Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 30


$$\int\frac{\sec^2 x}{\tan^2 x}dx=-\cot x+C$$

Work Step by Step

$$A=\int\frac{\sec^2 x}{\tan^2 x}dx$$ Let $u=\tan x$. We would have $du=\sec^2 xdx$. Substitute into $A$, we have $$A=\int\frac{1}{u^2}du$$ $$A=\int u^{-2}du$$ $$A=\frac{u^{-1}}{-1}+C$$ $$A=-\frac{1}{u}+C$$ $$A=-\frac{1}{\tan x}+C$$ $$A=-\cot x+C$$
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