## Calculus: Early Transcendentals (2nd Edition)

$$\ln \left( {\frac{{2 + \sqrt 3 }}{{\sqrt 3 }}} \right)$$
\eqalign{ & \int_{10/\sqrt 3 }^{10} {\frac{{dy}}{{\sqrt {{y^2} - 25} }}} \cr & y = 5\sec \theta ,{\text{ }}dy = 5\sec \theta \tan \theta d\theta \cr & {\text{new limits of integration}} \cr & \theta = {\sec ^{ - 1}}\left( {y/5} \right) \cr & y = 10,{\text{ }}\theta = \pi /3 \cr & y = 10/\sqrt 3 ,{\text{ }}\theta = \pi /6 \cr & {\text{we have}} \cr & \int_{10/\sqrt 3 }^{10} {\frac{{dy}}{{\sqrt {{y^2} - 25} }}} = \int_{\pi /6}^{\pi /3} {\frac{{5\sec \theta \tan \theta d\theta }}{{\sqrt {{{\left( {5\sec \theta } \right)}^2} - 25} }}} \cr & = \int_{\pi /6}^{\pi /3} {\frac{{5\sec \theta \tan \theta d\theta }}{{5\sqrt {{{\sec }^2}\theta - 25} }}} \cr & {\text{pythagorean identity}} \cr & = \int_{\pi /6}^{\pi /3} {\frac{{5\sec \theta \tan \theta d\theta }}{{5\sqrt {{{\tan }^2}\theta } }}} = \int_{\pi /6}^{\pi /3} {\sec \theta d\theta } \cr & {\text{integrating}} \cr & = \left[ {\ln \left| {\sec \theta + \tan \theta } \right|} \right]_{\pi /6}^{\pi /3} \cr & {\text{evaluate limits}} \cr & = \left[ {\ln \left| {\sec \left( {\frac{\pi }{3}} \right) + \tan \left( {\frac{\pi }{3}} \right)} \right|} \right] - \left[ {\ln \left| {\sec \left( {\frac{\pi }{6}} \right) + \tan \left( {\frac{\pi }{6}} \right)} \right|} \right] \cr & {\text{simplify}} \cr & = \ln \left| {2 + \sqrt 3 } \right| - \ln \left| {\frac{{2\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3}} \right| \cr & = \ln \left| {2 + \sqrt 3 } \right| - \ln \left| {\sqrt 3 } \right| \cr & = \ln \left( {\frac{{2 + \sqrt 3 }}{{\sqrt 3 }}} \right) \cr}