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: 65


$$\int\frac{\tan^2x}{\csc x}dx=\cos x+\frac{1}{\cos x}+C$$

Work Step by Step

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