# 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

## 5 comments

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