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: 36

Answer

$$\frac{1}{4}\sinh 2x - \frac{1}{2}x + C$$

Work Step by Step

$$\eqalign{ & \int {{{\sinh }^2}xdx} \cr & {\text{hyperbolic identity }}{\sinh ^2}x = \frac{{\cosh 2x - 1}}{2} \cr & = \int {\frac{{\cosh 2x - 1}}{2}} dx \cr & {\text{split the integrand}} \cr & = \int {\frac{{\cosh 2x}}{2}} dx - \int {\frac{1}{2}} dx \cr & = \frac{1}{4}\int {\cosh 2x\left( 2 \right)} dx - \int {\frac{1}{2}} dx \cr & {\text{find the antiderivative}} \cr & = \frac{1}{4}\sinh 2x - \frac{1}{2}x + C \cr} $$
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