How does Wolfram Alpha compute the indefinite integral $\int \frac{1}{\sqrt{2x-x^2}}dx$

Question

I want to find the indefinite integral of $1/\sqrt{2x-x^2}$.

I rearranged the expression to $\displaystyle \int \frac{1}{\sqrt{1-(x-1)^2}}$, which is equal to $\displaystyle \int \arcsin'(x-1)$. As far as I can tell, this should evaluate to $\arcsin(x-1)$, because $[\arcsin(x-1)]'=\arcsin'(x-1)\times1=\arcsin'(x-1)$.
When I evaluate this expression on Wolfram Alpha, I get $\displaystyle-2\arcsin(\sqrt{1-\frac{x}{2}})$. Differentiating this also yields $\arcsin'(x-1)$. I assume that these two expressions ($\arcsin(x-1)$ and $\displaystyle-2\arcsin(\sqrt{1-\frac{x}{2}})$ ) simply differ by some constant.
How exactly is Wolfram Alpha's output reached?

Answer 1

Note that \begin{align} \int \frac{1}{\sqrt{2x-x^2}}dx &=\frac12\int\frac1{\sqrt{\frac x2}\sqrt{1-\frac x2}}dx = 2\int\frac{d(\sqrt{\frac x2})} {\sqrt{1-\frac x2}}\\ &=-2\cos^{-1} \sqrt{\frac x2}=-2 \sin^{-1} \sqrt{1-\frac x2} \end{align}

Answer 2

$$ \begin{aligned} I & =\int \frac{1}{\sqrt{x}} \frac{1}{\sqrt{2-x}} d x \\ & =2 \int \frac{1}{\sqrt{2-(\sqrt{x})^2}} d \sqrt{x} \\ & =-2 \cos ^{-1}\left(\frac{\sqrt{x}}{\sqrt{2}}\right)+C\\&= -2 \sin ^{-1}\left(\frac{\sqrt{2-x}}{\sqrt{2}}\right)+C\\&= -2 \sin ^{-1}\left(\sqrt{1-\frac{x}{2}}\right)+C \end{aligned} $$