## Calculus: Early Transcendentals (2nd Edition)

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