Chapter 8: Techniques of Integration - Practice Exercises - Page 519: 61

$$\frac{\pi }{6}$$

$$\eqalign{ & \int_{ - \infty }^\infty {\frac{{dx}}{{4{x^2} + 9}}} \cr & {\text{By the definition of the improper integrals, we have }} \cr & \int_{ - \infty }^\infty {\frac{{dx}}{{4{x^2} + 9}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^c {\frac{{dx}}{{4{x^2} + 9}}} + \mathop {\lim }\limits_{b \to \infty } \int_c^b {\frac{{dx}}{{4{x^2} + 9}}} \cr & {\text{taking }}c = 0 \cr & \int_{ - \infty }^\infty {\frac{{dx}}{{4{x^2} + 9}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{dx}}{{4{x^2} + 9}}} + \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{dx}}{{4{x^2} + 9}}} \cr & \cr & {\text{Integrating }}\int {\frac{{dx}}{{4{x^2} + 9}}{\text{ by tables}}} \cr & \int {\frac{{dx}}{{4{x^2} + 9}} = \frac{1}{2}\int {\frac{2}{{{{\left( {2x} \right)}^2} + {{\left( 3 \right)}^2}}}} dx} \cr & {\text{Use the formula }}\int {\frac{{du}}{{{u^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr & \,\,\,\,\,\frac{1}{2}\int {\frac{2}{{{{\left( {2x} \right)}^2} + \left( 3 \right)}}} dx = \frac{1}{2}\left( {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{{2x}}{3}} \right)} \right) + C \cr & \,\,\,\,\,\frac{1}{2}\int {\frac{2}{{{{\left( {2x} \right)}^2} + \left( 3 \right)}}} dx = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + C \cr & \cr & {\text{Then}}{\text{,}} \cr & \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{dx}}{{4{x^2} + 9}}} = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{{2x}}{3}} \right)} \right)_a^0 \cr & = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{{2\left( 0 \right)}}{3}} \right) - \frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{{2\left( a \right)}}{3}} \right)} \right) \cr & = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{6}\left( 0 \right) - \frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{{2a}}{3}} \right)} \right) \cr & {\text{Evaluate the limit when }}a \to - \infty \cr & = - \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2\left( { - \infty } \right)}}{3}} \right) \cr & = - \frac{1}{6}\left( { - \frac{\pi }{2}} \right) \cr & = \frac{\pi }{{12}} \cr & \cr & \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{dx}}{{4{x^2} + 9}}} = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{{2x}}{3}} \right)} \right)_0^b \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{{2b}}{3}} \right) - \frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{{2\left( 0 \right)}}{3}} \right)} \right) \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{{2b}}{3}} \right) - \frac{1}{6}\left( 0 \right)} \right) \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{{2b}}{3}} \right)} \right) \cr & {\text{Evaluate the limit when }}b \to \infty \cr & = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2\left( \infty \right)}}{3}} \right) \cr & = \frac{1}{6}\left( {\frac{\pi }{2}} \right) \cr & = \frac{\pi }{{12}} \cr & \cr & {\text{Since both of these integrals converge}},{\text{ the improper integral is}} \cr & \int_{ - \infty }^\infty {\frac{{dx}}{{4{x^2} + 9}}} = \frac{\pi }{{12}} + \frac{\pi }{{12}} \cr & \int_{ - \infty }^\infty {\frac{{dx}}{{4{x^2} + 9}}} = \frac{\pi }{6} \cr} $$