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: 34

Answer

$\frac{1}{2}\sec^2\phi+C$

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.