Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 238: 14

Answer

a. $F(x)=-cot(x)+C$ b. $F(x)=cot(\frac{3x}{2})+C$ c. $F(x)=x+4cot(2x)+C$

Work Step by Step

a. Given $f(x)=csc^2x$, we have $F(x)=-cot(x)+C$ as its antiderivative, where $C$ is a constant. b. Given $f(x)=-\frac{3}{2}csc^2(\frac{3x}{2})$, we have $F(x)=cot(\frac{3x}{2})+C$ as its antiderivative, where $C$ is a constant. c. Given $f(x)=1-8csc^2(2x)$, we have $F(x)=x+4cot(2x)+C$ as its antiderivative, where $C$ is a constant.
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