Chapter 7 - Principles Of Integral Evaluation - Chapter 7 Review Exercises - Page 557: 7

$$\eqalign{ & 2{\sin ^{ - 1}}\left( {\sqrt {\frac{x}{2}} } \right) + C \cr & - 2{\sin ^{ - 1}}\left( {\sqrt {\frac{{2 - x}}{2}} } \right) + C \cr & {\sin ^{ - 1}}\left( {x - 1} \right) + C \cr} $$

$$\eqalign{ & \left( {\text{a}} \right)\int {\frac{1}{{\sqrt {2x - {x^2}} }}} dx \cr & {\text{Let }}u = \sqrt x ,{\text{ }}x = {u^2}{\text{ }}dx = 2udu \cr & {\text{substituting}} \cr & \int {\frac{1}{{\sqrt {2x - {x^2}} }}} dx = \int {\frac{{2udu}}{{\sqrt {2{u^2} - {u^4}} }}} = 2\int {\frac{{du}}{{\sqrt {2 - {u^2}} }}} \cr & {\text{Integrating, recall that }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }}} = {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr & 2\int {\frac{{du}}{{\sqrt {2 - {u^2}} }}} = 2{\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr & 2{\sin ^{ - 1}}\left( {\frac{{\sqrt x }}{{\sqrt 2 }}} \right) + C \cr & = 2{\sin ^{ - 1}}\left( {\sqrt {\frac{x}{2}} } \right) + C \cr & \cr & {\text{Let }}u = \sqrt {2 - x} ,{\text{ }}x = 2 - {u^2}{\text{ }}dx = - 2udu \cr & {\text{substituting}} \cr & \int {\frac{1}{{\sqrt {2x - {x^2}} }}} dx = \int {\frac{1}{{\sqrt {2\left( {2 - {u^2}} \right) - {{\left( {2 - {u^2}} \right)}^2}} }}} \left( { - 2u} \right)du \cr & = \int {\frac{1}{{\sqrt {2\left( {2 - {u^2}} \right) - {{\left( {2 - {u^2}} \right)}^2}} }}} \left( { - 2u} \right)du \cr & = - \int {\frac{{2u}}{{\sqrt {4 - 2{u^2} - 4 + 4{u^2} - {u^4}} }}} du \cr & = - \int {\frac{{2u}}{{\sqrt {2{u^2} - {u^4}} }}} du \cr & = - \int {\frac{2}{{\sqrt {2 - {u^2}} }}} du = - 2{\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr & = - 2{\sin ^{ - 1}}\left( {\frac{{\sqrt {2 - x} }}{{\sqrt 2 }}} \right) + C \cr & = - 2{\sin ^{ - 1}}\left( {\sqrt {\frac{{2 - x}}{2}} } \right) + C \cr & \cr & {\text{Completing the square}} \cr & \int {\frac{1}{{\sqrt {2x - {x^2}} }}} dx = \int {\frac{1}{{\sqrt {1 - {{\left( {x - 1} \right)}^2}} }}} dx \cr & {\text{Let }}u = x - 1,{\text{ }}du = dx \cr & = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr & = {\sin ^{ - 1}}\left( u \right) + C \cr & {\text{Back - substitute }}u = x - 1 \cr & = {\sin ^{ - 1}}\left( {x - 1} \right) + C \cr} $$