Calculus: Early Transcendentals (2nd Edition)

$$\ln \left( {\frac{{2 + \sqrt 3 }}{{\sqrt 2 + 1}}} \right)$$
\eqalign{ & \int_{8\sqrt 2 }^{16} {\frac{{dx}}{{\sqrt {{x^2} - 64} }}} \cr & x = 8\sec \theta ,{\text{ }}dx = 8\sec \theta \tan \theta d\theta \cr & {\text{New limits of integration}} \cr & \theta = {\sec ^{ - 1}}\left( {x/8} \right) \cr & x = 8\sqrt 2 ,{\text{ }}\theta = \pi /4 \cr & x = 16,{\text{ }}\theta = \pi /3 \cr & {\text{we have}} \cr & \int_{8\sqrt 2 }^{16} {\frac{{dx}}{{\sqrt {{x^2} - 64} }}} = \int_{\pi /4}^{\pi /3} {\frac{{8\sec \theta \tan \theta d\theta }}{{\sqrt {{{\left( {8\sec \theta } \right)}^2} - 64} }}} \cr & = \int_{\pi /4}^{\pi /3} {\frac{{8\sec \theta \tan \theta d\theta }}{{\sqrt {64{{\sec }^2}\theta - 64} }}} = \int_{\pi /4}^{\pi /3} {\frac{{8\sec \theta \tan \theta d\theta }}{{8\sqrt {{{\sec }^2}\theta - 1} }}} \cr & {\text{pythagorean identity}} \cr & = \int_{\pi /4}^{\pi /3} {\frac{{8\sec \theta \tan \theta d\theta }}{{8\sqrt {{{\tan }^2}\theta } }}} = \int_{\pi /4}^{\pi /3} {\sec \theta } d\theta \cr & {\text{integrating}} \cr & = \left[ {\ln \left| {\sec \theta + \tan \theta } \right|} \right]_{\pi /4}^{\pi /3} \cr & {\text{evaluate limits}} \cr & = \left[ {\ln \left| {\sec \left( {\pi /3} \right) + \tan \left( {\pi /3} \right)} \right|} \right] - \left[ {\ln \left| {\sec \left( {\pi /4} \right) + \tan \left( {\pi /4} \right)} \right|} \right] \cr & {\text{simplify}} \cr & = \left[ {\ln \left| {2 + \sqrt 3 } \right|} \right] - \left[ {\ln \left| {\sqrt 2 + 1} \right|} \right] \cr & = \ln \left( {\frac{{2 + \sqrt 3 }}{{\sqrt 2 + 1}}} \right) \cr}