Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 514: 19


$\int$$\frac{cos^4 x}{sin^6x}dx=-\frac{(\cot x)^5}{5}+C$

Work Step by Step

$\int$$\frac{cos^4 x}{sin^6x}dx$ First, split the integral. $\int$$\frac{cos^4 x}{sin^4x}\times\frac{1}{sin^2x}dx$ $\int$$cot ^4x \times csc^2x dx$ Then, Let $u= cot x$ and $du= -csc^2 x$ . $\int-u^4 du$ $-\frac{u^5}{5}$ Subtitute back $u=\cot x $ to form a final answer. $-\frac{(\cot x)^5}{5}+C$
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