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: 18

Answer

$$\ln \left| {\frac{{x + \sqrt {{x^2} - 49} }}{7}} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{\sqrt {{x^2} - 49} }}} \cr & {\text{substitute }}x = 7\sec \theta ,{\text{ }}dx = 7\sec \theta \tan \theta d\theta \cr & = \int {\frac{{7\sec \theta \tan \theta d\theta }}{{\sqrt {{{\left( {7\sec \theta } \right)}^2} - 49} }}} \cr & = \int {\frac{{7\sec \theta \tan \theta d\theta }}{{\sqrt {49{{\sec }^2}\theta - 49} }}} \cr & = \int {\frac{{7\sec \theta \tan \theta d\theta }}{{7\sqrt {{{\sec }^2}\theta - 1} }}} \cr & = \int {\frac{{7\sec \theta \tan \theta d\theta }}{{7\sqrt {{{\tan }^2}\theta } }}} \cr & = \int {\sec \theta d\theta } \cr & {\text{integrating}} \cr & = \ln \left| {\sec \theta + \tan \theta } \right| + C \cr & {\text{substituting tan}}\theta {\text{ and sec}}\theta \cr & = \ln \left| {\frac{x}{7} + \frac{{\sqrt {{x^2} - 49} }}{7}} \right| + C \cr & {\text{simplify}} \cr & = \ln \left| {\frac{{x + \sqrt {{x^2} - 49} }}{7}} \right| + C \cr} $$
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