Q&A

# Approximation of an elliptic integral

+1
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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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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? Derek Elkins‭ about 2 months 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? Joel Reyes Noche‭ about 1 month ago

@JoelReyesNoche‭ Putting b=0 makes it an indeterminate form, so, you need to put limits, on which, you get the exact values. theabbie‭ 29 days ago

Thanks for the response. I've deleted my comment (which contained a mistake). Joel Reyes Noche‭ 27 days 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. Adam‭ 21 days ago