Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 16


$$\frac{1}{2}[\tan x \sec x-\ln |\sec x+\tan x|]+C$$

Work Step by Step

Since \begin{aligned} \int \tan ^{2} x \sec x d x &=\int\left(\sec ^{2} x-1\right) \sec x d x \\ &=\int \sec ^{3} x-\sec x d x \\ &=\int \sec ^{3} x d x-\int \sec x d x \end{aligned} Use $$ \int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$ Then \begin{aligned} \int \tan ^{2} x \sec x d x &=\int \sec ^{3} x d x-\int \sec x d x \\ &=\frac{\tan x \sec ^{3-2} x}{3-1}+\frac{3-2}{3-1} \int \sec ^{3-2}-\int \sec x d x \\ &=\frac{\tan x \sec x}{2}+\frac{1}{2} \int \sec x d x-\int \sec x d x \\ &=\frac{1}{2} \tan x \sec x+\frac{1}{2} \ln |\sec x+\tan x|+C \\ &\left.=\frac{1}{2} \tan x \sec x-\frac{1}{2} \ln |\sec x+\tan x|\right]+C \\ &=\frac{1}{2}[\tan x \sec x-\ln |\sec x+\tan x|]+C \end{aligned}
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