Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises: 4

Answer

$\frac{\cos^2 x}{2}-\ln |\cos x|+C$

Work Step by Step

$\int\frac{\sin^3 x}{\cos x}dx$ $=\int\frac{\sin x\sin^2 x}{\cos x}dx$ $=\int\frac{\sin x(1-\cos^2 x)}{\cos x}dx$ Let $u=\cos x$. Then $du=-\sin x\ dx$, and $\sin x\ dx=-du$. $=\int\frac{1-u^2}{u}*(-1)du$ $=-\int(\frac{1}{u}-\frac{u^2}{u})du$ $=-\int(\frac{1}{u}-u)du$ $=-(\ln |u|-\frac{u^2}{2}+C)$ $=-(\ln |\cos x|-\frac{\cos^2 x}{2}+C)$ $=\boxed{\frac{\cos^2 x}{2}-\ln |\cos x|+C}$
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