How to calculate the integral <msubsup> &#x222B;<!-- ∫ --> <mrow class="MJX-TeXAtom-OR

How to calculate the integral ${\int }_{0.5}^{1}\frac{1}{\sqrt{2x-{x}^{2}}}$?
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Marlee Guerra
Note that
2x-x2=1-(1-2x+x2)=1-(1-x)2
Now you can use a trig substitution.
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minwaardekn
Given ${\int }_{0.5}^{1}\frac{1}{\sqrt{2x-{x}^{2}}}dx$
$=\int \frac{1}{\sqrt{1-\left(x-1{\right)}^{2}}}dx$
Apply u substitution u=x-1
$\int \frac{1}{\sqrt{1-{u}^{2}}}du=\mathrm{arcsin}\left(u\right)=\mathrm{arcsin}\left(x-1\right)$
Now apply the boundaries
$=\left[\mathrm{arcsin}\left(x-1\right){\right]}_{0.5}^{1}=\frac{\pi }{6}$