How to calculate the length of this curve Evaluate $L=\displaystyle \int\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx$ where, $y=\dfrac{h}{2}\left[1-\cos{2\pi}\dfrac{x}{l}\right]$.
 A: Hint:
$$y=\frac{\pi h}{l} \sin\left(\frac{2 \pi x}{l}\right),$$
and
$$L=\frac{l}{2 \pi} E\left(\frac{2 \pi x}{l} | - \frac{h^2 \pi^2}{l^2}\right),$$
where $E(x|m)$ is the incomplete elliptic integral of the second kind.
A: Picking the simplest possible case (i.e., $ h = 2, \ell = 2\pi$), so the integral becomes 
$$
\int \sqrt{1  + \sin^2(x)} ~ dx
$$
yields an elliptic integral of the second kind. Those are tabulated, of course, but the main message is "You're not going to get a nice answer for this." 
On the other hand, you have Excel, right? Well...just compute a numerical approximation to the integral by subdividing into, say, 200 points. That'll give you enough accuracy that you'll know the length within a tiny multiple of $\ell$, and hence will have a pretty good cost estimate unless your spring is going to be miles long. 
A: As @S. Maths answered, you have
$$L=\frac{l }{2 \pi }E\left(\frac{2 \pi  x}{l}|-\frac{h^2 \pi ^2}{l^2}\right)$$ For a better legibility, let $k=-\frac{h^2 \pi ^2}{l^2}$ and $X=\frac{2 \pi  x}{l}$.
If $X$ is small, you could use, for an approximation, the series expansion
$$L=\frac{l }{2 \pi }\left(X-\frac{5 X^3}{3}-\frac{13 X^5}{6}-\frac{979 X^7}{126}-\frac{166627 X^9}{4536}+O\left(X^{11}\right)   \right)$$
