The difference between two definite integrals related to elliptic arc length

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The difference between two definite integrals related to elliptic arc length

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Definite integral

elliptic integrals

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#1 h Sep 2, 2018, 2:03 AM • 3 Y

Y by RedFireTruck, Adventure10, Mango247

Proof that
$\int_0^{\frac{2}{3}\sqrt{6}}\sqrt{\frac{16-3x^2}{16-4x^2}}{\rm d}x-\int_{\frac{2}{3}\sqrt{6}}^2\sqrt{\frac{16-3x^2}{16-4x^2}}{\rm d}x=1$

This post has been edited 1 time. Last edited by methylbenzene, Sep 2, 2018, 2:05 AM
Reason: fix the title

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#2 h Sep 3, 2018, 9:57 AM • 3 Y

Y by methylbenzene, RedFireTruck, Adventure10

That's a nice problem. Let
$I:=\int_0^{\frac{2}{3}\sqrt{6}}\sqrt{\frac{16-3x^2}{16-4x^2}} \ dx, \ \ \ J:=\int_{\frac{2}{3}\sqrt{6}}^2\sqrt{\frac{16-3x^2}{16-4x^2}} \ dx.$ First let's make the intervals of both integrals the same so that we'll have one integral only.
In $I,$ put $x=\frac{2\sqrt{6}}{3}t$ to get
$I=2\int_0^1\sqrt{\frac{2-t^2}{3-2t^2}} \ dt.$ In $J,$ make the substitution $\frac{16-4x^2}{16-3x^2}=\frac{2}{3}t^2$ to get
$J=2\int_0^1 \frac{1}{(2-t^2)^2}\sqrt{\frac{2-t^2}{3-2t^2}} \ dt.$ Therefore
$I-J=\int_0^1\left(2-\frac{2}{(2-t^2)^2}\right)\sqrt{\frac{2-t^2}{3-2t^2}} \ dt. \ \ \ \ \ \ \ \ (*)$ But
$\left(2-\frac{2}{(2-t^2)^2}\right)\sqrt{\frac{2-t^2}{3-2t^2}}=\frac{6-8t^2+2t^4}{(2-t^2)^2}\sqrt{\frac{2-t^2}{3-2t^2}}$ $=\frac{(2-t^2)(3-2t^2)-t^2}{(2-t^2)^2}\sqrt{\frac{2-t^2}{3-2t^2}}$ $=\sqrt{\frac{3-2t^2}{2-t^2}}+t \cdot \frac{-t}{(2-t^2)^2}\sqrt{\frac{2-t^2}{3-2t^2}}=\frac{d}{dt}\left(t \sqrt{\frac{3-2t^2}{2-t^2}}\right)$ and so, by $(*),$
$I-J=\int_0^1\frac{d}{dt}\left(t \sqrt{\frac{3-2t^2}{2-t^2}}\right) dt =t \sqrt{\frac{3-2t^2}{2-t^2}} \Bigm|_0^1=1.$

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pi31415926535897

#3 h Sep 3, 2018, 7:00 PM • 3 Y

Y by RedFireTruck, Adventure10, Mango247

Is there a nice geometric proof?

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#5 h Sep 4, 2018, 12:06 AM • 2 Y

Y by RedFireTruck, Adventure10

Hint: Put $x=2\cos{a}$ in the first integral, $x=2\sin{a}$ in the second integral and the problem collapses

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#6 h Sep 4, 2018, 10:56 AM • 2 Y

Y by Adventure10, Mango247

Math1331Math wrote:

Hint: Put $x=2\cos{a}$ in the first integral, $x=2\sin{a}$ in the second integral and the problem collapses

