Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Practice Exercises - Page 518: 40


$$\frac{\sec ^{5} x}{5}-\frac{\sec ^{3} x}{3}+C $$

Work Step by Step

We evaluate the trigonometric integral as follows: \begin{align*} \int \tan ^{3} x \sec ^{3} x d x&=\int\left(\sec ^{2} x-1\right) \sec ^{2} x \cdot \sec x \cdot \tan x d x\\ &=\int \sec ^{4} x \cdot \sec x \cdot \tan x d x-\int \sec ^{2} x \cdot \sec x \cdot \tan x d x\\ &=\frac{\sec ^{5} x}{5}-\frac{\sec ^{3} x}{3}+C \end{align*} Where we used the fact that $\sec^2 x = \tan^2 x +1$
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.