Answer
$$\sec^{-1}12 -\sec^{-1}4 .$$
Work Step by Step
Since $(\sec^{-1}x)'=\frac{1}{x\sqrt{1-x^2}}dx $, we have
$$\int_4^{12} \frac{1}{x\sqrt{1-x^2}}dx =\sec^{-1}x|_4^{12}
=\sec^{-1}12 -\sec^{-1}4 .$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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