Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises - Page 525: 39

Answer

$$\ln | \csc x-\cot x|+c$$

Work Step by Step

Given $$ \int \csc x d x$$ Since \begin{align*} \int \csc x d x&=\int \frac{\csc x(\csc x-\cot x)}{\csc x-\cot x} d x\\ &=\int \frac{-\csc x \cot x+\csc ^{2} x}{\csc x-\cot x} d x \end{align*} Let $u= \csc x-\cot x\ \ \to\ \ du=-\csc x \cot x+\csc ^{2} x$, then \begin{align*} \int \csc x d x &=\int \frac{-\csc x \cot x+\csc ^{2} x}{\csc x-\cot x} d x \\ &=\int \frac{du}{u}\\ &=\ln u+c\\ &=\ln | \csc x-\cot x|+c \end{align*}
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