Approximate, within an error of $0.001$ units, the circumference of the ellipse given by the equation using integration: $$\left(\frac{x}{3}\right)^2+\left(\frac{y}{2}\right)^2=1$$
I have read about solving this problem in several different places and cannot wrap my mind around them. If someone could help, then I would greatly appreciate it.
In your case, the circumference has 4 symmetric pieces, so we will measure the curve length in the first quadrant and multiply by 4, so your final curve length will be $$ C = 4 \int_0^3 \sqrt{1 + \left[f'(x)\right]^2} dx $$ and $f(x) = 2 \sqrt{1 - (x/3)^2}$.
Can you take it from here?
The perimeter of a function $y=f(x)$ from $x_1$ to $x_2$ is given by $$P=\int_{x_1}^{x_2} \sqrt{1+\left(\frac{dy}{dx}\right)^2} \mathrm dx$$
For your question, $x_1=-3$ and $x_2=3$.
(Note that, if you integrate within given limits, the result will be semi perimeter since only upper half of the ellipse contributes to form 'function'.)
Can you proceed now?
This can be solved readily in the complex plane as follows. The arc length is given by
$$s=\int |\dot z|~du,\quad z=z(u)$$
For a generic ellipse we then have
$$ \begin{align} &z=a\cos\theta+ib\sin\theta,\quad \theta\in[0,2\pi]\\ &\dot z=-a\sin\theta+ib\cos\theta\\ &|\dot z|=\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}=b\sqrt{\left(\frac{a^2-b^2}{b^2}\right)\sin^2\theta+1}\\ \end{align} $$
We then find
$$ \begin{align} s &=\int_0^{2\pi}b\sqrt{A\sin^2\theta+1}~d\theta,\quad A=\left(\frac{a^2-b^2}{b^2}\right)\\ \\ &=4b\text{E}(-A)=4b\frac{\pi}{2}~ _2\text{F}_1\left(-\frac 12,\frac 12;1;-A \right) \end{align} $$
where $\text{E}$ is the complete elliptic integral of the second kind, expressed as $\text{E}(k^2)$ and $_2\text{F}_1$ is the Gauss hypergeometric funciton. For $a=3,\ b=2$ we find that
$$s\approx15.8654$$
