## Thomas' Calculus 13th Edition

$$\frac{1}{3}{\left( {\frac{v}{{\sqrt {1 - {v^2}} }}} \right)^3} + C$$
\eqalign{ & \int {\frac{{{v^2}dv}}{{{{\left( {1 - {v^2}} \right)}^{5/2}}}}} \cr & {\text{Use the trigonometric substitutions:}} \cr & {\text{ }}v = \sin \theta,{\text{ }}dv = \cos \theta d\theta \cr & {\text{With these substitutions}}{\text{, we have}} \cr & \int {\frac{{{v^2}dv}}{{{{\left( {1 - {v^2}} \right)}^{5/2}}}}} = \int {\frac{{{{\sin }^2}\theta \left( {\cos \theta d\theta } \right)}}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{5/2}}}}} \cr & {\text{Use the fundamental identity }}\cr & {\cos ^2}\theta + {\sin ^2}\theta = 1 \cr & = \int {\frac{{{{\sin }^2}\theta \cos \theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^{5/2}}}}} d\theta \cr & {\text{Simplifying, we get:}} \cr & = \int {\frac{{{{\sin }^2}\theta \cos \theta }}{{{{\cos }^5}\theta }}} d\theta \cr & = \int {\frac{{{{\sin }^2}\theta \cos \theta }}{{{{\cos }^4}\theta }}} d\theta \cr & = \int {\left( {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)} \left( {\frac{1}{{{{\cos }^2}\theta }}} \right)d\theta \cr & = \int {{{\tan }^2}\theta } {\sec ^2}\theta d\theta \cr & {\text{Integrate}} \cr & = \frac{1}{3}{\tan ^3}\theta + C \cr & {\text{Write in terms of }}x,{\text{ }}\tan \theta = \frac{v}{{\sqrt {1 - {v^2}} }} \cr & = \frac{1}{3}{\left( {\frac{v}{{\sqrt {1 - {v^2}} }}} \right)^3} + C \cr}