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: 55



Work Step by Step

We have $$\int^{1/2}_0\frac{ dx}{\sqrt{1-x^2}}=\sin^{-1}x|^{1/2}_0\\ =\sin^{-1}\frac{1}{2}-\sin^{-1}0=\frac{\pi}{6}-0=\frac{\pi}{6}.$$
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