University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.5 - Integral Tables and Computer Algebra Systems - Exercises - Page 451: 44


$$\int 2\sin^2 t\sec^4t dt=\frac{2}{3}\tan^3t+C$$

Work Step by Step

$$A=\int 2\sin^2 t\sec^4t dt$$ $$A=2\int\frac{\sin^2t}{\cos^2t}\sec^2tdt$$ $$A=2\int\tan^2t\sec^2tdt$$ We set $u=\tan t$, which means $$du=\sec^2tdt$$ Therefore, $$A=2\int u^2du=\frac{2u^3}{3}+C$$ $$A=\frac{2}{3}\tan^3t+C$$
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