Thomas' Calculus 13th Edition

$${\sin ^{ - 1}}\left( {x - 1} \right) + C$$
\eqalign{ & \int {\frac{{dx}}{{\sqrt {2x - {x^2}} }}} \cr & {\text{complete the square for }}2x - {x^2} \cr & 2x - {x^2} = - \left( {{x^2} - 2x + 1} \right) + 1 \cr & = - {\left( {x - 1} \right)^2} + 1 \cr & = 1 - {\left( {x - 1} \right)^2} \cr & {\text{then}} \cr & \int {\frac{{dx}}{{\sqrt {2x - {x^2}} }}} = \int {\frac{{dx}}{{\sqrt {1 - {{\left( {x - 1} \right)}^2}} }}} \cr & {\text{use the substitution method}}{\text{.}} \cr & u = x - 1,{\text{ so that }}du = dx \cr & \int {\frac{{dx}}{{\sqrt {1 - {{\left( {x - 1} \right)}^2}} }}} = \int {\frac{{dx}}{{\sqrt {1 - {u^2}} }}} \cr & {\text{integrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr & {\text{with }}a = 1 \cr & = {\sin ^{ - 1}}\left( {\frac{{x - 1}}{1}} \right) + C \cr & {\text{write in terms of }}y;{\text{ replace }}x - 1{\text{ for }}u \cr & = {\sin ^{ - 1}}\left( {x - 1} \right) + C \cr}