Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 69


$$-\frac{1}{2} (\ln( \cos^{-1}x))^2+c.$$

Work Step by Step

Let $u=\ln( \cos^{-1}x)$, then $du =-\frac{dx}{(\cos^{1-}x)\sqrt{1-x^2}}$. Now, we have $$\int \frac{\ln( \cos^{-1}x)dx}{(\cos^{1-}x)\sqrt{1-x^2}}=-\int udu\\ =-\frac{1}{2} u^2+c=-\frac{1}{2} (\ln( \cos^{-1}x))^2+c.$$
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