What is the length of a sine wave from $0$ to $2\pi$? What is the length of a sine wave from $0$ to $2\pi$? Physically I would plot $$y=\sin(x),\quad 0\le x\le {2\pi}$$ and measure line length.
I think part of the answer is to integrate this:
$$
\int_0^{2\pi} \sqrt{ 1 + (\sin(x))^2} \, {\rm d}x
$$
Any ideas?

\begin{align}
\int_0^{2\pi}\sqrt{1+\cos^2(x)} dx  
&= 4 \int_0^{\pi/2}\sqrt{1+\cos^2(x)}dx \\
&= 4 \int_0^{\pi/2}\sqrt{1+\dfrac{1+\cos(2x)}2 }dx\\
&= 4 \sqrt{\dfrac{3}2} \int_0^{\pi/2} \sqrt{1+\dfrac{\cos(2x)}3} dx \\
&= 2\sqrt6 \int_0^{\pi/2} \sum_{n=0}^\infty a_n (\dfrac{\cos(2x)}3)^n dx \tag1
\end{align}
where $$\sqrt{1+t}=\sum_{n=0}^\infty a_n t^n, \mbox{ and }\ a_n = \frac{(-1)^{n+1} (2n-3)!!}{n! 2^n}$$
Let $I_n = \int_0^{\pi/2} \cos^n(2x) dx$, then $I_n = 0\ $ if $\ n\ $ is odd, and $ I_n = \dfrac{\pi}2 b_{n/2} $ if $\ n\ $ is even,
where $b_k = \dfrac{(2k-1)!!}{k! 2^k}$. 
The equation $(1) 
= 2\sqrt6 \sum\limits_{n=0}^\infty \dfrac{a_n}{3^n} I_n 
= 2\sqrt6 \sum\limits_{k=0}^\infty \dfrac{a_{2k}}{3^{2k}} \dfrac{\pi}2 b_k 
= \sqrt6 \pi \sum\limits_{k=0}^\infty c_k \tag2$ 
where $c_k = a_{2k} b_k 3^{-2k} 
= -\dfrac{(4k-3)!!}{(2k)! 2^{2k}} \dfrac{(2k-1)!!}{k! 2^k} 3^{-2k}
= -\dfrac{(4k-3)!!}{(k!)^2 2^{4k} 3^{2k}}
= -\dfrac{\binom{4k-3}{2k-1}\binom{2k-1}{k}}{2^{6k-2} 3^{2k} k} $. 
Note $c_0 = -(-3)!! = -\dfrac{1(-1)(-3)!!}{1(-1)} = 1$.
Additionally, \begin{align}
\int_0^{2\pi}\sqrt{1+\cos^2(x)} dx & = 4 \int_0^{\pi/2}\sqrt{1+\cos^2(x)}dx = 4 \int_0^{\pi/2}\sqrt{2-\sin^2(x)}dx\\
& = 4 \sqrt2 \int_0^{\pi/2} \sqrt{1-\dfrac{\sin^2(x)}2} dx = 4 \sqrt2 E\left(\dfrac1{\sqrt2}\right) \tag3
\end{align}
where $$E(k) = \displaystyle \int_0^{\pi/2} \sqrt{1-k^2 \sin^2(x)} dx
= \dfrac{\pi}2 \sum_{n=0}^{\infty} \left(\dfrac{\dbinom{2n}n}{4^n} \right)^2 \dfrac{k^{2n}}{1-2n}$$ and is referred to as the complete elliptic integral of second kind.
$ E\left(\dfrac1{\sqrt2}\right) 
= \dfrac{\pi}2 \sum\limits_{n=0}^{\infty} d_n \tag4$ 
where $d_n 
= \left[ \dfrac{(2n-1)!!}{n!2^n} \right]^2 \dfrac{1}{(1-2n)2^n}$.
By $(2)$ and $(3)$, $E\left(\dfrac1{\sqrt2}\right)
= \dfrac{\sqrt3 \pi}4 \sum\limits_{n=0}^{\infty} c_n \tag5$
$(4) = (5)$.
Because $c_n$ and $d_n \to 0$ as $n\to \infty$, the convergence rate of $c_n$ and $d_n$ are given by 
\begin{align} 
\lim_{n\to\infty} \left| \dfrac{c_n}{c_{n-1}} \right| 
&= \lim_{n\to\infty} \dfrac{(4n-3)(4n-5)}{n^2 2^4 3^2} = \dfrac 1 9\
 \mbox{ and }\\
\lim_{n\to\infty} \left| \dfrac{d_n}{d_{n-1}} \right| 
&= \lim_{n\to\infty} \dfrac{(2n-1)(2n-3)}{n^2 8} = \dfrac 1 2,\ 
 \mbox{ respectively. }
\end{align}
We obtain the values of $(4)$ and $(5)$ by the power series expansion of 
$\sqrt{1-\dfrac{\sin^2(x)}2}$ and $\sqrt{1+\dfrac{\cos(2x)}3}$, respectively. Hence using $(5)$, we can obtain accurate estimates faster.
