Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3 - Page 286: 30


$\frac{1}{20}\tan^4 5x+C$

Work Step by Step

Let $u=\tan 5x$. Then $du=5\sec^2 5xdx$, so $\sec^2 5xdx=\frac{1}{5}du$. $\int \tan^3 5x\sec^2 5xdx$ $=\int u^3*\frac{1}{5}du$ $=\frac{1}{5}*\frac{u^4}{4}+C$ $=\frac{1}{20}u^4+C$ $=\frac{1}{20}\tan^4 5x+C$
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.