Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 72

Answer

$$\frac{\pi }{{120}}$$

Work Step by Step

$$\eqalign{ & \int_0^{6/5} {\frac{{dx}}{{25{x^2} + 36}}} \cr & {\text{substitute }}u = 5x,{\text{ }}du = 5dx \cr & {\text{express the limits in terms of }}u \cr & x = 0{\text{ implies }}u = 5\left( 0 \right) = 0 \cr & x = 6/5{\text{ implies }}u = 5\left( {6/5} \right) = 6 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^{6/5} {\frac{{dx}}{{25{x^2} + 36}}} = \frac{1}{5}\int_0^6 {\frac{{du}}{{{u^2} + {6^2}}}} \cr & {\text{find the antiderivative}} \cr & = \frac{1}{5}\left. {\left( {\frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{u}{6}} \right)} \right)} \right|_0^6 \cr & {\text{use the fundamental theorem}} \cr & = \frac{1}{5}\left( {\frac{1}{6}{{\tan }^{ - 1}}\left( {\frac{6}{6}} \right)} \right) - \frac{1}{5}\left( {\frac{1}{6}{{\tan }^{ - 1}}\left( 0 \right)} \right) \cr & {\text{simplify}} \cr & = \frac{1}{5}\left( {\frac{\pi }{{24}}} \right) \cr & = \frac{\pi }{{120}} \cr} $$
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