Answer
$\frac{1}{3}tan^3x+C$
Work Step by Step
Let u=tanx$=>du=sec^2x$dx
=$\int tan^2x sec^2xdx$
=$\int u^2du$
=$\frac{1}{3}u^3+C$
=$\frac{1}{3}tan^3x+C$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 24
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
