University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 60


$$\int\frac{1}{r}\csc^2(1+\ln r)dr=-\cot(1+\ln r)+C$$

Work Step by Step

$$A=\int\frac{1}{r}\csc^2(1+\ln r)dr$$ We set $a=1+\ln r$, which means $$da=\frac{1}{r}dr$$ Therefore, $$A=\int\csc^2ada $$ $$A=-\cot a+C$$ $$A=-\cot(1+\ln r)+C$$
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