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 404: 79

Answer

$$\int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$

Work Step by Step

Given $$\int \sec ^{m} x d x =\int \sec ^{m-2} x\sec^2 x d x $$ Let \begin{align*} u&= \sec ^{m-2}x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\sec^2 x d x\\ du&= (m-2)\sec^{m-2}x\tan x \ \ \ \ \ \ \ v=\tan x \end{align*} Then \begin{align*} \int \sec ^{m} x d x&= \tan x \sec ^{m-2}x-(m-2)\int \sec^{m-2}x\tan^2 xdx\\ &= \tan x \sec ^{m-2}x-(m-2)\int \sec^{m-2 }x(\sec^2 x-1)dx\\ &= \tan x \sec ^{m-2}x-(m-2)\int \sec^{m }x dx +(m-2)\int \sec^{m-2 }xdx\\ \end{align*} Hence $$\int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$
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