I am trying to find the arc length for the lemniscate $r^2 = a^2cos(2x)$ using the equation $$\int \sqrt{r^2 + (\frac{dr}{d\theta})^2} \, d\theta$$
However, I end up with the integration of $\sqrt{sec(x)}$ which is too difficult to integrate. Is there any different path that I can following using the same formula to find the length of the curve? Is there a way to make this easier?
Any suggestion would be appreciated.
If $r^2 = a^2 \cos 2x$ then \begin{align} 2\frac{d r}{d \theta}r &= -2a^2 \sin 2x \\ \implies \left(\frac{d r}{d \theta}\right)^2 r^2&=a^4 \sin^2 2x \\ \implies \left(\frac{d r}{d \theta}\right)^2 &= a^2 \frac{\sin^2 2x}{\cos 2x} \\ \implies r^2 + \left(\frac{d r}{d \theta}\right)^2 &= \ldots \end{align} once you get to this stage, noting that $$\cos 2\theta = \frac{r^2}{2a^2}$$ one can write $$(\frac{d\theta}{dr})^2 = \frac{r^2}{4a^4 - r^4}$$
from which you should find \begin{align}s &= \int_{0}^{r'}\sqrt{\frac{4a^4}{4a^4 - r^4}}dr \\ &= \int_{0}^{r'}\frac{2a^2}{\sqrt{4a^4 - r^4}}dr \end{align}
Can you take it from here?