## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 8: Techniques of Integration - Section 8.4 - Trigonometric Substitutions - Exercises 8.4 - Page 467: 31

#### Answer

$$\frac{1}{2}\ln \left| {\sqrt {{x^2} - 1} } \right| + \frac{1}{2}{x^2} + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{{x^3}dx}}{{{x^2} - 1}}} \cr & {\text{Use the trigonometric substitutions:}} \cr & {\text{ }}x = \sec \theta,{\text{ }}dt = \sec \theta \tan \theta d\theta \cr & {\text{With these substitutions}}{\text{, we have}} \cr & \int {\frac{{{x^3}dx}}{{{x^2} - 1}}} = \int {\frac{{{{\sec }^3}\theta \left( {\sec \theta \tan \theta d\theta } \right)}}{{{{\sec }^2}\theta - 1}}} \cr & = \int {\frac{{{{\sec }^4}\theta \tan \theta }}{{{{\sec }^2}\theta - 1}}d\theta } \cr & {\text{Use the fundamental identity }}{\sec ^2}\theta = 1 + {\tan ^2}\theta \cr & = \int {\frac{{{{\sec }^4}\theta \tan \theta }}{{{{\tan }^2}\theta }}d\theta } \cr & = \int {\frac{{{{\sec }^4}\theta }}{{\tan \theta }}d\theta } \cr & = \int {\frac{{{{\sec }^2}\theta }}{{\tan \theta }}{{\sec }^2}\theta d\theta } \cr & = \int {\left( {\frac{{1 + {{\tan }^2}\theta }}{{\tan \theta }}} \right){{\sec }^2}\theta d\theta } \cr & = \int {\left( {\frac{1}{{\tan \theta }} + \tan \theta } \right){{\sec }^2}\theta d\theta } \cr & {\text{Integrate }} \cr & = \ln \left| {\tan \theta } \right| + \frac{{{{\tan }^2}\theta }}{2} + C \cr & {\text{Write in terms of }}x,\tan \theta = \sqrt {{x^2} - 1} \cr & = \ln \left| {\sqrt {{x^2} - 1} } \right| + \frac{{\left( {\sqrt {{x^2} - 1} } \right)}}{2} + C \cr & = \frac{1}{2}\ln \left| {\sqrt {{x^2} - 1} } \right| + \frac{{{x^2} - 1}}{2} + C \cr & = \frac{1}{2}\ln \left| {\sqrt {{x^2} - 1} } \right| + \frac{1}{2}{x^2} - \frac{1}{2} + C \cr & {\text{Combine constants}} \cr & = \frac{1}{2}\ln \left| {\sqrt {{x^2} - 1} } \right| + \frac{1}{2}{x^2} + C \cr}

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