Numerically, the calculation of $x(t)$ is a table with two rows, $t$, and $x$, and it simultaneously calculates $t(x)$: calculating $t(x)$ from is simply a matter of reading the table inversely. Estimates on $dx/dt$ will allow you to estimate how large your table will need to be according to your accuracy needs.

Numerically, the calculation of $x(t)$ is a table with two rows, $t$, and $x$, and it simultaneously calculates $t(x)$: calculating $t(x)$ from is simply a matter of reading the table inversely. Estimates on $dx/dt$ will allow you to estimate how large your table will need to be according to your accuracy needs. The integrand you have written is wrong; you should use $t_0$ instead of $t$, like so:

$$E(t_0) = 2\pi x(t_0)\int_0^{x(t_0)}\frac{dx}{t(x)}$$

In $E(t_0)$, the numerator of the integrand is constant. You're merely integrating $\int^{x(t_0)}_0\frac{dx}{t(x)}$, which, from the table, is approximated by the sum $\displaystyle\frac{x(t_0)}{N}\sum_{k=1}^N \frac{1}{t(x(t_0)\frac{k}{N})}$. The value for $N$ will depend on your estimates of $dx/dt$, where in general, small values of $dx/dt$ will require larger $N$ and/or splitting the integration interval in many intervals. See the inverse function rule on Wikipedia.