Prove that these two curves have the same length My midterms are approaching, and I was going through some of our past Calculus midterms when I stumbled upon this question from 1996:

Show that these two curves, 
$$(\Gamma) : \frac {x^2}{4a^2} + \frac{y^2}{a^2} -1 = 0 $$  $$(\Omega): r=a\sin2\theta$$
have the same length.

Now, what I've done so far is find the corresponding parametric equations and use the following method to compute the lengths:
$$ \int_a^{b} \sqrt{x'(t)^2 +y'(t)^2}dt  $$
And the results are:
Length of $(\Gamma)=\int_0^{2\pi} |a|\sqrt{4\sin^2\theta + \cos^2\theta}d\theta$
Length of $(\Omega)=\int_0^{2\pi} 2|a| \sqrt{ \sin^6\theta + \cos^6\theta}d\theta $
But this is where I reached an impass. I tried to compute the integrals myself, to no avail. Then I tried to compute them using an online integrator, and it turns out that they have the same value, so I think I'm on the right track. Can anyone suggest a way to prove they are equal that doesn't involve a calculator at the end (as they were banned during exams back in 1996, and I don't think that many students had access to them anyway) or a whole different method to get the lengths?
I really appreciate any help you can provide.
 A: $S$ is term inside square root.
$S_1=1+3 \sin^2 x$
$S_2=\sin^4 x+\cos^4 x-\sin^2 x \cos^2 x$
$=1-3\sin^2 x \cos^2 x$
Bring that 2 inside
$=4-3\sin^2 2x$
Put $2x=t \implies 2dx=dt$
Limit will go to $4\pi$ but $dt/2$ shall counter it due to periodicity.
$=4-3\sin^2 t$
$=1+3\cos^2t$
Plotting both functions in a rough graph will lead to believe that they have the same area.
A: I do not know how much this could help $$4 \sin ^2(t)+\cos ^2(t)-4 \left(\sin ^6(t)+\cos ^6(t)\right)=-\frac{3}{2} (\cos (2 t)+\cos (4 t))$$
A: The length element for a polar curve $\theta\mapsto r(\theta)$ is $\mathrm{d}\ell=\sqrt{r(\theta)^2+r'(\theta)^2}\mathrm{d}\theta$. This fact is easy to prove as follows (and should be a known result): from
$$\mathrm{d}(r\vec e_r)=(\mathrm{d}r)\vec e_r+r\mathrm{d}\vec e_r=r'(\theta)\vec e_r\mathrm{d}\theta+r(\theta)\vec e_\theta\mathrm{d}\theta$$
we obtain
$$\mathrm{d}\ell=\bigl\lvert\mathrm{d}(r\vec e_r)\bigr\rvert=\sqrt{r(\theta)^2+r'(\theta)^2}\mathrm{d}\theta$$
since $(\vec e_r,\vec e_\theta)$ is an orthonormal frame.
Now the length of $(\Omega)$ is:
$$\mathscr{L}=\int_{(\Omega)}\mathrm{d}\ell=\int_0^{2\pi}\sqrt{r(\theta)^2+r'(\theta)^2}\,\mathrm{d}\theta=\int_0^{2\pi}\sqrt{a^2\sin^2(2\theta)+4a^2\cos^2(2\theta)}\,\mathrm{d}\theta$$.
Now a straightforward substitution ($t=2\theta$) yields:
$$\mathscr{L}=\int_0^{4\pi}\sqrt{a^2\sin^2(t)+4a^2\cos^2(t)}\,\frac{\mathrm{d}t}2$$
Because we're integrating a function that is periodic of period $2\pi$ we also have:
$$\mathscr{L}=2\int_0^{2\pi}\sqrt{a^2\sin^2(t)+4a^2\cos^2(t)}\,\frac{\mathrm{d}t}2.$$
Simplify the $2$'s and you obtain the length of the curve $(\Gamma)$.
