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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?

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  • $\begingroup$ This maybe irrelevant but I felt like it needed attention here $$\,$$ Only now does one truly realise the importance of the integration constant; $+C$. Integrals that lead to trigonometric functions like this have multiple answers. With $\arcsin(1-x), -2\arcsin(\sqrt{1 - \dfrac{x}{2}})$ another possible answer could be $\arccos(1 - x)$. Each differing with a constant of $\dfrac{\pi}{2}$. While all these are correct options, in future, remember that each of these graphs have a different range (although your question did not require this info now). (and don't forget your $+C$!). $\endgroup$
    – Dstarred
    Commented Jun 6, 2023 at 13:59

2 Answers 2

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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}

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    $\begingroup$ Thank you for the response. Could you please clarify what you mean by $d(\sqrt{\frac{x}{2}})$? I have not seen that notation before. $\endgroup$
    – Frightened of Sinusoids
    Commented Jun 6, 2023 at 13:26
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    $\begingroup$ @FrightenedofSinusoids - Merely a shortcut for the substitution $t= \sqrt{\frac x2}$, i.e. $dt = d(\sqrt{\frac x2})=(\sqrt{\frac x2})’dx$. $\endgroup$
    – Quanto
    Commented Jun 6, 2023 at 13:30
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    $\begingroup$ Is this a "u-substitution" with "$u=\sqrt{\frac{x}{2}}$" and "$2du=\frac{1}{2\sqrt{\frac{x}{2}}}dx$"? $\endgroup$
    – Frightened of Sinusoids
    Commented Jun 6, 2023 at 13:31
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    $\begingroup$ ahh. Thank you! $\endgroup$
    – Frightened of Sinusoids
    Commented Jun 6, 2023 at 13:32
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$$ \begin{aligned} I & =\int \frac{1}{\sqrt{x}} \frac{1}{\sqrt{2-x}} d x \\ & =2 \int \frac{1}{\sqrt{2-(\sqrt{x})^2}} d \sqrt{x} \\ & =-2 \cos ^{-1}\left(\frac{\sqrt{x}}{\sqrt{2}}\right)+C\\&= -2 \sin ^{-1}\left(\frac{\sqrt{2-x}}{\sqrt{2}}\right)+C\\&= -2 \sin ^{-1}\left(\sqrt{1-\frac{x}{2}}\right)+C \end{aligned} $$

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