Calculate the Arc length of a Polar curve $r=1+\cos(2\theta)$ 
Given the polar equation, $r=1+\cos(2\theta)$,  prove that the length of the curve corresponding to $0 \leq\theta \leq2\pi$ is $8+\frac{4}{\sqrt{3}}\log(2+\sqrt{3})$.

Applying integration to find arc length, $l$,  in polar form:
$\int_\alpha^\beta\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta$
$r^2=(1+\cos(2\theta))^2=1+2\cos(2\theta)+\cos^2(2\theta)$,
and, $\frac{dr}{d\theta}=-2\sin(2\theta)\Rightarrow (\frac{dr}{d\theta})^2=4\sin^2(2\theta)=4-4\cos^2(2\theta)$
$\therefore r^2+(\frac{dr}{d\theta})^2=5+2\cos(2\theta)-3\cos^2(2\theta)$
Completing the square, $5+2\cos(2\theta)-3\cos^2(2\theta)=3\left[\frac{16}{9}-(\cos(2\theta)-\frac{1}{3})^2 \right]$
Just doing the integration using (1), but without limits to get a function only, I get:
$\int\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta=\int\sqrt{3\left[\left(\frac{4}{3}\right)^2-(\cos(2\theta)-\frac{1}{3})^2 \right]}d\theta=\sqrt{3}\int\sqrt{\left[\left(\frac{4}{3}\right)^2-(\cos(2\theta)-\frac{1}{3})^2 \right]}d\theta$
The integral has the general form: $\int\sqrt{a^2-x^2}dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\arcsin\frac{x}{a}$
Ignoring factor of $\sqrt{3}$ the integral is:
$\int\sqrt{\left[\left(\frac{4}{3}\right)^2-(\cos(2\theta)-\frac{1}{3})^2 \right]}d\theta=\frac{((\cos(2\theta)-\frac{1}{3})\sqrt{(\frac{4}{3})^2-(\cos(2\theta)-\frac{1}{3})^2}}{2}+ \frac{\left(\frac{4}{3}\right)^2}{2}\arcsin\frac{(\cos(2\theta)-\frac{1}{3})}{\frac{4}{3}}$
No log term in the numerical solution.
Another idea I had was to try using $\sqrt{1+\cos(t)}=\frac{\sin(t)}{\sqrt{1-\cos(t)}}$, but this leads to a substitution which doesn't give the log term?
I'm a bit stuck with this one. Most of the other examples I've read, e.g. for $r=1+\cos(\theta)$ seem to be OK. It's the $\cos(2\theta)$ causing the problem for me.

$r = 1 + \cos2\theta = 2 \cos^2\theta$
$r_{\theta} = - 4 \cos\theta \sin\theta$
$ \displaystyle l = 4 \int_0^{\pi/2}\sqrt{4\cos^4\theta + 16 \cos^2\theta \sin^2\theta} ~ d\theta$
$ \displaystyle l = 8 \int_0^{\pi/2} \cos\theta \sqrt{1 + 3 \sin^2\theta} ~ d\theta$
Now substitute $t = \sin\theta$. Can you take it from here?
