Equality of tw0 arclengths

Question

We have the two curves: $$\alpha :[0, 2\pi] \to \mathbb{R}, \space \space \alpha(t) = (2sin(t), cos(t))$$ and $$\beta :[0, 2\pi] \to \mathbb{R}, \space \space \beta(t) = (sin(2t)cos(t), sin(2t)sin(t))$$
I want to show that both have the same arclength, I tried using the definition:
$\int_a^b |g'(x)|dx =$ arclength from a to b. However for the first one I get:
$$4\int_0^\frac{\pi}{2} \sqrt{4cos^2+sin^2}$$
And for the second one after simplifiyng:
$$4\int_0^\frac{\pi}{2} \sqrt{8sin^6-9sin^4+sin^2+1}$$
I might did a mistake there, but I don't see any.. What do I do now? Or is there an easier way to show the equality?

EDIT: I just got it guys. The second one is equal to:

$$4\int_0^\frac{\pi}{2} \sqrt{4cos^2(2t)+sin^2(2t)}$$
Therefore for $x = 2t$ we get the same arclength.

Answer 1

HINT: $$sin(2t)cos(t) = \frac{1}{2}(sin(3t)+sin(t))$$ $$sin(2t)sin(t) = \frac{1}{2}(cos(t)-cos(3t))$$

Answer 2

Parametric problems like this can be readily solved in the complex plane, where the arc length is given by

$$s=\int |\dot z|~dt$$

Thus we have

$$ z=2\sin t+i\cos t\\ \dot z=2\cos t-i\sin t\\ |\dot z|=\sqrt{4\cos^ t+\sin^2 t}=\sqrt{1+3\cos^2 t} $$

and

$$ w=\sin 2t~\cos t+i\sin 2t~\sin t=\sin2t~e^{it}\\ \dot w=(2\cos2t+i\sin 2t)e^{it}\\ |\dot w|=\sqrt{4\cos^ 2t+\sin^2 2t}=\sqrt{1+3\cos^2 2t} $$

Now,

$$ s_1=\int_0^{2\pi}\sqrt{1+3\cos^2 t}~dt\\ \begin{align} s_2 &=\int_0^{2\pi}\sqrt{1+3\cos^2 2t}~dt\\ &=\frac{1}{2}\int_0^{4\pi}\sqrt{1+3\cos^2 u}~du,\quad u=2t\\ &=\int_0^{2\pi}\sqrt{1+3\cos^2 u}~du \end{align} $$

where the last integral is a conseqeunce of the periodicity.

So both integrals are the the same and, in fact,

$$s=\int_0^{2\pi}\sqrt{1+3\cos^2 t}~dt=8\text{E}\left(\frac{3}{4} \right)\approx 9.68845$$

where $\text{E}$ is the complete elliptic integral of the second kind, with $k^2=3/4$.