Chapter 7 - Review - Exercises - Page 538: 59

$$ - \frac{{\sqrt {{a^2} - {u^2}} }}{u} - {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C$$

$$\eqalign{ & {\text{Formula 33: }}\int {\frac{{\sqrt {{a^2} - {u^2}} }}{{{u^2}}}} du = - \frac{1}{u}\sqrt {{a^2} - {u^2}} - {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr & \cr & *{\text{Verify by differentiation}} \cr & \frac{d}{{dx}}\left[ { - \frac{{\sqrt {{a^2} - {u^2}} }}{u} - {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C} \right] \cr & = \frac{d}{{dx}}\left[ {\frac{{ - \sqrt {{a^2} - {u^2}} }}{u}} \right] - \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}\left( {\frac{u}{a}} \right)} \right] + \frac{d}{{dx}}\left[ C \right] \cr & = \frac{{ - u\left( {\frac{{ - 2u}}{{2\sqrt {{a^2} - {u^2}} }}} \right) + \sqrt {{a^2} - {u^2}} }}{{{u^2}}} - \frac{{1/a}}{{\sqrt {1 - {{\left( {u/a} \right)}^2}} }} + 0 \cr & {\text{Simplifying}} \cr & = \frac{{\frac{{{u^2}}}{{\sqrt {{a^2} - {u^2}} }} + \sqrt {{a^2} - {u^2}} }}{{{u^2}}} - \frac{1}{{\sqrt {{a^2} - {u^2}} }} \cr & = \frac{{{u^2} + {a^2} - {u^2}}}{{{u^2}\sqrt {{a^2} - {u^2}} }} - \frac{1}{{\sqrt {{a^2} - {u^2}} }} \cr & = \frac{{{a^2}}}{{{u^2}\sqrt {{a^2} - {u^2}} }} - \frac{1}{{\sqrt {{a^2} - {u^2}} }} \cr & = \frac{{{a^2} - {u^2}}}{{{u^2}\sqrt {{a^2} - {u^2}} }} \cr & = \frac{{\sqrt {{a^2} - {u^2}} }}{{{u^2}}} \cr & \cr & *{\text{Using trigonometric substitution}} \cr & {\text{Let }}u = a\sin \theta ,{\text{ }}du = a\cos \theta d\theta \cr & \int {\frac{{\sqrt {{a^2} - {u^2}} }}{{{u^2}}}} du = \int {\frac{{\sqrt {{a^2} - {{\left( {a\sin \theta } \right)}^2}} }}{{{{\left( {a\sin \theta } \right)}^2}}}} \left( {a\cos \theta } \right)d\theta \cr & = \int {\frac{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}{{{a^2}{{\sin }^2}\theta }}} \left( {a\cos \theta } \right)d\theta \cr & = \int {\frac{{a\sqrt {1 - {{\sin }^2}\theta } }}{{{a^2}{{\sin }^2}\theta }}} \left( {a\cos \theta } \right)d\theta \cr & = \int {\frac{{\sqrt {{{\cos }^2}\theta } }}{{{{\sin }^2}\theta }}} \left( {\cos \theta } \right)d\theta \cr & = \int {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} d\theta \cr & = \int {\left( {{{\cot }^2}\theta } \right)} d\theta \cr & {\text{Use pythagorean identity}} \cr & = \int {\left( {{{\csc }^2}\theta - 1} \right)} d\theta \cr & = - \cot \theta - \theta + C \cr & = - \frac{{\cos \theta }}{{\sin \theta }} - \theta + C \cr & {\text{Where cos}}\theta = \frac{{\sqrt {{a^2} - {u^2}} }}{a},{\text{ sin}}\theta = \frac{u}{a},{\text{ then}} \cr & = - \left( {\frac{{\sqrt {{a^2} - {u^2}} }}{a}} \right)\left( {\frac{a}{u}} \right) - {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr & = - \frac{{\sqrt {{a^2} - {u^2}} }}{u} - {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr} $$