Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 55


$\int\frac{cosx-1}{x}dx=c+\Sigma_{n=1}^{\infty}\frac{(-1)^{n}x^{2n}}{2n. (2n)!}$ $R=\infty$

Work Step by Step

$\frac{cosx-1}{x}=-\frac{1}{x}+\frac{cosx}{x}$ $=-\frac{1}{2!}x+\frac{1}{4!}x^{3}-\frac{1}{6!}x^{5}+......+\frac{(-1)^{n}x^{2n-1}}{2n!}x+...$ Now, $\int\frac{cosx-1}{x} dx=\int(-\frac{1}{2!}x+\frac{1}{4!}x^{3}-\frac{1}{6!}x^{5}+...+\frac{(-1)^{n}x^{2n-1}}{2n!}x+...)dx$ $=c+\Sigma_{n=1}^{\infty}\frac{(-1)^{n}x^{2n}}{2n. (2n)!}$ Hence, $\int\frac{cosx-1}{x}dx=c+\Sigma_{n=1}^{\infty}\frac{(-1)^{n}x^{2n}}{2n. (2n)!}$ $R=\infty$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.