Calculus (3rd Edition)

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

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 61


$\frac{1}{27} \tan(9t^3+1)+c.$

Work Step by Step

Since $(\tan(9t^3+1))'=27t^2\sec^2(9t^3+1)$, then we have \begin{align*} \int t^2 \sec^2(9t^3+1) d t &=\frac{1}{27}\int 27t^2 \sec^2(9t^3+1) d t \\ &=\frac{1}{27} \tan(9t^3+1)+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.