(possibly) unsolvable integral

Ok, so I was bored during AP Calculus after the test on integrals, and decided to entertain myself with a seemingly innocent math problem...

if you take a semi-circle represented by the equation

$\displaystyle \sqrt[2]{r^2-x^2}$

and divide the semi-circle into fourths, you end up with one-eight of a circle, with an area of

$\displaystyle \frac{\Pi*r^2}{8}$

I want to know when the integral of the equation, represented by

$\displaystyle \frac{x}{2}*\sqrt[2]{r^2-x^2}+\frac{r^2}{2}*\arcsin{\frac{x}{r}}$ from -R to X is equal to $\displaystyle \frac{\Pi*r^2}{8}$.

Basically, I need the following equation solved for X:

$\displaystyle \frac{\Pi*r^2}{8}=\frac{x}{2}*\sqrt[2]{r^2-x^2}+\frac{r^2}{2}*\arcsin{\frac{x}{r}}$

BTW, this problem will get me some SERIOUS extra credit points