Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions - Exercises - Page 415: 36


$$\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{8} \cosh x \sinh x+\frac{3}{8} x+C$$

Work Step by Step

Given $$\int \cosh ^{4} x d x$$ Use $$\int \cosh ^{n} x d x=\frac{1}{n} \cosh ^{n-1} x \sinh x+\frac{n-1}{n} \int \cosh ^{n-2} x d x$$ Then for $n=4$ \begin{aligned} \int \cosh ^{4} x d x &=\frac{1}{4} \cosh ^{4-1} x \sinh x+\frac{4-1}{4} \int \cosh ^{4-2} x d x \\ &=\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{4} \int \cosh ^{2} x d x \\ &=\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{4}\left[\frac{1}{2} \cosh ^{2-1} x \sinh x+\frac{2-1}{2} \int \cosh ^{2-2} x d x\right] \\ &=\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{4}\left[\frac{1}{2} \cosh x \sinh x+\frac{1}{2} \int 1 \, d x\right] \\ &=\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{8} \cosh x \sinh x+\frac{3}{8} x+C \end{aligned}
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