Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3 - Page 286: 27


$\frac{1}{2}\tan (x^2)+C$

Work Step by Step

Let $u=x^2$. Then $du=2xdx$, so $xdx=\frac{du}{2}$. $\int x\sec^2 (x^2) dx$ $=\int \sec^2(x^2)xdx$ $=\int \sec^2 u \frac{du}{2}$ $=\frac{1}{2}\tan u+C$ $=\frac{1}{2}\tan (x^2)+C$
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