Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 485: 34



Work Step by Step

$\int\frac{\sin\phi}{\cos^3\phi}\ d\phi$ Let $u=\cos \phi$. Then $du=-\sin\phi\ d\phi$, and $\sin\phi\ d\phi=-du$. $=\int\frac{-1}{u^3}\ du$ $=-\int u^{-3}\ du$ $=-\frac{u^{-2}}{-2}+C$ $=\frac{1}{2u^2}+C$ $=\frac{1}{2\cos^2 \phi}+C$ $=\boxed{\frac{1}{2}\sec^2\phi+C}$
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