# Prove: $\int_{0}^{\pi/2}\frac{dx}{\sqrt{(1-m \cos^2 x)(1+ m \sin^2x)}}=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-m^2 \sin^2 x}}$ for $0\le m<1$

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.

• Trigonometric identities are not enough, and coordinate change is not enough, but perhaps both combined?... Nov 2, 2017 at 16:19
• 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. Nov 2, 2017 at 18:37
• I updated my answer to get rid of any unnecessary details. You may have second look at it. Nov 2, 2017 at 21:59

## 1 Answer

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 $(4)$.

• (+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. Nov 2, 2017 at 19:18
• @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. Nov 2, 2017 at 19:50
• @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. Nov 2, 2017 at 22:03
• Very nice, I would give you an extra (+1), but I cannot :) Nov 2, 2017 at 22:04