Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 27



Work Step by Step

Since $(\sin(\theta/3))'=(1/3)3\cos(\theta/3)$, we have $$\int_{0 }^{\pi/2 } \cos(\theta/3)d \theta= 3\sin(\theta/3) |_{0 }^{\pi /2}\\ = 3\sin(\pi/6)- 3\sin(0)=\frac{3}{2}.$$
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