Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises: 34

Answer

$$ - \frac{{{{\left( {\coth x} \right)}^3}}}{3} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\coth }^2}x{{\operatorname{csch} }^2}x} dx \cr & {\text{substitute }}u = \coth x,{\text{ }}du = - {\operatorname{csch} ^2}xdx{\text{ }} \cr & \int {{{\coth }^2}x{{\operatorname{csch} }^2}x} dx = \int {{u^2}} \left( { - du} \right) \cr & = - \int {{u^2}du} \cr & {\text{find the antiderivative by the power rule}} \cr & = - \frac{{{u^3}}}{3} + C \cr & {\text{ with}}\,\,\,u = \coth x \cr & = - \frac{{{{\left( {\coth x} \right)}^3}}}{3} + C \cr} $$
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