alpha, Beta, Gammow.

One substitution and you're done!! Substitute $\large x=\sin^2\theta$ such that $\large\mathrm{d}x=\underbrace{\sin 2\theta}_{2\sin \theta\cos \theta}\mathrm{d}\theta$ and thus:

$\large\alpha=\displaystyle\int_0^{\pi/2}\dfrac{2\cancel{\sin \theta\cos \theta}\mathrm{d}\theta}{\cancel{\sin \theta \cos \theta} }$

$\large =2\left|\theta\right|_0^{\pi/2}=\pi\approx 3.14$

I completed the square at the bottom so that the expression under the root became (1/4 - (x-1/2)^2) this allowed me to sub in u= x-1/2. Then I got a nice standard integral resulting in Arcsin (2u).

Using Beta function we get $\alpha = B(1/2,1/2) = \frac{\Gamma(1/2) \cdot \Gamma(1/2)}{\Gamma(1)} = \pi \approx 3.14$

$\begin{aligned}\int_{0}^{1}\frac{1}{\sqrt{x}\sqrt{1-x}}\,dx &= \int_{0}^{1} \frac{2}{\sqrt{1-u^2}}\,du \quad\quad\quad\quad\quad{u = \sqrt{x}; \space du = \frac{1}{2u}dx} \\&= 2\int_{0}^{1}\frac{1}{\sqrt{1-u^2}} \quad\quad\quad\quad\quad\quad{\int \frac{1}{\sqrt{1-u^2}}\,dx = \sin^{-1}(u)+C} \\&= \left[2 \sin^{-1}(\sqrt{x}) \right]_{0}^{1} \\&= \pi \space \square \end{aligned}$

$\LARGE \text{ADIOS!!!}$