Solving Integration Problems for EIy and ymax

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Homework Statement



Two questions are of the same problem (these are taken from a textbook):

EIdy/dx = - (Fx^2)/4 + A

The constant A can be obtained from the boundary conditions: slope dy/dx = 0, x = 1/2L. Thus A = (FL^2)/16 hence

EIdy/dx = - (Fx^2)/4 + (FL^2)/16

My problem 1: I have absolutely no idea why A = (FL^2)/16. If someone could please show me in a few small steps how A was determined, I should be very grateful!

Integrating again (with respect to x), the second constant, B, equals zero thus

EIy = - (Fx^2)/4 + (FL^2)/16 + B

B = 0

y = (Fx/48EI) (3L^2 - 4x^2)

x = 1/2 L thus

ymax = (FL^3)/48EI

My problem 2: I have no idea how the solution for ymax was determined. I Also have no idea why y = (Fx/48EI) (3L^2 - 4x^2) when EIy = - (Fx^2)/4 + (FL^2)/16 + B. Is the equation for y simply the equation for EIy transposed for y? Because when I tried them I ended up with values which did not fit. (Did I make a simple error in my calculating, or is something else going on?)

Please could you explain to me what is going on in the above equations?
 
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This is a scan of the book from whence the equations originated:

http://img16.imageshack.us/img16/9725/beamproblembookscan.png
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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