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: 50



Work Step by Step

$$A=\int\sec\theta\tan\theta\sqrt{1+\sec\theta}d\theta$$ We set $a=1+\sec\theta$, which means $$da=\sec\theta\tan\theta d\theta$$ Therefore, $$A=\int \sqrt ada=\int a^{1/2}da$$ $$A=\frac{a^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{2a^{3/2}}{3}+C$$ $$A=\frac{2(1+\sec\theta)^{3/2}}{3}+C$$
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