Find $\int \dfrac{dx}{\sqrt{2x-x^{2}}}$.

Solution The integrand contains the quadratic expression $2x-x^{2}$, so we complete the square. $$\begin{eqnarray*} 2x-x^{2}&=&-x^{2}+2x=-(x^{2}-2x) =-(x^{2}-2x+1)+1\\ &=&-( x-1) ^{2}+1=1-(x-1)^{2} \end{eqnarray*}$$

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*}$$