Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 252: 34


$\int sec(x)tan(x)dx=sec(x)+C$

Work Step by Step

This is an identity: $\int sec(x)tan(x)dx=sec(x)+C$ However, here it is arithmetically: $\int sec(x)tan(x)dx$ $=\int\frac{sin(x)}{cos^2(x)}dx;$ $u=cos(x);$ $du=-sin(x)dx$ $=-\int\frac{du}{u^2}$ $=\frac{1}{u}+C$ $=sec(x)+C$
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