University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 14


$$\int 4x\sec^22xdx=2x\tan2x+\ln|\cos2x|+C$$

Work Step by Step

$$A=\int 4x\sec^22xdx$$ Set $u=4x$ and $dv=\sec^22xdx$ Then we have $du=4dx$ and $v=\frac{1}{2}\tan2x$ Using the formula $\int udv= uv-\int vdu$: $$A=4x\times\frac{1}{2}\tan2x-\int\frac{1}{2}\tan2x\times4dx$$ $$A=2x\tan2x-2\int\tan2xdx$$ $$A=2x\tan2x-2\int\frac{\sin2x}{\cos2x}dx$$ Take $a=\cos2x$, then $$da=-2\sin2xdx$$ That means $$\sin2xdx=-\frac{1}{2}da$$ Therefore, $$A=2x\tan2x-2\Big(-\frac{1}{2}\Big)\int\frac{da}{a}$$ $$A=2x\tan2x+1(\ln|a|)+C$$ $$A=2x\tan2x+\ln|\cos2x|+C$$
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