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?
 A: It is given by $$I = \int_{0}^{2 \pi} \sqrt{ 1 + (\cos{x})^{2}} \ \rm{dx}$$ and I think this is an elliptic integral of the second kind. (That's what Wolfram says.)
A: The length of A sin(x) from 0 to 2$\pi$ is
$$4 \sqrt{A^2+1}  E\left(\frac{A^2}{A^2+1}\right)$$
Where $E(m)$ is the elliptic integral of the second kind.
So if your corrugated sheet is 10cm thick and has 20cm between peaks $A = \frac{10/2}{20/2\pi} = \pi/2$ so the length is $$\frac{20\text{ cm}}{2\pi} \times 4 \sqrt{\pi^2/4+1}  E\left(\frac{\pi^2/4}{\pi^2/4+1}\right) = 29.3\text{ cm}$$
A: The given integral equals
$$ 4\int_{0}^{\pi/2}\sqrt{1+\sin^2 x}\,dx = 4\int_{0}^{1}\frac{\sqrt{1+x^2}}{\sqrt{1-x^2}}\,dx = 4\int_{0}^{1}\frac{1+x^2}{\sqrt{1-x^4}}\,dx \\= \int_{0}^{1}\left(x^{-3/4}+x^{-1/4}\right)(1-x)^{-1/2}\,dx = B\left(\tfrac{1}{4},\tfrac{1}{2}\right)+B\left(\tfrac{3}{4},\tfrac{1}{2}\right)\tag{A}$$
where $B$ is Euler's Beta function. In terms of the $\Gamma$ function this length equals
$$ L=\frac{1}{\sqrt{2\pi}}\,\Gamma\left(\tfrac{1}{4}\right)^2 +4\pi\sqrt{2\pi}\,\Gamma\left(\tfrac{1}{4}\right)^{-2}.\tag{B}$$
On the other hand $\Gamma\left(\tfrac{1}{4}\right)$, the value of the complete elliptic integral of the first kind $K(m)$ at $m=\frac{1}{2}$ and the lemniscate constant are all related (see some special values for the $\Gamma$ function). Additionally, the AGM mean provides a very efficient numerical technique for the evaluation of a complete elliptic integral of the first kind. We may write $(B)$ as
$$ L = \frac{2\pi}{\text{AGM}(1,\sqrt{2})}+2\,\text{AGM}\left(1,\sqrt{2}\right) \tag{C}$$
hence this is an efficient algorithm for the numerical evaluation of the wanted length:


*

*Initialize $a\leftarrow 1$, $b\leftarrow\sqrt{2}$

*Repeat $a\leftarrow\frac{a+b}{2}$, $b\leftarrow\sqrt{ab}$ until $a-b$ is smaller than the wanted accuracy

*Return $2\sqrt{ab}+\frac{2\pi}{\sqrt{ab}}$.


We get $L\approx 7.640395578055424$ with very few steps.
A: \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.
A: On MATLAB:
$$t  = 0:0.001:(2\pi);$$
$$st = \sin(t);$$
$$\text{sum( sqrt( diff(st).^2 + diff(t).^2 ) )}$$
$$\text{ans} = 7.6401$$
You will need $21.6$% more paint to paint the corrugated roof ;).
A: I'm nowhere near a computer with elliptic integrals handy, so I'll give the explicit evaluation of
$$\int_0^{2 \pi} \sqrt{1+\cos^2 x}\,\mathrm dx$$
Note that an entire sine wave can be cut up into four congruent arcs; we can thus consider instead the integral
$$4\int_0^{\pi/2} \sqrt{1+\cos^2 x}\,\mathrm dx$$
(alternatively, one can split the integral into four "chunks" and find that those four chunks can be made identical; I'll leave that manipulation to somebody else.)
Now, after some Pythagorean manipulation:
$$4\int_0^{\pi/2} \sqrt{1+\cos^2 x}\,\mathrm dx=4\int_0^{\pi/2} \sqrt{2-\sin^2 x}\,\mathrm dx$$
and then a bit of algebraic massage:
$$4\sqrt{2}\int_0^{\pi/2} \sqrt{1-\frac12\sin^2 x}\,\mathrm dx$$
we then recognize the complete elliptic integral of the second kind $E(m)$
$$E(m):=\int_0^{\pi/2}\sqrt{1-m\sin^2u}\mathrm du$$
(where $m$ is a parameter):
$$4\sqrt{2}E\left(\frac12\right)$$
As Robert notes in a comment, different computing environments have different argument conventions for elliptic integrals; Maple for instance uses the modulus $k$ (thus, $E(k)$) instead of the parameter $m$ as input (as used by Mathematica and MATLAB), but these conventions are easy to translate to and from: $m=k^2$. So, using the modulus, the answer is then $4\sqrt{2}E\left(\frac1{\sqrt 2}\right)$.

