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The integral -dz (81-22)를 can be reduced to the form k sin" (0) cos" (0) de, with the substitution z(e) = 9*sin(theta) so that, in terms of 0, V81 – 22 9*cos(theta) 2 0, k = 1/9 m =3 and n =-4 The substitution u(0): reduces the integral to S R(u) du, where the rational function R(u) = The partial fraction expansion of R(u) is The integral S R(u) du = from which we see that (81 – 2) = rp

The integral -dz (81-22)를 can be reduced to the form k sin" (0) cos" (0) de, with the substitution z(e) = 9*sin(theta) so that, in terms of 0, V81 – 22 9*cos(theta) 2 0, k = 1/9 m =3 and n =-4 The substitution u(0): reduces the integral to S R(u) du, where the rational function R(u) = The partial fraction expansion of R(u) is The integral S R(u) du = from which we see that (81 – 2) = rp

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The integral dz (81 – 22) can be reduced to the form k sin" (0) cos" (0) de, with the substitution a(@) =9*sin(theta) so that, in terms of 0, V81 – a? 9*cos(theta) 2 0, k 1/9 and n =|-4 m =3 The substitution u(0) reduces the integral to f R(u) du, where the rational function R(u) =| The partial fraction expansion of R(u) is The integral S R(u) du = from which we see that dr (81 – 2)
Transcribed Image Text:The integral dz (81 – 22) can be reduced to the form k sin" (0) cos" (0) de, with the substitution a(@) =9*sin(theta) so that, in terms of 0, V81 – a? 9*cos(theta) 2 0, k 1/9 and n =|-4 m =3 The substitution u(0) reduces the integral to f R(u) du, where the rational function R(u) =| The partial fraction expansion of R(u) is The integral S R(u) du = from which we see that dr (81 – 2)
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