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: 56


$c+\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{4n+3}}{(4n+3)(2n+1)}$ $R=1$

Work Step by Step

$arctan(x^{2})=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{4n+2}}{(2n+1)}$ $=x^{2}-\frac{1}{3}x^{6}+\frac{1}{5}x^{10}-\frac{1}{7}x^{14}+......+\frac{(-1)^{n}x^{4n+2}}{2n+1}+...$ Now, $\int arctan(x^{2})dx=(x^{2}-\frac{1}{3}x^{6}+\frac{1}{5}x^{10}-\frac{1}{7}x^{14}+......+\frac{(-1)^{n}x^{4n+2}}{2n+1}+...)dx$ $=c+\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{4n+3}}{(4n+3)(2n+1)}$ $R=1$
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