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Comments on Approximation of an elliptic integral

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Approximation of an elliptic integral

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I used some nice guesswork to get this formula
$$\int_0^{\frac{\pi}{2}}\sqrt{a^2\sin^2x+b^2\cos^2x}dx=\frac{ab\pi}{\left(a+b\right)\sin\left(\frac{a\pi}{a+b}\right)}$$
The comparision for some values of a and b are
$a=a,b=0, value=a, exact=a$
$a=1,b=5, value=5.236, exact= 5.2525$
$a=b, value=\frac{\pi.a}{2}, exact=\frac{\pi.a}{2} $
You can also check it gives consistent result. Am i right?

https://www.desmos.com/calculator/g0w1nkeotm

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1 comment thread

General comments (6 comments)
General comments
Derek Elkins‭ wrote almost 4 years ago

Are you right about what? That your expression "approximates" the integral? What criteria are you using to decide that something is an "approximation"? On what domain are you considering?

JRN‭ wrote almost 4 years ago

Your formula is incorrect. It uses the "equals" symbol when you yourself state that the equality doesn't hold when a=1, b=5. Perhaps you meant to use an "approximately equal to" symbol?

Skipping 1 deleted comment.

theabbie‭ wrote almost 4 years ago

@JoelReyesNoche‭ Putting b=0 makes it an indeterminate form, so, you need to put limits, on which, you get the exact values.

JRN‭ wrote almost 4 years ago

Thanks for the response. I've deleted my comment (which contained a mistake).

Adam‭ wrote over 3 years ago

It's equal to $\frac{a|b|\pi}{(|a|+|b|)\sin(a\pi/(|a|+|b|))}$. You can prove it with series expansions for $\sec$ and the elliptic integral.

Skipping 1 deleted comment.

Incnis Mrsi‭ wrote over 3 years ago

Did @Adam‭ intend “|ab|π” for the numerator?