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: 40


$$\frac{1}{4} \tan x \sec ^{3} x-\frac{1}{8} \tan x \sec x-\frac{1}{8} \ln |\sec x+\tan x|+C$$

Work Step by Step

\begin{align*} \int \tan^2x\sec^3xdx&= \int (\sec^2x-1)\sec^3 xdx\\ &=\int (\sec^5x-\sec^3 x)dx\\ \end{align*} 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{align*} \int \tan ^{2} x \sec ^{3} x d x&=\frac{1}{4} \tan x \sec ^{3} x+\frac{3}{4} \int \sec ^{3} x d x-\int \sec ^{3} x d x\\ &=\frac{1}{4} \tan x \sec ^{3} x-\frac{1}{4} \int \sec ^{3} x d x\\ &= \frac{1}{4} \tan x \sec ^{3} x-\frac{1}{4}\left(\frac{1}{2} \tan x \sec x+\frac{1}{2} \int \sec x d x\right)\\ &=\frac{1}{4} \tan x \sec ^{3} x-\frac{1}{8} \tan x \sec x-\frac{1}{8} \ln |\sec x+\tan x|+C \end{align*}
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