Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises: 33

Answer

$$\frac{{x\sqrt {4 + {x^2}} }}{2} - 2\ln \left( {x + \sqrt {4 + {x^2}} } \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{x^2}}}{{\sqrt {4 + {x^2}} }}dx} \cr & {\text{substitute }}x = 2\tan \theta ,{\text{ }}dx = 2{\sec ^2}\theta d\theta \cr & = \int {\frac{{{x^2}}}{{\sqrt {4 + {x^2}} }}dx} = \int {\frac{{4{{\tan }^2}\theta }}{{\sqrt {4 + 4{{\tan }^2}\theta } }}} \left( {2{{\sec }^2}\theta d\theta } \right) \cr & = \int {\frac{{8{{\tan }^2}\theta {{\sec }^2}\theta }}{{2\sqrt {1 + {{\tan }^2}\theta } }}} d\theta \cr & {\text{pythagorean identity}} \cr & = \int {\frac{{4\left( {{{\sec }^2}\theta - 1} \right){{\sec }^2}\theta d\theta }}{{\sqrt {{{\sec }^2}\theta } }}} = 4\int {\left( {{{\sec }^2}\theta - 1} \right)\sec \theta d\theta } \cr & = 4\int {\left( {{{\sec }^3}\theta - \sec \theta } \right)d\theta } \cr & {\text{integrating}} \cr & = 4\left( {\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| - \ln \left| {\sec \theta + \tan \theta } \right|} \right) + C \cr & {\text{simplify}} \cr & = 4\left( {\frac{1}{2}\sec \theta \tan \theta - \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right|} \right) + C \cr & = 2\sec \theta \tan \theta - 2\ln \left| {\sec \theta + \tan \theta } \right| \cr & {\text{with sec}}\theta = \frac{{\sqrt {4 + {x^2}} }}{2}{\text{ and tan}}\theta = \frac{x}{2} \cr & = 2\left( {\frac{{\sqrt {4 + {x^2}} }}{2}} \right)\left( {\frac{x}{2}} \right) - 2\ln \left| {\frac{{\sqrt {4 + {x^2}} }}{2} + \frac{x}{2}} \right| + C \cr & {\text{simplify}} \cr & = \frac{{x\sqrt {4 + {x^2}} }}{2} - 2\ln \left| {\frac{{x + \sqrt {4 + {x^2}} }}{2}} \right| + C \cr & = \frac{{x\sqrt {4 + {x^2}} }}{2} - 2\ln \left( {x + \sqrt {4 + {x^2}} } \right) + C \cr} $$
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