Arc Length of an Ellipse An ellipse has parametric representation $x = a\cos t$, $y = b \sin t$ for $0 ≤ t ≤ 2\pi$. Can you
write a formula for its total length? Do not waste your time trying to calculate it.
The way I was thinking to approach is basically gives me the formula for any ellipse
$$F(a,b)=\int_0^{2\pi}\!\sqrt{a^2\sin^2(t)+b^2\cos^2(t)}\,dt$$
but I don't think this is the right approach 
 A: If you read carefully the wikipedia page, you'll find an expression for the complete elliptic integral of the second kind that reads :
$$E(k) = \int_0^{\pi/2} \! \sqrt{1-k^2 \sin^2(t)} \, \mathrm{d}t$$
First remark: limits of integration.
You can slice your integral from $0$ to $2\pi$ into four parts that will describe the same arc of ellipse. Your are then left with the following integral :
$$F(a,b) = 4 \int_0^{\pi/2} \! \sqrt{a^2 \sin(t) + b^2 \cos(t)} \, \mathrm{d}t $$
Second remark : parametrization and relation between $a$ and $b$.
You must first check whether $a \ge b$ or $b < a$. Your parametrization of the ellipse will depend on it, especially if you want to fall back on the same form of $E(k)$ as the one given above.
In the general case, we suppose $a$ to be the semi-major axis, therefore $a \ge b$. In that case, your parametrization won't give you the expected result right away, you'll have to use a change of variable $s = \pi/2-t$. This amounts to the same as using the following parametrization:
$$
\begin{cases}
x = a \sin(t)\\
y = b \cos(t)
\end{cases}
$$ 
Then you'll get (using $\cos^2(x)=1-\sin^2(x)$) :
$$
F(a,b) = 4a \, E(k)
$$
with $k^2 = \dfrac{a^2-b^2}{a^2}=e^2$ with $e$, the eccentricity. The values of $E(k)$ are computed numerically.
