Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 52

Answer

$$\int \frac{\cot ^{3} t}{\csc t} d t =-\csc t-\sin t+c$$

Work Step by Step

$$ \int \frac{\cot ^{3} t}{\csc t} d t $$ Since \begin{align*} \int \frac{\cot ^{3} t}{\csc t} d t&= \int \frac{(\csc ^{2} t-1)\cot t}{\csc t} d t\\ &=\int (\csc t\cot t-\cos t)dt\\ &=-\csc t-\sin t+c \end{align*}
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