Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 5


$$\frac{1}{6} \tan \theta \sec ^{5} \theta-\frac{7}{24} \tan \theta \sec ^{3} \theta+\frac{1}{16} \tan \theta \sec \theta+\frac{1}{16} \ln |\sec \theta+\tan \theta|+C$$

Work Step by Step

Since \begin{align*} \int \sec ^{3} \theta \tan ^{4} \theta d \theta&=\int \sec ^{3} \theta(\sec ^{2}\theta-1) ^{2} d \theta\\ &=\int\left(\sec ^{7} \theta-2 \sec ^{5} \theta+\sec ^{3} \theta\right) d \theta \end{align*} Since $$\int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$ then \begin{aligned} \int \sec ^{3} \theta d \theta &=\frac{\tan \theta \sec \theta}{2}+\frac{1}{2} \int \sec \theta d \theta \\ &=\frac{1}{2} \tan \theta \sec \theta+\frac{1}{2} \ln |\sec \theta+\tan \theta|+C \end{aligned} and \begin{aligned} \int \sec ^{5} \theta d \theta &=\frac{\tan \theta \sec ^{3} \theta}{4}+\frac{3}{4} \int \sec ^{3} \theta d \theta \\ &=\frac{1}{4} \tan \theta \sec ^{3} \theta+\frac{3}{4}\left[\frac{1}{2} \tan \theta \sec \theta+\frac{1}{2} \ln |\sec \theta+\tan \theta|\right]+C \\ &=\frac{1}{4} \tan \theta \sec ^{3} \theta+\frac{3}{8} \tan \theta \sec \theta+\frac{3}{8} \ln |\sec \theta+\tan \theta|+C \end{aligned} and \begin{aligned} \int \sec ^{7} \theta d \theta &=\frac{\tan \theta \sec ^{5} \theta}{6}+\frac{5}{6} \int \sec ^{5} \theta d \theta \\ &=\frac{\tan \theta \sec ^{5} \theta}{6}+\frac{5}{6}\left[\frac{1}{4} \tan \theta \sec ^{3} \theta+\frac{3}{8} \tan \theta \sec \theta+\frac{3}{8} \ln |\sec \theta+\tan \theta|\right]+C \\ &=\frac{1}{6} \tan \theta \sec ^{5} \theta+\frac{5}{24} \tan \theta \sec ^{3} \theta+\frac{15}{48} \tan \theta \sec \theta+\frac{15}{48} \ln |\sec \theta+\tan \theta|+C \end{aligned} Hence \begin{align*} \int \sec ^{3} \theta \tan ^{4} \theta d \theta&=\int \sec ^{3} \theta(\sec ^{2}\theta-1) ^{2} d \theta\\ &=\int\left(\sec ^{7} \theta-2 \sec ^{5} \theta+\sec ^{3} \theta\right) d \theta\\ &= \frac{1}{6} \tan \theta \sec ^{5} \theta-\frac{7}{24} \tan \theta \sec ^{3} \theta+\frac{1}{16} \tan \theta \sec \theta+\frac{1}{16} \ln |\sec \theta+\tan \theta|+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.