Evaluate $\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy}$ I'm a little rusty with my integrals, how may I evaluate the following:
$$
\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy}
$$
I've tried:
$$
\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy} = \int_0^1{\sqrt{\frac{y}{1-y}} dy} 
$$
Make the substitution z = 1-y
$$
= \int_0^1{\sqrt{\frac{1-z}{z}} dz} = \int_0^1{\sqrt{\frac{1}{z} - 1} dz}
$$
Which seems simpler but I have not managed to evaluate it. All help appreciated (just a few pointers would be appreciated just as much as the full solution).
Thank you
EDIT:
Solution following Ng Hong Wai suggestion:
At 
$$
\int_0^1{\sqrt{\frac{y}{1-y}} dy} 
$$
We make the substitution $y = sin^2(\theta)$, so $dy = 2sin(\theta)cos(\theta)d\theta$. This results in:
$$
 \int_0^\frac{\pi}{2}{2 sin^2(\theta)d\theta} = \int_0^\frac{\pi}{2}{[1-cos(2\theta)]d\theta} = [x-\frac{1}{2}sin(2\theta)]_0^\frac{\pi}{2} = \frac{\pi}{2}
$$
Solution according to M. Strochyk's suggestion:
$$
\int_0^1{\sqrt{\frac{y}{1-y}} dy} = 2 \int_0^\infty{\frac{u^2}{(1+u^2)^2}du}
$$
By decomposing into partial fractions this becomes:
$$
2 \int_0^\infty{\frac{1}{1+u^2}du} - 2 \int_0^\infty{\frac{1}{(1+u^2)^2}du}
$$
The first term integrates directly to $arctan(u)$ while the second can be evaluated by making the substitution $u=tan(x)$ and then applying the double-angle formula for cosine. The result is (unsurprisingly):
$$
 2[arctan[u]]_0^\infty -[x + \frac{1}{2}sin(2x)]_0^\frac{\pi}{2} = \frac{\pi}{2}
$$

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.
