Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 46


$$\frac{1}{2}\tan^2 (\ln t) -\ln |\sec (\ln t) |+c$$

Work Step by Step

Given $$\int \frac{\tan^3 \ln t}{t}dt $$ Let $$ u =\ln t \ \ \ \to \ \ \ du=\frac{dt}{t}$$ Then \begin{align*} \int \tan^3 udu&= \int \tan^2 u \tan udu\\ &= \int (\sec^2 u-1) \tan udu\\ &=\frac{1}{2}\tan^2 u -\ln |\sec u |+c\\ &=\frac{1}{2}\tan^2 (\ln t) -\ln |\sec (\ln t) |+c \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.