Evaluate $\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy}$

At $$\int_{0}^{1}{\sqrt{\frac{y}{1-y}}}dy,$$ you can use substitution $y=sin^2\theta$

By using beta function $$\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy}=B(3/2,1/2)=\frac{\Gamma(3/2)\Gamma(1/2)}{\Gamma(3/2+1/2)}=\frac{\Gamma^2(1/2)}{2}=\frac{\pi}{2}$$

Q.E.D.

Substitute $u=\sqrt{\dfrac{y}{1-y}}.$ Then $$\dfrac{y}{1-y}=\dfrac{y-1+1}{1-y}=-1+\dfrac{1}{1-y}=u^2, \\ ( {0}< {y} <{1} \Leftrightarrow {0}< {u}<{+\infty}),\\ \dfrac{1}{1-y}=1+u^2, \\ 1-y=\dfrac{1}{1+u^2}, \\ y=1-\dfrac{1}{1+u^2}, \\ dy=\dfrac{2u}{(1+u^2)^2}\, du, $$ so $$\int\limits_0^1{\sqrt{\dfrac{y}{1-y}} dy}=\int\limits_0^{+\infty}{\dfrac{2u^2}{(1+u^2)^2}}\, du=2\int\limits_0^{+\infty}{\dfrac{u^2}{(1+u^2)^2}}\, du$$ Last integral can be calculated by decomposition into partial fractions.