Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.4 Trigonometric Substitutions - Exercises Set 7.4 - Page 513: 26

Answer

$$\frac{{37\sqrt 3 }}{{72}}$$

Work Step by Step

$$\eqalign{ & \int_0^3 {\frac{{{x^3}}}{{{{\left( {3 + {x^2}} \right)}^{5/2}}}}} dx \cr & {\text{substitute }}x = \sqrt 3 \tan \theta ,{\text{ }}\,\,\,dx = \sqrt 3 {\sec ^2}\theta d\theta \,\,\,\,\theta = {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) \cr & \,\,\,x = 3 \to \theta = {\tan ^{ - 1}}\left( {\frac{3}{{\sqrt 3 }}} \right) = \frac{\pi }{3} \cr & \,\,\,x = 0 \to \theta = {\tan ^{ - 1}}\left( {\frac{0}{{\sqrt 3 }}} \right) = 0 \cr & \cr & {\text{write the integral in terms of }}\theta \cr & \int_0^3 {\frac{{{x^3}}}{{{{\left( {3 + {x^2}} \right)}^{5/2}}}}} dx = \int_0^{\pi /3} {\frac{{3\sqrt 3 {{\tan }^3}\theta }}{{{{\left( {3 + 3{{\tan }^2}\theta } \right)}^{5/2}}}}\left( {\sqrt 3 {{\sec }^2}\theta } \right)d\theta } \cr & = \int_0^{\pi /3} {\frac{{3\sqrt 3 {{\tan }^3}\theta }}{{{3^{5/2}}{{\left( {1 + {{\tan }^2}\theta } \right)}^{5/2}}}}\left( {\sqrt 3 {{\sec }^2}\theta } \right)d\theta } \cr & \cr & {\text{use the pythagorean identity }}{\tan ^2} + 1 = {\sec ^2}\theta \cr & = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {\frac{{{{\tan }^3}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^{5/2}}}}\left( {{{\sec }^2}\theta } \right)d\theta } \cr & = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {\frac{{{{\tan }^3}\theta }}{{{{\sec }^5}\theta }}d\theta } \cr & = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {{{\sin }^3}\theta d\theta } \cr & = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {{{\sin }^2}\theta \sin \theta d\theta } \cr & = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {\left( {1 - {{\cos }^2}\theta } \right)\sin \theta d\theta } \cr & {\text{Integrate}} \cr & = \frac{{\sqrt 3 }}{3}\left[ { - \cos \theta + \frac{{{{\cos }^3}\theta }}{3}} \right]_0^{\pi /3} \cr & \cr & {\text{evaluate the limits}} \cr & = \frac{{\sqrt 3 }}{3}\left[ { - \cos \left( {\frac{\pi }{3}} \right) + \frac{1}{3}{{\cos }^3}\left( {\frac{\pi }{3}} \right)} \right] - \frac{{\sqrt 3 }}{3}\left[ { - \cos \left( 0 \right) + \frac{1}{3}{{\cos }^3}\left( 0 \right)} \right] \cr & = \frac{{\sqrt 3 }}{3}\left[ { - \frac{1}{2} + \frac{1}{{24}}} \right] - \frac{{\sqrt 3 }}{3}\left[ { - 1 + \frac{1}{3}} \right] \cr & = - \frac{{11\sqrt 3 }}{{72}} + \frac{{2\sqrt 3 }}{3} \cr & = \frac{{37\sqrt 3 }}{{72}} \cr} $$
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