University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 435: 64


$$\int\frac{\sin^3x}{\cos^4 x}dx=\frac{1}{3\cos^3x}-\frac{1}{\cos x}+C$$

Work Step by Step

$$A=\int\frac{\sin^3x}{\cos^4 x}dx=\int\frac{\sin^2x}{\cos^4x}(\sin xdx)$$ $$A=-\int\frac{1-\cos^2x}{\cos^4x}d(\cos x)$$ We set $u=\cos x$ $$A=-\int\frac{1-u^2}{u^4}du$$ $$A=-\int\Big(\frac{1}{u^4}-\frac{1}{u^2}\Big)du$$ $$A=-\int(u^{-4}-u^{-2})du$$ $$A=-\Big(\frac{u^{-3}}{-3}-\frac{u^{-1}}{-1}\Big)+C$$ $$A=\frac{u^{-3}}{3}-u^{-1}+C$$ $$A=\frac{1}{3u^3}-\frac{1}{u}+C$$ $$A=\frac{1}{3\cos^3x}-\frac{1}{\cos x}+C$$
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