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 - Page 508: 65


$$\int \frac{sin2x}{1+cos^{4}x}dx=-arctan(cos^{2}x)+C$$

Work Step by Step

Observe that: $$sin2x=2sinxcosx={(-cos^{2}x)}'$$ Hence: $$\int \frac{sin2x}{1+cos^{4}x}dx=-\int \frac{1}{1+(cos^{2}x)^{2}}d(cos^{2}x)$$ $$=-arctan(cos^{2}x)+C$$
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