I have to find the arc length of a lemniscate with polar equation $r=2(\cos(2\theta))^{1/2}$.
So far I got like
$\sqrt{4\cos(2\theta)+\left(-2\frac{\sin(2\theta)}{\sqrt{\cos(2\theta)}}\right)}$.
I don't know how to proceed, I can't seem to figure out how to approach this.
Best Answer
Since $$ \frac{\mathrm{d}}{\mathrm{d}\theta}2(\cos(2\theta))^{1/2} =-2(\cos(2\theta))^{-1/2}\sin(2\theta) $$ we get $$ \begin{align} r^2+\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^2 &=4\cos(2\theta)+\frac{4\sin^2(2\theta)}{\cos(2\theta)}\\ &=\frac4{\cos(2\theta)} \end{align} $$ Doubling the length of one leaf, we get $$ \begin{align} 2\int_{-\pi/4}^{\pi/4}\frac{2\,\mathrm{d}\theta}{\sqrt{\cos(2\theta)}} &=4\sqrt2\operatorname{K}\left(\frac1{\sqrt2}\right)\\ &=\sqrt{\frac2\pi}\,\Gamma\left(\frac14\right)^2\\ &=\frac{4\pi}{\operatorname{AGM}\left(1,\sqrt2\right)}\\[6pt] &=10.488230217168479242 \end{align} $$ where $\operatorname{K}$ is the Complete Elliptic Integral of the First Kind and, as Jack D'Aurizio comments, $\operatorname{AGM}$ is the Arithmetic-Geometric Mean.
A Derivation of the Integral
I've stated several equivalent forms of the integral above. Here is the derivation of one of them. $$ \begin{align} 2\int_{-\pi/4}^{\pi/4}\frac{2\,\mathrm{d}\theta}{\sqrt{\cos(2\theta)}} &=4\int_0^{\pi/4}\frac{\mathrm{d}2\theta}{\sqrt{\cos(2\theta)}}\tag{1}\\ &=4\int_0^1\frac{\mathrm{d}u}{\sqrt{u\vphantom{u^2}}\sqrt{1-u^2}}\tag{2}\\ &=2\int_0^1v^{-3/4}(1-v)^{-1/2}\mathrm{d}v\tag{3}\\ &=2\,\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac34\right)}\tag{4}\\ &=2\frac{\sqrt\pi\,\Gamma\left(\frac14\right)^2}{\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)}\tag{5}\\ &=\sqrt{\frac2\pi}\,\Gamma\left(\frac14\right)^2\tag{6} \end{align} $$ Explanation:
$(1)$: exploit the symmetry of $\cos(x)$
$(2)$: substitute $u=\cos(2\theta)$
$(3)$: substitute $v=u^2$
$(4)$: use the Beta Function
$(5)$: $\Gamma\left(\frac12\right)=\sqrt\pi$
$(6)$: $\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)=\pi\csc\left(\frac\pi4\right)=\pi\sqrt2$ using Euler's Reflection Formula