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: 33



Work Step by Step

Substitute $u= tan x$ so that $dx=\frac{du}{sec^{2}x}$ in the given integral. Then, $\int sec^{2}xtanxdx=\int sec^{2}x\, u\frac{du}{sec^{2}x}$ $=\int udu= \frac{u^{2}}{2}+C$ Undoing substitution, we get $\int sec^{2}x tanxdx= \frac{1}{2}tan^{2}x+C$
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