Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 32

Answer

$\frac{sin^3x}{3} - \frac{sin^5x}{5} +C$

Work Step by Step

Find the indefinite integral $\int \frac{tan^2x}{sec^5x}dx$, Convert tangents to secants $\int \frac{sec^2x -1}{sec^5x}dx$ $\int[\frac{1}{sec^3x}- \frac{1}{sec^5x}]dx$ $\int[cos^3x- cos^5x]dx$ $\int cos^3xdx- \int cos^5xdx$, Break into two integrals $\int (1-sin^2x)cosxdx- \int (1-sin^2x)^2 cosxdx$ $\int (1-sin^2x)cosxdx - \int (1-2sin^2x+sin^4x)cosxdx$ Use u-substitution, let $u=sinx$, $du=cosxdx$ $\int (1-u^2)du - \int (1-2u^2 +u^4)du$, Integrate $-\frac{u^3}{3}+ \frac{2u^3}{3}- \frac{u^5}{5} +C$ $\frac{sin^3x}{3} - \frac{sin^5x}{5} +C$, Resubstitute
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