Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 84


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

Work Step by Step

Let $u=4x+1$, then $du =4dx$. Now, we have $$\int \tan(4x+1)dx=\frac{1}{4}\int \tan udu\\ =\frac{1}{4}\int\frac{\sin u}{\cos u}du=-\frac{1}{4}\ln(\cos u)+c\\ =-\frac{1}{4}\ln(\cos (4x+1))+c.$$
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