Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 92


$-\frac{1}{4} \ln|\cos (4x+1)|+c $

Work Step by Step

Let $ u=4x+1$ and hence $ du=4dx $, then we have $$ \int \tan(4x+1)dx=\frac{1}{4}\int \tan u du\\ =\frac{1}{4}\int \frac{\sin u}{\cos u} du=-\frac{1}{4} \ln|\cos u|+c\\ =-\frac{1}{4} \ln|\cos (4x+1)|+c $$ where we used the facts that $(\cos u)'=-\sin u $ and $\int \frac{u'}{u}=\ln u $.
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