Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 35


(1/4)$sec^{3}$x·tanx+(3/8)ln|tanx + secx|-(5/8)tanx·secx +C

Work Step by Step

$\int$$(tan^{4}$x) (secx)dx Step 1: Use the identity: $tan^{2}$x =$sec^{2}$x -1 $\int$$(tan^{4}$x) (secx)dx=$\int$$(tan^{2}x)^{2}$secx dx=$\int$$(sec^{2}x-1)^{2}$ secx dx Expand $(sec^{2}x-1)^{2}$ secx : the expression= $\int$$sec^{5}$x dx + $\int$secx dx -2 $\int$$sec^{3}$x dx = $\int$$sec^{5}$x dx + ln|secx + tanx|-2 $\int$$sec^{3}$x dx (1) Step 2: In expression (1), $\int$$sec^{3}$x dx=$\int$secx ($sec^{2}$x) dx Let u=secx & dv= ($sec^{2}$x) dx, so v=tanx $\int$$sec^{3}$x dx=secx(tanx)-$\int$$(tan^{2}$x) (secx)dx As $tan^{2}$x =$sec^{2}$x -1, the expression =secx(tanx)-$\int$$(sec^{2}$x-1) (secx)dx Split the integral: the expression $\int$$sec^{3}$x dx= secx(tanx)-$\int$$sec^{3}$xdx+$\int$$sec$x dx ==> $\int$$sec^{3}$x dx =secx(tanx)-$\int$$sec^{3}$xdx+ln|tanx + secx| (2) Solve for $\int$$sec^{3}$x dx in equation (2): $\int$$sec^{3}$x dx=(1/2)(ln|tanx + secx|+ tanx·secx) Step 3: In expression (1), $\int$$sec^{5}$x dx=$\int$($sec^{3}$x) ($sec^{2}$x) dx Let u=$sec^{3}$x & dv= ($sec^{2}$x) dx, so v=tanx $\int$$sec^{5}$x dx=$sec^{3}$x(tanx)-3$\int$$(tan^{2}$x) ($sec^{3}$x)dx Use the identity $tan^{2}$x =$sec^{2}$x -1, The expression = $sec^{3}$x(tanx)-3$\int$$(sec^{2}$x-1)($sec^{3}$x)dx And split the integral: $\int$$sec^{5}$x dx =$sec^{3}$x(tanx)-3$\int$$sec^{5}$x dx+3$\int$$sec^{3}$x dx From step (2) we get $\int$$sec^{3}$x dx=(1/2)(ln|tanx + secx|+ tanx·secx) So $\int$$sec^{5}$x dx =$sec^{3}$x(tanx)-3$\int$$sec^{5}$x dx+(3/2)(ln|tanx + secx|+ tanx·secx) (3) Solve for $\int$$sec^{5}$x dx in equation (3): $\int$$sec^{5}$x dx=(1/4)$sec^{3}$x·tanx+(3/8)ln|tanx+secx|+(3/8)tanx·secx Thus, combining expressions: (1/4)$sec^{3}$x·tanx+(3/8)ln|tanx + secx|-(5/8)tanx·secx +C
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