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: 58

Answer

$$\frac{1}{5}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{5}} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{{x^2} - 6x + 34}}} \cr & {\text{completing the square}} \cr & = \int {\frac{{dx}}{{{x^2} - 6x + 9 + 25}}} = \int {\frac{{dx}}{{\left( {{x^2} - 6x + 9} \right) + 25}}} \cr & = \int {\frac{{dx}}{{{{\left( {x - 3} \right)}^2} + 25}}} \cr & u = x - 3,{\text{ }}du = dx \cr & = \int {\frac{{du}}{{{u^2} + 25}}} \cr & {\text{substitute }}u = 5\tan \theta ,{\text{ }}du = 5{\sec ^2}\theta d\theta \cr & \int {\frac{{du}}{{{u^2} + 25}}} = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{{{\left( {5\tan \theta } \right)}^2} + 25}}} \cr & = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{25{{\tan }^2}\theta + 25}}} = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{25\left( {{{\tan }^2}\theta + 1} \right)}}} \cr & {\text{pythagorean identity}} \cr & = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{25\left( {{{\sec }^2}\theta } \right)}}} = \frac{1}{5}\int {d\theta } \cr & {\text{integrating}} \cr & = \frac{1}{5}\theta + C \cr & u = 5\tan \theta \cr & = \frac{1}{5}{\tan ^{ - 1}}\left( {\frac{u}{5}} \right) + C \cr & u = x - 3 \cr & = \frac{1}{5}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{5}} \right) + C \cr} $$
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