University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 89


$$\int^{0}_{-\pi/3}\sec x\tan xdx=-1$$

Work Step by Step

$$A=\int^{0}_{-\pi/3}\sec x\tan xdx$$ $$A=\sec x\Big]^{0}_{-\pi/3}=\frac{1}{\cos x}\Big]^{0}_{-\pi/3}$$ $$A=\frac{1}{\cos0}-\frac{1}{\cos(-\pi/3)}$$ $$A=1-\frac{1}{\frac{1}{2}}$$ $$A=1-2=-1$$
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