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: 58


$$\int\frac{\tan(\ln v)}{v}dv=-\ln|\cos(\ln v)|+C$$

Work Step by Step

$$A=\int\frac{\tan(\ln v)}{v}dv$$ We set $a=\ln v$, which means $$da=\frac{1}{v}dv$$ Therefore, $$A=\int\tan ada$$ $$A=\int\frac{\sin a}{\cos a}da$$ We set $u=\cos a$, which means $$du=-\sin ada$$ $$\sin ada=-du$$ Therefore, $$A=-\int\frac{1}{u}du$$ $$A=-\ln|u|+C$$ $$A=-\ln|\cos a|+C$$ $$A=-\ln|\cos(\ln v)|+C$$
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