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

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The discussion centers on a challenging integral from John Bohn's "Analytical Mechanics," specifically related to the action of a simple pendulum. The integral in question is ##\int_{u}^{v} \sqrt{a-bx^2} {}dx##, which the user finds complex due to its messy appearance and numerous constants. A suggested approach involves recognizing that the integral can be transformed into a more manageable form, specifically ##\sqrt{a} \int\sqrt{1-k^2\phi^2} d\phi##, using the substitution ##\phi=\frac{1}{k}\sin \theta##. This method not only aids in solving the integral but also serves as valuable practice for both calculus and trigonometric identities. Understanding these transformations is crucial for mastering the integral and its application in mechanics.
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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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