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: 64



Work Step by Step

Since $u=4x$, then $du= 4dx$, and hence when $x:1/4 \to 1/2$ then $u:1\to 2$. Now, we have $$\int_{1/4}^{1/2}\frac{ dx}{x\sqrt{16x^2-1}}=\frac{1}{4}\int \frac{ du}{u\sqrt{u^2-1}}\\=\sec^{-1}u|_1^2 =\sec^{-1}2-\sec^{-1}1=\frac{\pi}{3}-0=\frac{\pi}{3}.$$
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