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: 45

Answer

$$ - \frac{1}{{\sqrt {{x^2} - 1} }} - {\sec ^{ - 1}}x + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{x{{\left( {{x^2} - 1} \right)}^{3/2}}}}} \cr & {\text{substitute }}x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta d\theta \cr & \int {\frac{{dx}}{{x{{\left( {{x^2} - 1} \right)}^{3/2}}}}} = \int {\frac{{\sec \theta \tan \theta d\theta }}{{\sec \theta {{\left( {{{\sec }^2}\theta - 1} \right)}^{3/2}}}}} \cr & {\text{pythagorean identity}} \cr & \int {\frac{{\tan \theta d\theta }}{{{{\left( {{{\tan }^2}\theta } \right)}^{3/2}}}}} \cr & = \int {\frac{{\tan \theta d\theta }}{{{{\tan }^3}\theta }}d\theta } = \int {{{\cot }^2}\theta d\theta } \cr & \int {\left( {{{\csc }^2}\theta - 1} \right)d\theta } \cr & {\text{integrating}} \cr & = - \cot \theta - \theta + C \cr & {\text{with }}\cot \theta = \frac{1}{{\sqrt {{x^2} - 1} }} \cr & = - \frac{1}{{\sqrt {{x^2} - 1} }} - {\sec ^{ - 1}}x + C \cr} $$
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