Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.4 Exercises - Page 539: 46

$\ln \left( {2 + \sqrt 3 } \right) - \frac{{\sqrt 3 }}{2}$

$$\eqalign{ & \int_4^8 {\frac{{\sqrt {{x^2} - 16} }}{{{x^2}}}} dx \cr & {\text{Refer to the triangle below}} \cr & \sec \theta = \frac{x}{4},{\text{ }}dx = 4\sec \theta \tan \theta d\theta \cr & {\text{Substitute}} \cr & \int {\frac{{\sqrt {{x^2} - 16} }}{{{x^2}}}dx = } \int {\frac{{4\tan \theta }}{{{{\left( {4\sec \theta } \right)}^2}}}} \left( {4\sec \theta \tan \theta } \right)d\theta \cr & = \int {\frac{{16\sec \theta {{\tan }^2}\theta }}{{16{{\sec }^2}\theta }}} d\theta \cr & = \int {\frac{{{{\tan }^2}\theta }}{{\sec \theta }}} d\theta = \int {\frac{{{{\sin }^2}\theta }}{{\cos \theta }}} d\theta \cr & = \int {\frac{{1 - {{\cos }^2}\theta }}{{\cos \theta }}} d\theta \cr & = \int {\left( {\sec \theta - \cos \theta } \right)} d\theta \cr & {\text{Integrating}} \cr & = \ln \left| {\sec \theta + \tan \theta } \right| - \sin \theta + C \cr & {\text{Substituting using the triangle}} \cr & = \ln \left| {\frac{x}{4} + \frac{{\sqrt {{x^2} - 16} }}{4}} \right| - \frac{{\sqrt {{x^2} - 16} }}{x} + C \cr & = \ln \left| {\frac{{x + \sqrt {{x^2} - 16} }}{4}} \right| - \frac{{\sqrt {{x^2} - 16} }}{x} + C \cr & \int_4^8 {\frac{{\sqrt {{x^2} - 16} }}{{{x^2}}}} dx = \left[ {\ln \left| {\frac{{x + \sqrt {{x^2} - 16} }}{4}} \right| - \frac{{\sqrt {{x^2} - 16} }}{x}} \right]_4^8 \cr & {\text{Evaluating}} \cr & = \left[ {\ln \left| {\frac{{8 + \sqrt {{8^2} - 16} }}{4}} \right| - \frac{{\sqrt {{8^2} - 16} }}{8}} \right] - \left[ {\ln \left| {\frac{{4 + \sqrt 0 }}{4}} \right| - \frac{{\sqrt 0 }}{4}} \right] \cr & = \ln \left( {\frac{{8 + \sqrt {48} }}{4}} \right) - \frac{{\sqrt {48} }}{8} \cr & {\text{Simplifying}} \cr & = \ln \left( {\frac{{8 + 4\sqrt 3 }}{4}} \right) - \frac{{4\sqrt 3 }}{8} \cr & = \ln \left( {2 + \sqrt 3 } \right) - \frac{{\sqrt 3 }}{2} \cr & \approx 0.450932 \cr & \cr & \left( {\text{b}} \right){\text{ Change the limits of integration}} \cr & {\text{sec}}\theta = \frac{x}{4},{\text{ }}\theta = {\sec ^{ - 1}}\left( {\frac{x}{4}} \right) \cr & x = 8 \to {\sec ^{ - 1}}\left( {\frac{8}{4}} \right) = \frac{\pi }{3} \cr & x = 4 \to {\sec ^{ - 1}}\left( {\frac{4}{4}} \right) = 0 \cr & \int_4^8 {\frac{{\sqrt {{x^2} - 16} }}{{{x^2}}}} dx = \int_0^{\pi /3} {\left( {\sec \theta - \cos \theta } \right)} d\theta \cr & {\text{Integrating}} \cr & = \left[ {\ln \left| {\sec \theta + \tan \theta } \right| - \sin \theta } \right]_0^{\pi /3} \cr & = \left[ {\ln \left| {\sec \left( {\frac{\pi }{3}} \right) + \tan \left( {\frac{\pi }{3}} \right)} \right| - \sin \left( {\frac{\pi }{3}} \right)} \right] - \left[ {\ln \left| {\sec \left( 0 \right) + \tan \left( 0 \right)} \right| - 0} \right] \cr & = \ln \left( {2 + \sqrt 3 } \right) - \frac{{\sqrt 3 }}{2} \cr & \approx 0.45093 \cr} $$