Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.4 - Trigonometric Substitutions - Exercises 8.4 - Page 467: 11

Answer

$$\sqrt {{y^2} - 49} - 7{\sec ^{ - 1}}\left( {\frac{y}{7}} \right) + C $$

Work Step by Step

$$\eqalign{ & \int {\frac{{\sqrt {{y^2} - 49} }}{y}} dy \cr & = \int {\frac{{\sqrt {{y^2} - {7^2}} }}{y}} dy \cr & {\text{We set}}{\text{, }}y = 7\sec \theta,{\text{ }}dy = 7\sec \theta \tan \theta d\theta,{\text{ }}0 \leqslant \theta \leqslant \frac{\pi }{2} \cr & {\text{With these substitutions}}{\text{, we have}} \cr & = \int {\frac{{\sqrt {{{\left( {7\sec \theta } \right)}^2} - {7^2}} }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr & = \int {\frac{{\sqrt {49{{\sec }^2}\theta - 49} }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr & = \int {\frac{{\sqrt {49\left( {{{\sec }^2}\theta - 1} \right)} }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr & {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr & = \int {\frac{{\sqrt {49{{\tan }^2}\theta } }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr & {\text{simplifying}} \cr & = \int {\frac{{\sqrt {49{{\tan }^2}\theta } }}{{7\sec \theta }}} \left( {7\sec \theta \tan \theta } \right)d\theta \cr & = 7\int {{{\tan }^2}\theta } d\theta \cr & = 7\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr & {\text{integrate}} \cr & = 7\tan \theta - 7\theta + C \cr & {\text{with tan }}\theta = \frac{{\sqrt {{y^2} - 49} }}{7}{\text{ and }}\theta = {\sec ^{ - 1}}\left( {\frac{y}{7}} \right) \cr & = 7\left( {\frac{{\sqrt {{y^2} - 49} }}{7}} \right) - 7{\sec ^{ - 1}}\left( {\frac{y}{7}} \right) + C \cr & = \sqrt {{y^2} - 49} - 7{\sec ^{ - 1}}\left( {\frac{y}{7}} \right) + C \cr} $$
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