Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 12



Work Step by Step

$\int^{\frac{\pi}{0}}_{0}4\frac{sin u}{cos^2u}du$ $=\frac{4}{cosu}]^{\frac{\pi}{3}}_0$ =$\frac{4}{\frac{1}{2}}-\frac{4}{1}$ =4
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