Taking Residues of Infinity of Square Roots. I am looking for worked out exercises of real valued integrals with square roots where ideas of residues at infinity are used. 
I was hoping that from $$\int_0^1 \sqrt{x} \thinspace dx $$ I could figure out a more complicated integral specifically $$ \int_0^1 \sqrt{x} \sqrt{1-x} \thinspace dx$$ on the basis of real valued integrals using ideas of residues of infinity. 
I believe $$\int_0^1 \sqrt{x} dx = \int_0^1 x^{\frac{1}{2}} dx = \int_0^1 \frac{1}{w^{\frac{1}{2}}} dw.$$ I am not sure what happens to the $dw$ but I believe it becomes $$\frac{1}{w}$$ so that I get something like $$\int_0^1 \frac{1}{w^{\frac{1}{2}}} \frac{1}{w}.....$$ or something.. but I can't quite piece it together since I haven't seen such examples worked out in any of the textbooks that I've read through and online there is a rarity of solutions regarding the real valued integrals.
I'm very confused and am merely looking for some direction in solving residues at infinity and in applying the concept to other such problems.
When I try to check my answer using Residues at infinity, I do not get the answer that I should. For $$ \int_0^1 \sqrt{x} \sqrt{1-x} \thinspace dx$$ the answer should be $\frac{\pi}{8}$ which is what I get using calculus and u substition specifically $t= \text{sin} u$ . But the correct answer doesn't match for when I use Residue's at Infinity. 
So this is why I ask a more realistic question which is $$\int_0^1 \sqrt{x} \thinspace dx .$$ Using Residue's at Infinity.
 A: For $\int_0^1 \sqrt{x}\sqrt{1-x}\; dx$, you consider the function
$$ f(z) = z \sqrt{1-1/z}$$
which (using the principal branch of the square root) has a branch cut on the real interval $[0,1]$.  For $0 < x < 1$, $\lim_{y \to 0+} f(x+iy) = i \sqrt{x (1-x)}$ and $\lim_{y \to 0-} f(x+iy) = - i \sqrt{x(1-x)}$.  Thus we should have
$$ \int_0^1 \sqrt{x} \sqrt{1-x}\; dx = -\frac{1}{2i} \oint_C f(z)\; dz$$
where $C$ is a simple positively oriented closed contour with $[0,1]$ inside it.  Now $$ \oint_C f(z)\; dz = - 2 \pi i \; \text{res}(f; \infty)$$
and since the residue of $f(z)$ at $\infty$ is $1/8$, we find 
$$ \int_0^1 \sqrt{x} \sqrt{1-x}\; dx = \frac{\pi}{8}$$
A: We can define $\sqrt{z(1-z)}$ unambiguously on $\mathbb{C}\setminus[0,1]$ by defining
$$
\log\left(\sqrt{z(1-z)}\right)=\frac12\int\left(\frac1z+\frac1{z-1}\right)\mathrm{d}z\tag1
$$
If we circle the singularities at both $0$ and $1$, the sum of the residues inside the contour is $2\pi i$, and so $\log\left(\sqrt{z(1-z)}\right)$ increases by $2\pi i$; and therefore, $\sqrt{z(1-z)}$ returns to the same value.
Next, define $\sqrt{z(1-z)}$ to be the positive real value on the upper side of $[0,1]$. Near $z=\infty$, we have
$$
\begin{align}
\sqrt{z(1-z)}
&=-iz\sqrt{1-\tfrac1z}\\
&=-iz\left(1-\tfrac1{2z}-\tfrac1{8z^2}+O\!\left(\tfrac1{z^3}\right)\right)\\
&=-iz+\frac i2\color{#C00}{+\frac i{8z}}+O\!\left(\tfrac1{z^2}\right)\tag2
\end{align}
$$
Thus, the residue near $\infty$ is $\frac i8$, so if we circle once clockwise, the integral is $\frac\pi4$. This counts the integral above $[0,1]$ and below. Thus, the integral above is
$$
\int_0^1\sqrt{z(1-z)}\mathrm{d}z=\frac\pi8\tag3
$$
