Arc Lengths of Sine Curve I am dealing with the following problem:

In this sketch d and e are arc lengths.
Now I want to find the value of a for which:
$\dfrac{d}{2e}=0.05$
when the value of r is given.
I know that:
$a\sin(2\pi fb)=r$
$a\sin(2\pi fc)=r$
$b=\dfrac{1}{2f}-c$
$d=\int_b^c\sqrt{1+4\pi^2f^2a^2\cos^2(2\pi fx)}\mathrm{d}x$
$e=\int_0^\frac{1}{2f}\sqrt{1+4\pi^2f^2a^2\cos^2(2\pi fx)}\mathrm{d}x$
I have got Matlab installed on my computer, which can be helpful in this case I suppose...
 A: Arc length integrals can't be calculated in terms of elementary functions. Result will need to use elliptic integrals.
Straight up calculation with computer shows that equation $\frac{d}{2e} = 0.05$ can be reduced to a form:
$$ \frac{E(\pi-\arcsin(\frac{r}{a}),z)-E(\arcsin(\frac{r}{a}),z)}{4E(z)} = 0.05$$,
where $z = 1- \frac{1}{1+(2af\pi)^2}$, $E(\phi,m)$ is elliptic integral of second kind (EllipticE in Mathematica) and $E(m) = E(\frac{\pi}{2},m)$ is complete elliptic integral.
I wasn't able to reduce this further, because I don't have any prior experience using elliptic integrals. Maybe someone else can continue on this.
Solving this equation agrees with my numerical calculations. For example if $f=1$ and $r=1$, solving the equation gives $a \approx 1.08713$, which is a correct answer.
A: To avoid confusion, let me call your function $g(x)=a\sin(2\pi fx)$. Then using standard arc length formulas from calculus,
\begin{align}
d=&\int_b^c \sqrt{1+[g'(x)]^2}dx\\
e=&\int_{1\over 2f}^{1\over f} \sqrt{1+[g'(x)]^2}dx,
\end{align}
so you want to solve ${d\over 2e}=0.05$. 
Are you looking for a numerical answer? If so, we need to know the parameters $a$, $b$, $c$, and $f$.
