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How to prove the following identity? For $0\le m<1$, $$\int_{0}^{\pi/2}\frac{dx}{\sqrt{(1-m \cos^2 x)(1+ m \sin^2x)}}=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-m^2 \sin^2 x}}$$ Then it will be an elliptic integral.

Firstly, the objects in the squre root are not equal, so it cann't be solved just by Trigonometric Identities.

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  • $\begingroup$ Trigonometric identities are not enough, and coordinate change is not enough, but perhaps both combined?... $\endgroup$
    – Martigan
    Nov 2, 2017 at 16:19
  • $\begingroup$ One approach is to expand the integrand as a power series in $m$ and integrate term by term and then compare the coefficients on both sides. But this is bit tricky to handle when dealing with the integrand in LHS. $\endgroup$
    – Paramanand Singh
    Nov 2, 2017 at 18:37
  • $\begingroup$ I updated my answer to get rid of any unnecessary details. You may have second look at it. $\endgroup$
    – Paramanand Singh
    Nov 2, 2017 at 21:59

1 Answer 1

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Let $$I(a, b) =\int_{0}^{\pi/2}\frac{dx}{\sqrt{a^{2}\cos^{2}x+b^{2}\sin^{2}x}}\tag{1}$$ then it is easy to observe that $$cI(ac, bc)=I(a, b) \tag{2}$$ Using the substitution $b\tan x=t$ in $(1)$ we see that $$I(a, b) =\int_{0}^{\infty}\frac{dt}{\sqrt{(t^{2}+a^{2})(t^{2}+b^{2})}}\tag{3}$$ The integral on the right hand side of the identity in question is a complete elliptic integral of first kind and is commonly denoted by $K(m)$. It is almost obvious from equation $(1)$ that $$K(m) =I(1, \sqrt{1-m^{2}})\tag{4}$$ Using substitution $\tan x=t$ we can see that the left hand side of the identity in question is equal to $$\int_{0}^{\infty}\frac{dt}{\sqrt{(t^{2}+1-m)(1+(1+m)t^{2})}}=\frac{I(1/\sqrt{1+m},\sqrt{1-m})}{\sqrt{1+m}}=I(1,\sqrt{1-m^{2}})$$ (via equations $(2)$ and $(3)$). The proof is now complete via equation $(4)$.

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  • $\begingroup$ (+1) Beautiful answer (as expected). I would add that OP's claim "cannot be solved by trigonometric substitution" is not true, strictly speaking, since $K(m)=\frac{\pi}{2\,\text{AGM}(1,\sqrt{1-m^2})}$ can be proved through Lagrange's identity and a suitable substitution. $\endgroup$ Nov 2, 2017 at 19:18
  • $\begingroup$ @JackD'Aurizio: I was in a hurry so I gave the link to my blog. The identity can be proved just via trigonometric substitution. I will update my answer to fix this a bit later. Now is sleep time here. $\endgroup$
    – Paramanand Singh
    Nov 2, 2017 at 19:50
  • $\begingroup$ @JackD'Aurizio: I could not resist the temptation to fix my answer as soon as possible for me. The updated answer just uses basic trigonometric substitutions and nothing more. $\endgroup$
    – Paramanand Singh
    Nov 2, 2017 at 22:03
  • $\begingroup$ Very nice, I would give you an extra (+1), but I cannot :) $\endgroup$ Nov 2, 2017 at 22:04

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