Now to address the noted equivalence for negative parameter and a parameter in the interval $(0,1)$ by Henry, there is what's called the "imaginary modulus transformations"; the DLMF link gives the transformation for the incomplete case, but I'll explicitly do the complete case here for reference since it's not too gnarly to do (all you have to remember are the symmetries of the trigonometric functions):
Letting $E(-1)=\int_0^{\pi/2}\sqrt{1+\sin^2 u}\,\mathrm du$, we then go this way:
$$\int_0^{\pi/2}\sqrt{1+\sin^2 u}\,\mathrm du=\int_{-\pi/2}^0\sqrt{1+\sin^2 u}\,\mathrm du$$
$$=\int_0^{\pi/2}\sqrt{1+\sin^2\left(u-\frac{\pi}{2}\right)}\,\mathrm du=\int_0^{\pi/2} \sqrt{1+\cos^2 u}\,\mathrm du$$
from which I've shown what you're supposed to do earlier.

Computationally, the complete elliptic integral of the second kind isn't too difficult to evaluate, thanks to the arithmetic-geometric mean. Usually, this method is used for computing the complete elliptic integral of the first kind, but the iteration is easily hijacked to compute the integral of the second kind as well.
Here's some C(-ish) code for computing $E(m)$:
#include <math.h>

double ellipec(double m)
{
    double f, pi2, s, v, w;

    if (m == 1.0)
        return 1.0;

    pi2 = 2.0 * atan(1.0);

    v = 0.5 * (1.0 + sqrt(1 - m));
    w = 0.25 * m / v;
    s = v * v;
    f = 1.0;

    do {
        v = 0.5 * (v + sqrt((v - w) * (v + w)));
        w = 0.25 * w * w / v;
        f *= 2.0;
        s -= f * w * w;
    } while (abs(v) + abs(w) != abs(v))

    return pi2 * s / v;
}

(make sure either your compiler does not (aggressively) optimize out the while (abs(v) + abs(w) != abs(v)) portion, or you'll have to use a termination criterion of the form abs(w) < tinynumber.)

Finally,
"I am also puzzled: a circle's circumference is $2\pi r$ and yet an ellipse's is an infinite series - why?"
My belief is that we are actually very lucky that the arclength function for a circle is remarkably simple compared to most other curves, the symmetry of the circle (and thus also the symmetry properties of the trigonometric functions that can parametrize it) being one factor. The reduction in symmetry in going from a circle to an ellipse means that you will have to compensate for those "perturbations", and that's where the series comes in...
A: The arc length of the graph of a function $f$ between $x=a$ and $x=b$ is given by $ \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx$. So, if you're considering $f(x)=\sin(x)$ then the correct integral is  $\int_{0}^{2\pi} \sqrt { 1 + [\cos(x)]^2 }\, dx$. Unfortunately, this integral cannot be expressed in elementary terms. This is quite common for arc-length integrals. However, the definite integral might be expressible in elementary terms; Wolfram Alpha says it cannot.
A: An improved analytical approximation to the one I gave before (user375743) is:
$$
l(x) = \frac{121}{100}x + \frac{1}{10} \mathrm{sin(2x)}
$$ 
It has a maximum error of less than 1.5% and an error of 0.5% for the range 0 to $2\pi$.
A: Dr Math did give a much better response than anyone else in this forum did, but no one seemed able to understand Dr Math and Dr Math is no longer available for anything like this anymore, so here is a different answer taken from an elementary Calculus I book...
A 4.2" Corrugated Metal Panel has a period of 10.67 cm (distance from the top of one crest to the next one) and an amplitude of 1.35 cm (height from the mid-point of the wave to the top of a crest). Modeling the sinewave gives us:
y = 1.35 sin 0.589x
...where 1.35 is the amplitude and
the period is 0.589 (or 2π/10.67)
The arc length of the curve is:
$\displaystyle \int_{a}^{b}$$\sqrt{1+\left(\frac{dy}{dx}\right)^2}$dx
Simple calculus tells us if $\frac{dy}{dx}$ = $0.795 * cos(0.589x)$
and since $a=0$ and $b=10.67$
$\displaystyle \int_{0}^{10.67}$$\sqrt{1+\left(0.795 * cos(0.589x)\right)^2}$dx
Then the answer should be 12.196 cm per wavelength.
That wasn't so hard now, was it?
