How to calculate the integral $\int_{0.5}^{1} \frac{1}{\sqrt{2x-x^2}}$? I have the following integration problem:
$$ \int_{0.5}^{1} \frac{1}{\sqrt{2x-x^2}} $$
And I can see I should probably be completing the square here. I may be missing something extremely obvious, but wouldn't this mean I have to take the negative outside the expression to do this? I need the $x^2$ term by itself, don't I?
Any help would be greatly appreciated!

There's no hyperbolic function here.
Set
$$
t=\sqrt{\frac{x}{2-x}}
$$
so $(2-x)t^2=x$ and
$$
x=\frac{2t^2}{1+t^2}=\frac{2t^2+2-2}{1+t^2}=2-\frac{2}{1+t^2}
$$
whence
$$
dx=\frac{4t}{(1+t^2)^2}
$$
and the integral becomes
\begin{align}
\int_{1/2}^1\frac{1}{2x-x^2}\,dx
&=\int_{1/2}^1\frac{1}{x}\sqrt{\frac{x}{2-x}}\,dx \\[4px]
&=\int_{1/\sqrt{3}}^{1}\frac{1+t^2}{2t^2}t\frac{4t}{(1+t^2)^2}\,dt \\[4px]
&=2\int_{1/\sqrt{3}}^1\frac{1}{1+t^2}\,dt \\[4px]
&=2\Bigl[\arctan t\Bigr]_{1/\sqrt{3}}^1 \\[4px]
&=2\frac{\pi}{4}-2\frac{\pi}{6} \\[4px]
&=\frac{\pi}{6}
\end{align}
