## Thomas' Calculus 13th Edition

$$- \frac{{\sqrt {1 - {v^2}} }}{v} - {\sin ^{ - 1}}v + C$$
\eqalign{ & \int {\frac{{\sqrt {1 - {v^2}} }}{{{v^2}}}} dv \cr & {\text{Using Trigonometric Substitutions}} \cr & {\text{ }}v = \sin \theta ,{\text{ }}dv = \cos \theta d\theta \cr & {\text{With these substitutions}}{\text{, we have}} \cr & \int {\frac{{\sqrt {1 - {v^2}} }}{{{v^2}}}} dv = \int {\frac{{\sqrt {1 - {{\sin }^2}\theta } }}{{{{\left( {\sin \theta } \right)}^2}}}} \left( {\cos \theta } \right)d\theta \cr & {\text{Use the fundamental identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \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 {{{\cot }^2}\theta } d\theta \cr & \cr & {\text{Use the identity co}}{{\text{t}}^2}x = {\csc ^2}x - 1 \cr & = \int {\left( {{{\csc }^2}\theta - 1} \right)} d\theta \cr & {\text{Integrating}} \cr & = - \cot \theta - \theta + C \cr & {\text{Write in terms of }}\theta ,\,\,{\text{ }}v = \sin \theta ,\,\,\,\theta = {\sin ^{ - 1}}v{\text{ and cot}}\theta = \frac{{\sqrt {1 - {v^2}} }}{v} \cr & = - \frac{{\sqrt {1 - {v^2}} }}{v} - {\sin ^{ - 1}}v + C \cr}