Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 634: 66

Answer

a) $\tan^{-1} x=\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \gt 1$ b) $\tan^{-1} x=-\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \lt -1$

Work Step by Step

a) For $ x \gt 1$ : $\int_x^{\infty}\dfrac{1}{1+t^2}dt= \int_x^{\infty} (\dfrac{1} {t^2}-\dfrac{1} {t^4}+\dfrac{1} {t^6}+.....) dt\\ [\tan^{-1} t]_x^{\infty}=[-\dfrac{1} {t}+\dfrac{1} {3t^3}-\dfrac{1} {5t^5}+.....]_x^{\infty}\\ \tan^{-1} x=\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \gt 1$ (b) For $ x \lt -1$ : $\int_{-\infty}^ {x} \dfrac{1}{1+t^2}dt= \int_{-\infty}^ {x} (\dfrac{1} {t^2}-\dfrac{1} {t^4}+\dfrac{1} {t^6}+.....) \space dt \space or, \\ [\tan^{-1} t]_{-\infty}^ {x} =[-\dfrac{1} {t}+\dfrac{1} {3t^3}-\dfrac{1} {5t^5}+.....]\int_{-\infty}^ {x} \space or, \tan^{-1} x=-\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \lt -1$
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