Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 58

Answer

$$ \frac{1}{2}[\sin x \sinh x-\cos x \cosh x]+C$$

Work Step by Step

Given $$ \int(\sin x)(\cosh x) d x $$ Let \begin{align*} u&=\sin x \ \ \ \ \ \ \ \ \ \ dv= \cosh x d x \\ du&=\cos xdx\ \ \ \ \ \ \ \ \ v=\sinh x \end{align*} Then \begin{align*} \int(\sin x)(\cosh x) d x&= \sin x\sinh x-\int\cos x \sinh xdx \end{align*} Let \begin{align*} u&=\cos x \ \ \ \ \ \ \ \ \ \ dv= \sinh x d x \\ du&=-\sin xdx\ \ \ \ \ \ \ \ \ v=\cosh x \end{align*} Then \begin{aligned} \int \sin x \cosh x \, d x &=\sin x \cdot \sinh x-\int \sinh x \cdot \cos x \, d x \\ &=\sin x \cdot \sinh x-\left[\cos x \cdot \cosh x-\int \cosh x \cdot(-\sin x) d x\right] \\ &=\sin x \cdot \sinh x-\left[\cos x \cdot \cosh x+\int \sin x \cosh x \, d x\right] \\ &=\sin x \sinh x-\cos x \cosh x-\int \sin x \cosh x \, d x \end{aligned} Hence $$\int \sin x \cosh x \, d x=\frac{1}{2}[\sin x \sinh x-\cos x \cosh x]+C$$
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