Perhaps you are being a little optimistic.
After variable substitution in your Hint, the intervals of both integrals are still different.
From my perspective, the method of ysharifi really is a more skillful way.
But I wonder how to obtain the variable substitution in ysharifi's method.
$I:=\int_0^{\frac{2}{3}\sqrt{6}}\sqrt{\frac{16-3x^2}{16-4x^2}}{\rm d}x, \ \ \ J:=\int_{\frac{2}{3}\sqrt{6}}^2\sqrt{\frac{16-3x^2}{16-4x^2}}{\rm d}x.$ $\begin{align*} I:=\int_0^{\frac{2}{3}\sqrt{6}}\sqrt{\frac{16-3x^2}{16-4x^2}}{\rm d}x\quad &\overset{x\,=2\cos\theta}{\overline{\overline{\hspace{2cm}}}}\quad2\int_{\arccos\left(\frac{\sqrt{6}}{3}\right)}^{\frac{\pi}{2}}\sqrt{1-\frac{3}{4}\cos^2\theta}\,{\rm d}\theta\\ &\overset{x\,=2\sin\theta}{\overline{\overline{\hspace{2cm}}}}\quad2\int_{0}^{\arcsin\left(\frac{\sqrt{6}}{3}\right)}\sqrt{1-\frac{3}{4}\sin^2\theta}\,{\rm d}\theta \end{align*}$ $\begin{align*} J:=\int_{\frac{2}{3}\sqrt{6}}^{2}\sqrt{\frac{16-3x^2}{16-4x^2}}{\rm d}x\quad &\overset{x\,=2\cos\theta}{\overline{\overline{\hspace{2cm}}}}\quad2\int_{0}^{\arccos\left(\frac{\sqrt{6}}{3}\right)}\sqrt{1-\frac{3}{4}\cos^2\theta}\,{\rm d}\theta\\ &\overset{x\,=2\sin\theta}{\overline{\overline{\hspace{2cm}}}}\quad2\int_{\arcsin\left(\frac{\sqrt{6}}{3}\right)}^{\frac{\pi}{2}}\sqrt{1-\frac{3}{4}\sin^2\theta}\,{\rm d}\theta \end{align*}$

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#7 h Jan 11, 2021, 1:27 PM

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$\begin{align*} \int_0^{a\cdot\sqrt{\frac{a}{a+b}}}\sqrt{\frac{a^2-e^2x^2}{a^2-x^2}}{\mathrm{d}}x -\int_{a\cdot\sqrt{\frac{a}{a+b}}}^{a}\sqrt{\frac{a^2-e^2x^2}{a^2-x^2}}{\mathrm{d}}x&=a-b\\ \\ \int_0^{2\cdot\sqrt{\frac{2}{2+1}}}\sqrt{\frac{2^2-\big(\tfrac{\sqrt{3}}{2}\big)^2x^2}{2^2-x^2}}{\mathrm{d}}x -\int_{2\cdot\sqrt{\frac{2}{2+1}}}^{2}\sqrt{\frac{2^2-\big(\tfrac{\sqrt{3}}{2}\big)^2x^2}{2^2-x^2}}{\mathrm{d}}x&=2-1\\ \\ \int_0^{\frac{2}{3}\sqrt{6}}\sqrt{\frac{16-3x^2}{16-4x^2}}{\mathrm{d}}x-\int_{\frac{2}{3}\sqrt{6}}^2\sqrt{\frac{16-3x^2}{16-4x^2}}{\mathrm{d}}x&=1 \end{align*}$ Multiple addition theorem of elliptic integral:
$\begin{align*} \begin{split} \color{red}{\int_0^u\sqrt{\frac{1-k^2t^2}{1-t^2}}{\mathrm{d}}t=\int_u^w\sqrt{\frac{1-k^2t^2}{1-t^2}}{\mathrm{d}}t+k^2u^2w}\\ \\ \color{red}{w\,\,\colon\!=\frac{2u\sqrt{\big(1-u^2\big)\big(1-k^2u^2\big)}}{1-k^2u^4}\qquad\qquad} \end{split} \end{align*}$
$\begin{align*} \operatorname{arc}\left(BX_0\right)-\operatorname{arc}\left(X_0A\right)&=\int_0^{a\cdot\sqrt{\frac{a}{a+b}}}\sqrt{\frac{a^2-e^2x^2}{a^2-x^2}}{\mathrm{d}}x -\int_{a\cdot\sqrt{\frac{a}{a+b}}}^{a}\sqrt{\frac{a^2-e^2x^2}{a^2-x^2}}{\mathrm{d}}x\\ &=a\int_{0}^{\sqrt{\frac{a}{a+b}}}\sqrt{\frac{1-e^2t^2}{1-t^2}}{\mathrm{d}}t -a\int_{\sqrt{\frac{a}{a+b}}}^{1}\sqrt{\frac{1-e^2t^2}{1-t^2}}{\mathrm{d}}t\\ &=a\cdot e^2\cdot \left(\sqrt{\frac{a}{a+b}}\,\right)^2\cdot1=a\cdot\dfrac{a^2-b^2}{a^2}\cdot\dfrac{a}{a+b}\\ &=a-b \end{align*}$

This post has been edited 2 times. Last edited by methylbenzene, Jan 11, 2021, 1:36 PM

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