WKB boundary conditions [duplicate]

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Technically, this is a math problem but I think it is better here than in the math stacks. Consider the differential equation

$$y'' = (x^4 - E)y$$

The boundary conditions are the usual $y(\pm\infty)=0$. I want to find the eigenvalues, $E$, using WKB. According to my textbook (Bender and Orszag), we have the guessed form

$$y=e^{S_0(x)}$$

where

$$S_0'(x) = \pm\sqrt{x^4-E}$$

One picks the sign such that the boundary conditions are obeyed.

Now, let us consider the "classical region" where $E>x^4$. After lots of work, the book reaches the eigenvalue condition $$\int_{-E^{1/4}}^{E^{1/4}} \sqrt{E-x^4}dx =\left(n+\frac{1}{2}\right)\pi$$

Notice, that in this condition, the $\pm$ disappeared and the sign under the square root flips. The way this condition was obtained was fairly complicated, patching together solutions at infinity with an Airy function in between the turning points but can someone explain, in a simple way, how the boundary conditions forbade

$$\int_{-E^{1/4}}^{E^{1/4}}-\sqrt{E-x^4}dx=\left(n+\frac{1}{2}\right)\pi~?$$