I have a parametric curve (elipse) defined as follows
$$\begin{aligned} x(t) &= \cos(t)\\ y(t) &= 2 \sin(t)\end{aligned}$$
and need to calculate the surface area of the ellipsoid produced by rotating this curve around the $x$-axis. I know the formula
$$S=2\pi \int_{\alpha}^{\beta}{[y(t)]\sqrt{x(t)'^2 +y(t)'^2}} \, \mathrm d t$$
I've managed to calculate the squares of derivatives:
$$x(t)'^2 = \sin^2(t)$$
$$y(t)'^2 = 4\cos^2(t)$$
Then, I've noticed that
$$\sqrt{\sin^2(t)+4\cos^2(t)}=\sqrt{1+3\cos^2(t)}$$
Therefore, I have to calculate definite integral
$$\int_{\alpha}^{\beta}{2\sin(t)\sqrt{1+3\cos^2(t)}} \mathrm d t$$
I have no idea how to solve this integral, and even what the boundaries should be in this case. I need help here.
Hint: Use the substitution $u=\sin{(t)}$. The bounds would be from $0$ to $\pi$.