Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises: 39


$$\int\csc xdx=-\ln|\csc x+\cot x|+C$$

Work Step by Step

$$A=\int\csc xdx$$ This problem is actually very hard to deal with if you do not have any ideas where to start. Thankfully, if you do read the textbook, there is an example where $\int\sec xdx$ has been solved that you can utilize its way of doing here. We would multiply both numerator and denominator by $\csc x+\cot x$, which means $$A=\int\csc x\frac{\csc x+\cot x}{\csc x+\cot x}dx$$ $$A=\int\frac{\csc^2 x+\csc x\cot x}{\csc x+\cot x}dx$$ Now we apply the Substitution Rule. Let $u=\csc x+\cot x$ That means $du=(\csc x+\cot x)'dx=(-\csc x\cot x-\csc^2 x)dx=-(\csc x\cot x+\csc^2 x)dx$ Therefore, $$A=-\int\frac{1}{u}du$$ $$A=-\ln|u|+C$$ $$A=-\ln|\csc x+\cot x|+C$$
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