University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 37



Work Step by Step

Plug in: t=tan x so that $dx=\frac{dt}{sec^{2}x}$ in the given integral. Then $\int sec^{2}x tan^{2}xdx= \int sec^{2}x t^{2}\frac{dt}{sec^{2}x}$ $=\int t^{2}dt= \frac{t^{3}}{3}+C$ Undoing substitution, we have $\int sec^{2}x tan^{2}xdx=\frac{tan^{3}x}{3}+C$
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