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: 24


$2\tan \sqrt{x}+C$

Work Step by Step

Let $u=\sqrt{x}$. Then $u=x^\frac{1}{2}$, so $du=\frac{1}{2}x^{-\frac{1}{2}}dx=\frac{1}{2\sqrt{x}}dx$. This means that $\frac{1}{\sqrt{x}}dx=2du$. $\int\frac{\sec^2\sqrt{x}}{\sqrt{x}}dx$ $=\int\sec^2\sqrt{x}*\frac{1}{\sqrt{x}}dx$ $=\int\sec^2u*2 du$ $=2\tan u+C$ $=2\tan \sqrt{x}+C$
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