Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 550: 68

$\frac{\pi }{4}$

$$\eqalign{ & \int_2^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx \cr & {\text{Expressing the integral as a sum of improper integrals type 1}} \cr & {\text{and type 2 as follows:}} \cr & \int_2^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx = \int_2^4 {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx + \int_4^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx \cr & \cr & {\text{*Calculating }}\int_2^4 {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx \cr & {\text{Using the definition of improper integrals }} \cr & \int_2^4 {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx = \mathop {\lim }\limits_{a \to {2^ + }} \int_a^4 {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx \cr & {\text{Integrating}}{\text{, use the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}{{\sec }^{ - 1}}\frac{u}{a} + C} \cr & = \mathop {\lim }\limits_{a \to {2^ + }} \left[ {\frac{1}{2}{{\sec }^{ - 1}}\frac{x}{2}} \right]_a^4 \cr & = \mathop {\lim }\limits_{a \to {2^ + }} \left[ {\frac{1}{2}\left( {\frac{\pi }{3}} \right) - \frac{1}{2}{{\sec }^{ - 1}}\frac{a}{2}} \right] \cr & = \mathop {\lim }\limits_{a \to {2^ + }} \left[ {\frac{\pi }{6} - \frac{1}{2}{{\sec }^{ - 1}}\frac{a}{2}} \right] \cr & = \underbrace {\mathop {\lim }\limits_{a \to {2^ + }} \left[ {\frac{\pi }{6}} \right]}_{{\text{Tends to }}\frac{\pi }{6}} - \frac{1}{2}\underbrace {\mathop {\lim }\limits_{a \to {2^ + }} \left[ {{{\sec }^{ - 1}}\frac{a}{2}} \right]}_{{\text{Tends to 0}}} \cr & = \frac{\pi }{6} + 0 \cr & = \frac{\pi }{6} \cr & \cr & {\text{*Calculating }}\int_4^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx \cr & {\text{Using the definition of improper integrals }} \cr & \int_4^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx = \mathop {\lim }\limits_{b \to \infty } \int_4^b {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx \cr & = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{2}{{\sec }^{ - 1}}\frac{x}{2}} \right]_4^b \cr & = \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\sec }^{ - 1}}\frac{b}{2} - {{\sec }^{ - 1}}\frac{4}{2}} \right] \cr & = \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\sec }^{ - 1}}\frac{b}{2} - \frac{\pi }{3}} \right] \cr & = \frac{1}{2}\underbrace {\mathop {\lim }\limits_{b \to \infty } \left[ {{{\sec }^{ - 1}}\frac{b}{2}} \right]}_{{\text{Tends to }}\frac{\pi }{2}} - \frac{1}{2}\underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{\pi }{3}} \right]}_{{\text{Tends to }}\frac{\pi }{3}} \cr & = \frac{1}{2}\left( {\frac{\pi }{2}} \right) - \frac{1}{2}\left( {\frac{\pi }{3}} \right) \cr & = \frac{\pi }{{12}} \cr & \cr & {\text{Therefore}}{\text{,}} \cr & \int_2^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx = \int_2^4 {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx + \int_4^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx \cr & \int_2^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx = \frac{\pi }{6} + \frac{\pi }{{12}} \cr & \int_2^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx = \frac{\pi }{4} \cr} $$