Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 35

Answer

$arctan(x^{2})=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(x)}^{(4n+2)}}{2n+1}$

Work Step by Step

$f(x)=arctan(x^{2})=tan^{-1}(x^{2})$ Since, $\int \frac{1}{1+t^{2}}dt=tan^{-1}t$ $\frac{1}{1+t^{2}}=\frac{1}{1-(-t^{2})}$ $=\Sigma_{n=0}^{\infty}(-t^{2})^{n}$ $=\Sigma_{n=0}^{\infty}(-1)^{n}(t^{2n})$ $tan^{-1}t=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(t)}^{(2n+1)}}{2n+1}$ Thus, $arctan(x^{2})=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(x)}^{(4n+2)}}{2n+1}$
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