Help with this integral on page 34 of Analytical Mechanics by John Bohn

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SUMMARY

The discussion focuses on solving the integral ##\int_{u}^{v} \sqrt{a-bx^2} {}dx## from John Bohn's "Analytical Mechanics." The integral is recognized as complex but manageable with the right substitutions. Specifically, it can be transformed into the form ##\sqrt{a} \int\sqrt{1-k^2\phi^2} d\phi##, where $$k=\sqrt\frac{b}{a}$$. The recommended substitution for simplification is $$\phi=\frac{1}{k}\sin \theta$$, which aids in performing integration by substitution effectively.

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Students of physics and mathematics, particularly those studying classical mechanics and integral calculus, will benefit from this discussion. It is also valuable for anyone looking to enhance their problem-solving skills in complex integrals.

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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!)
BohnPage34.jpg
 
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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).
 
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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}##.
 
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You will also need to know the relation between ##E## and ##\phi_0##.
 
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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.
 
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