## Calculus: Early Transcendentals (2nd Edition)

$$\sqrt 3 - 1 - \pi /12$$
\eqalign{ & \int_{\sqrt 2 }^2 {\frac{{\sqrt {{x^2} - 1} }}{x}dx} \cr & x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta d\theta \cr & {\text{new limits of integration}} \cr & \theta = {\sec ^{ - 1}}\left( x \right) \cr & x = 2,{\text{ }}\theta = \pi /3 \cr & x = \sqrt 2 ,{\text{ }}\theta = \pi /4 \cr & {\text{we have}} \cr & \int_{\sqrt 2 }^2 {\frac{{\sqrt {{x^2} - 1} }}{x}dx} = \int_{\pi /4}^{\pi /3} {\frac{{\sqrt {{{\sec }^2}\theta - 1} }}{{\sec \theta }}} \left( {\sec \theta \tan \theta d\theta } \right) \cr & {\text{pythagorean identity}} \cr & = \int_{\pi /4}^{\pi /3} {\frac{{\sqrt {{{\tan }^2}\theta } }}{{\sec \theta }}} \left( {\sec \theta \tan \theta d\theta } \right) = \int_{\pi /4}^{\pi /3} {{{\tan }^2}\theta } d\theta \cr & = \int_{\pi /4}^{\pi /3} {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr & {\text{integrating}} \cr & = \left[ {\tan \theta - \theta } \right]_{\pi /4}^{\pi /3} \cr & {\text{evaluate limits}} \cr & = \left[ {\tan \left( {\frac{\pi }{3}} \right) - \frac{\pi }{3}} \right] - \left[ {\tan \left( {\frac{\pi }{4}} \right) - \frac{\pi }{4}} \right] \cr & {\text{simplify}} \cr & = \left[ {\sqrt 3 - \frac{\pi }{3}} \right] - \left[ {1 - \frac{\pi }{4}} \right] \cr & = \sqrt 3 - 1 - \pi /12 \cr}