How to solve $t$ in $x = t - \sin{(t)}$ The question pretty much says it.
I need to solve $t$ in this equation:
$$ x = t - \sin{(t)} $$
Either I've forgotten how to do it, or I am just blind, etc. Anyway, I'm completely stuck at this.
Actually, I need to solve a vector:
$$(\; x \; , \; y \;) = (\; t - \sin{(t)} \; , \; 1 - \cos{(t)} \;)$$
Inverse of $y$ is trivial: $t = \cos^{-1}{(1 - y)}$. But that doesn't help me much further on.
 A: The length of the curve is the following: 
$$
\int_0^{2\pi}\sqrt{(x'(t))^2+(y'(t))^2}dt \\ 
=\int_0^{2\pi} \sqrt{(1-\cos t)^2 + (\sin t)^2}dt \\
=\int_0^{2\pi}\sqrt{1-2\cos t + \cos^2 t + \sin^2 t}dt \\
=\int_0^{2\pi}\sqrt{2-2\cos t}dt \\
=\int_0^{2\pi}\sqrt{2}\sqrt{1-\cos t}dt. 
$$
Now recall one of the half-angle formulas: $\sin^2 u = \dfrac{1}{2}-\dfrac{1}{2}\cos (2u)$. Plug in $t = 2u$ to obtain 
$$
\sin^2 \left(\frac{t}{2}\right) = \dfrac{1}{2}-\dfrac{1}{2}\cos (t), 
$$
which is the same as 
$$
2 \sin^2 \left(\frac{t}{2}\right) =  1- \cos (t). 
$$
Returning back to our integral and making appropriate substitutions, we obtain 
$$
\int_0^{2\pi}\sqrt{2}\sqrt{2 \sin^2\left(\frac{t}{2}\right)}dt \\
= \int_0^{2\pi} 2 \sin \left( \dfrac{t}{2}\right) dt\\
= 2\int_0^{2\pi}\sin\left( \frac{t}{2}\right) dt. \\
$$
Finally, we finish by making a substitution: let $v = t/2$.  Then $dv = dt/2$. This is called a $u$-substitution but in order to avoid confusion, I'm using the letter $v$ instead. 
Thus, we conclude 
$$
2 \int_0^{\pi}\sin v (2dv) \\
= 4 \int_0^{\pi} \sin v dv \\ 
= -4 \cos v |_0^{\pi} \\ 
 = -4(-1-1) = -4 (-2) = 8. 
$$
A: To solve for t you would read this http://en.wikipedia.org/wiki/Kepler_equation. It is related to the Kepler equation. It can be done as a series, which may converge fast depending on the values.
