Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 76


$$\sec^{-1}12 -\sec^{-1}4 .$$

Work Step by Step

Since $(\sec^{-1}x)'=\frac{1}{x\sqrt{1-x^2}}dx $, we have $$\int_4^{12} \frac{1}{x\sqrt{1-x^2}}dx =\sec^{-1}x|_4^{12} =\sec^{-1}12 -\sec^{-1}4 .$$
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