## Calculus: Early Transcendentals (2nd Edition)

$$\frac{{\sqrt 2 }}{6}$$
\eqalign{ & \int_0^{1/3} {\frac{{dx}}{{{{\left( {9{x^2} + 1} \right)}^{3/2}}}}} \cr & x = \frac{1}{3}\tan \theta ,{\text{ }}dx = \frac{1}{3}{\sec ^2}\theta d\theta \cr & {\text{new limits of integration}} \cr & \theta = {\tan ^{ - 1}}\left( {3x} \right) \cr & x = 0,{\text{ }}\theta = 0 \cr & x = 1/3,{\text{ }}\theta = \pi /4 \cr & {\text{we have}} \cr & \int_0^{1/3} {\frac{{dx}}{{{{\left( {9{x^2} + 1} \right)}^{3/2}}}}} = \int_0^{\pi /4} {\frac{{\left( {1/3} \right){{\sec }^2}\theta d\theta }}{{{{\left( {9{{\left( {1/3\tan \theta } \right)}^2} + 1} \right)}^{3/2}}}}} \cr & = \frac{1}{3}\int_0^{\pi /4} {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {{{\tan }^2}\theta + 1} \right)}^{3/2}}}}d\theta } \cr & {\text{pythagorean identity}} \cr & = \frac{1}{3}\int_0^{\pi /4} {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^{3/2}}}}d\theta } = \frac{1}{3}\int_0^{\pi /4} {\frac{1}{{\sec \theta }}} d\theta \cr & = \frac{1}{3}\int_0^{\pi /4} {\cos \theta } d\theta \cr & {\text{integrating}} \cr & = \frac{1}{3}\left[ {\sin \theta } \right]_0^{\pi /4} \cr & {\text{evaluate limits}} \cr & = \frac{1}{3}\left[ {\sin \left( {\frac{\pi }{4}} \right) - \sin \left( 0 \right)} \right] \cr & {\text{simplify}} \cr & = \frac{1}{3}\left[ {\frac{{\sqrt 2 }}{2}} \right] \cr & = \frac{{\sqrt 2 }}{6} \cr}