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$\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{a^2 \cos^2 \theta+b^2 \sin^2 \theta}} d \theta$

$ = \int_0^{\frac{\pi}{2}} \frac{1}{a}\sec \theta \frac{1}{ \sqrt{1+(b/a)^2 \tan^2 \theta}} d \theta$

But i know $d(\tan \theta) = \sec^2 \theta$. But this is not in that form. Can you tell me how to proceed further?

Magneto
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    This integral leads to an elliptic one. – Dr. Sonnhard Graubner Aug 12 '18 at 07:58
  • @mr_e_man i need to find Arthmetic geometric mean of 2 sequences. But when i searched, it gave this mean interms of this integral. I donot knw how to calculate the value of integral. – Magneto Aug 12 '18 at 08:40
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    @Magneto: you got it the wrong way: since the AGM is a fast-convergent iteration (quadratic convergence), it provides an efficient way for the numerical evaluation of the complete elliptic integral of the first kind. – Jack D'Aurizio Aug 13 '18 at 05:36

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$$I(a, b) =\int_0^\frac{\pi}{2} \frac{1}{\sqrt{a^2 \sin^2 t + b^2 \cos^2 t}}dt$$ $$ = \int_0^\frac{\pi}{2} \frac{1}{\sqrt{b^2-(b^2-a^2)\sin^2 t}}dt=\frac1b \int_0^\frac{\pi}{2} \frac{1}{\sqrt{1-\left(1-\frac{a^2}{b^2}\right)\sin^2 t}}dt=\frac1b K\left(1-\frac{a^2}{b^2}\right)$$ Or this can be written as: $I(a, b)=\frac{\pi}{2M(a,b)},$ where $M(a,b)$ is the arithmetic-geometric mean of $a,b$ defined by $M(a,b):=\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n$ where $$a_0=a, \ b_0=b, \ a_{n+1}=\frac{a_n+b_n}{2}, \ b_{n+1}=\sqrt{a_nb_n}$$

Zacky
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  • I have already seen this in wiki. But can u tell me what is value of this M(a,b)? – Magneto Aug 12 '18 at 08:41
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    @Magneto -- An elliptic integral is not an elementary function. You just have to learn the properties of this new function $K(x)$, like you learned $\cos(x)$ or $e^x$. – mr_e_man Aug 12 '18 at 08:43
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    Im not sure what you mean by its value, you can use a calculator in order to aproximate it, where you can use this relation: $$M(a,b)=\frac{\pi}{2b}\frac1{K\left(1-\frac{a^2}{b^2}\right)}$$ As said above, you would need to learn new properties, see: https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_first_kind – Zacky Aug 12 '18 at 08:47
  • @Zacky How about $$ \int_0^\frac{\pi}{2} \frac{\cos(A \cos(t))}{a^2 \sin^2 t + b^2 \cos^2 t}dt?$$ – Dinesh Shankar Aug 16 '18 at 20:16
  • @Zacky Here is a discussion about this integral. – Dinesh Shankar Aug 19 '18 at 11:16