Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 42

Answer

$= \frac{\pi}{3}$

Work Step by Step

$\int^{\frac{1}{\sqrt 2}}_{\frac{1}{2}} \frac{4}{\sqrt {1-x^{2}}} dx$ $= 4\int^{\frac{1}{\sqrt 2}}_{\frac{1}{2}} \frac{1}{\sqrt {1-x^{2}}} dx$ $= 4 [\arcsin x |^{\frac{1}{\sqrt 2}}_{\frac{1}{2}}]$ $= 4[\arcsin (\frac{1}{\sqrt 2}) - \arcsin (\frac{1}{2})]$ $= 4[(\frac{\pi}{4}) - (\frac{\pi}{6})]$ $= 4[(\frac{6\pi}{24}) - (\frac{4\pi}{24})]$ $= 4[(\frac{2\pi}{24}) ]$ $= 4(\frac{\pi}{12})$ $= \frac{4\pi}{12}$ $= \frac{\pi}{3}$
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