Why are the two calculations of $ \int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x $ give two distinct answers? One of the calculations:
$$
\begin{aligned}
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x &=\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{\sqrt{x}} \mathrm{~d} x \\
&=2 \int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \mathrm{~d}(\sqrt{x})=2 \arcsin \sqrt{x}+C .
\end{aligned}
$$
The other:
$$ 
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x = \int \frac{d\left(x-\frac{1}{2}\right)}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}} = \arcsin \frac{x-1 / 2}{1 / 2}+C=\arcsin(2x-1)+C
$$

The answers are actually the same.
$\begin{align*}
\theta = 2\arcsin \sqrt x &\implies \sin \frac \theta 2 = \sqrt x \\
&\implies \sin^2 \frac \theta 2 = x \\
&\implies \frac{1-\cos \theta}{2} = x \\
&\implies \cos \theta = 1-2x \\
&\implies \sin(\theta+\pi/2) = 2x-1 \\
&\implies \theta = \arcsin(2x-1) - \pi/2
\end{align*}$
And $\pi/2$ gets absorbed into $+C$
