Action integral for a simple pendulum in Bohn's Analytical Mechanics

[RahSuh]

TL;DR Summary: Am stuck on an integral at the bottom of page 34

Hi - I am working thru (by myself) the small textbook by Bohn on Analytical Mechanics. Its very good but am stuck on Page 34, at the bottom. It concerns the "action" of a simple pendulum - I understand the math concept of action as Bohn . I just dont understand how he gets the integral works. ie in the snip attached how he gets from the first line of the integral for A to the second line. The integral looks really, really messy. Any help appreciated. (the book is very, very good - so far!)

 

[dextercioby]

It's the integral $\int_{u}^{v} \sqrt{a-bx^2} {}dx$ which is even handled in high-school mathematics (at least where I live).

 

[Orodruin]

That it has a lot of constants doesn’t really make it much more difficult. In the end, the integrand is on the form $\sqrt{A - B\phi^2}$.

 

[TSny]

You will also need to know the relation between $E$ and $\phi_0$.

 

[Delta2]

here is the result of indefinite integral according to wolfram. It is quite messy and its gonna be even messier if you replace constants a and b with the combined constants you have in your expression.

https://www.wolframalpha.com/input?i=integral+of+\sqrt(a-b\phi^2)+d\phi

if you want to understand the inner workings of calculating this integral, first see that it is the same as $$\sqrt{a} \int\sqrt{1-k^2\phi^2} d\phi$$ for $$ k=\sqrt\frac{b}{a}$$ and then use the substitution $$\phi=\frac{1}{k}\sin \theta$$ to do integration by substitution.

That is gonna be a good practice for you, not only in calculus but in trigonometric identities too.