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: 31


$$\int x\sec x^2dx=\frac{1}{2}\ln|\sec x^2+\tan x^2|+C$$

Work Step by Step

$$A=\int x\sec x^2dx$$ We set $a=x^2$, which means $$da=2xdx$$ $$xdx=\frac{1}{2}da$$ Therefore, $$A=\frac{1}{2}\int \sec ada$$ $$A=\frac{1}{2}\ln|\sec a+\tan a|+C$$ $$A=\frac{1}{2}\ln|\sec x^2+\tan x^2|+C$$
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