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 - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 47


$$\frac{1}{4} \tan ^{4} x-\frac{1}{2} \tan ^{2} x+\ln |\sec x|+C $$

Work Step by Step

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