Find \(\int \dfrac{dx}{\sqrt{2x-x^{2}}}\).
Then \[ \begin{eqnarray*} \int\! \frac{dx}{\sqrt{2x-x^{2}}}&=&\!\int\! \frac{dx}{\sqrt{1-(x-1)^{2}}}&&\underset{\underset{\underset{\color{#0066A7}{\hbox{\(du=dx\)}}}{\color{#0066A7}{\hbox{\(u=x-1\)}}}}{\color{#0066A7}{\uparrow }}}{=}\!\int\! \frac{du}{\sqrt{ 1-u^{2}}}=\sin ^{-1}u+C=\sin ^{-1}(x-1)+C \\[-11pt] \end{eqnarray*} \]