no solution to how to get to there. If I try to use the sine substitution, I get stuck at changing the integral variables, and the integral becomes a mess. I saw that the complete elliptic integral of 1st order appears, but I don't know what to do with that.

Can anyone help with this clue? How to get that Taylors expansion?

Thank you!

 

[dingo_d]

Anyone?

 

[tomcue]

Hello! Thank you for reaching out with your question about the mathematical pendulum. It looks like you have made good progress so far in using energy to prove the expression for the period, and now you are trying to use substitutions and Taylor series to expand the integral. This can be a challenging problem, but I can offer some guidance to help you continue.

First, let's start with the substitution \cos\theta=1-2\sin^2(\theta/2). This is a good choice, as it will help us to simplify the integral and make it easier to work with. However, it is important to keep in mind that whenever we make a substitution, we also need to change the limits of integration accordingly. In this case, since we are integrating from 0 to \theta_0, we need to change our limits to 0 and \sin^{-1}(\sqrt{1-\cos\theta_0}). This will ensure that we are still integrating over the same range of values for \theta.

Next, let's consider the substitution \sin x =\sin(\theta/2)/\sin(\theta_0/2). This is a bit more complicated, as it involves changing the variable of integration. In order to do this, we can use the identity \sin 2x = 2\sin x \cos x to rewrite the integral as:

2\sqrt{\frac{l}{g}}\int_0^{\sin^{-1}(\sqrt{1-\cos\theta_0})}\frac{\cos x}{\sqrt{\sin^2(\theta_0/2)-\sin^2(x)}}dx

Now, we can use the Taylor series expansion of \sqrt{1-x} to expand the denominator of the integrand. This will give us a series of terms involving powers of \sin x. We can then use the identity \sin^2 x = \frac{1-\cos 2x}{2} to simplify these terms and rewrite the integral as a series of integrals involving only \cos x. This will bring us closer to the desired Taylor series expansion.

Finally, we can use the power series expansion of \cos x to expand each term in the integral. This will give us a series of integrals involving powers of \cos x, which can be evaluated using the power rule for integration. After evaluating these integrals and simplifying, we should be left with an