Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 409: 36


$$\int^{\pi/4}_0(\sec \theta\tan \theta)d\theta=\sqrt 2-1$$

Work Step by Step

$$A=\int^{\pi/4}_0(\sec \theta\tan \theta)d\theta$$ According to Table 1, we have $$\int \sec x\tan xdx=\sec x+C$$ Therefore, $$A=(\sec\theta)\Bigg]^{\pi/4}_0$$ $$A=(\sec\frac{\pi}{4}-\sec0)$$ $$A=\sqrt 2-1$$
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