How to integrate $ x = \int \frac{ky^2}{\sqrt{1-k^2y^4}} dy $?

Question

The problem is to make the following integral stationary: $$ \int_{x_1}^{x_2} \frac{\sqrt{1+y'^2}}{y^2}dx $$ to simplify the Euler equation, I tried to change the independent variable: $$ \int_{y_1}^{y_2} \frac{\sqrt{1+x'^2}}{y^2}dy, \: \: \: \: \: \: \: \: y=y\left( x \right),\: y'=\frac{dy}{dx} $$ with the correspondent Euler equation: $$ \frac{d}{dy}\frac{\partial F}{\partial x'}-\frac{\partial F}{\partial x}=0 $$ thus $$ \begin{aligned} \frac{\partial F}{\partial x}=0 \Rightarrow \frac{\partial F}{\partial x'} &= k\\ \frac{x'}{y^2 \sqrt{1+x^2}} &= k\\ \frac{dx}{dy} = x' &= \frac{ky^2}{\sqrt{1-k^2y^4}} \end{aligned} $$ and I get: $$ x = \int \frac{ky^2}{\sqrt{1-k^2y^4}} dy $$ Now, can I change $ ky^2 $ to a new single arbitrary variabel to simplify the integrand? Or are there a more effective method?

Answer 1

Do it, substitute $u=ky^2,\frac{du}{dy}=2\sqrt{ku}$ (assuming $k>0$): $$x=\frac1{2\sqrt k}\int\sqrt{\frac u{1-u^2}}\,du$$ This is an elliptic integral. Say we're integrating from $0$ to $u$, then Byrd and Friedman 235.06 gives $$x=\sqrt{\frac2k}\left(E(\varphi,m)-F(\varphi,m)/2-\sqrt{\frac{u(1-u)}{2(1+u)}}\right)+C$$ where $\sin^2\varphi=\frac{2u}{1+u}$ and $m=1/2$. You will have to use numerical methods to get $y$ in terms of $x$.

Answer 2

As said in comments and answer, using $\frac{u}{\sqrt{k}}$ you will end with $$\int \frac{ky^2}{\sqrt{1-k^2y^4}}\, dy=\frac{E\left(\left.\sin ^{-1}(u)\right|-1\right)-F\left(\left.\sin ^{-1}(u)\right|-1\right)}{\sqrt{k}}$$

If you need to solve for $u$ the equation $$f(u)=E\left(\left.\sin ^{-1}(u)\right|-1\right)-F\left(\left.\sin ^{-1}(u)\right|-1\right)=a$$ you could use the series expansion $$f(u)=\sum_{n=0}^\infty b_n \,u^{4n+3}$$ where the $b_n$'s form the sequence $$\left\{\frac{1}{3},\frac{1}{14},\frac{3}{88},\frac{1}{48},\frac{35}{2432},\frac{63}{5 888},\frac{77}{9216},\frac{429}{63488},\frac{1287}{229376},\frac{935}{196608},\cdots\right\}$$ Using the above coefficients, the fit is quite good : the absolute error is

• $<0.1$% if $u \lt 0.921$
• $<0.01$% if $u \lt 0.874$
• $<0.001$% if $u \lt 0.829$

Using series reversion $$u=t \left(1-\sum_{n=1}^\infty c_n \,t^{4n} \right)\qquad \text{where} \qquad t=\sqrt[3]{3a}$$ the first $c_n$'s forming the sequence $$\left\{\frac{1}{14},\frac{15}{4312},\frac{61}{181104},\frac{85325}{2119641216},\frac{ 1214019}{227508157184}\right\}$$

For a test example, using $a=0.5$ the above would give as estimate $u=0.99010375$ while the "exact" solution is $u=0.99010360$