University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 551: 66


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) When $ 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 $ or, $[\tan^{-1} t]_x^{\infty}=[-\dfrac{1} {t}+\dfrac{1} {3t^3}-\dfrac{1} {5t^5}+.....]_x^{\infty}$ So, $\tan^{-1} x=\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \gt 1$ b) When $ 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}+.....) dt $ or, $[\tan^{-1} t]_{-\infty}^ {x} =[-\dfrac{1} {t}+\dfrac{1} {3t^3}-\dfrac{1} {5t^5}+.....]\int_{-\infty}^ {x} $ So, $\tan^{-1} x=-\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \lt -1$
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