## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.8 Hyperbolic Functions And Hanging Cables - Exercises Set 6.8 - Page 481: 32

#### Answer

$$= - \frac{1}{3}\coth 3x + C$$

#### Work Step by Step

\eqalign{ & \int {{{\operatorname{csch} }^2}\left( {3x} \right)} dx \cr & {\text{substitute }}u = 3x,{\text{ }}du = 3dx \cr & = \int {{{\operatorname{csch} }^2}\left( {3x} \right)} dx = \int {{{\operatorname{csch} }^2}u\left( {\frac{1}{3}du} \right)} \cr & = \frac{1}{3}\int {{{\operatorname{csch} }^2}udu} \cr & {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{see page 476}}} \right) \cr & = \frac{1}{3}\left( { - \coth u} \right) + C \cr & = - \frac{1}{3}\coth u + C \cr & {\text{write in terms of }}x \cr & = - \frac{1}{3}\coth 3x + C \cr}

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