Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 22



Work Step by Step

We have the Maclaurin series for $\dfrac{1}{ (1-x)}$ as follows $$\dfrac{1}{ (1-x)}=1+x+x^2+x^3+...$$ Replace $\sin x$ for $x$ in the above form. Therefore, we have: $$\dfrac{1}{1+\sin x}=1-\sin x+\sin^2 x +... \\=1-(x-\dfrac{x^3}{6}) +(x-\dfrac{x^3}{6})^2 +...\\ =1-x+x^2-\dfrac{5x^3}{6}+...$$ So, the four terms of the Maclaurin series for $f(x)=\dfrac{1}{1+\sin x}$ are: $$1-x+x^2-\dfrac{5x^3}{6}+...$$
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