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hw-7-sec-8.4-solns

# hw-7-sec-8.4-solns - 642 CHAPTER 8 TECHNIQUES OF...

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Unformatted text preview: 642 CHAPTER 8 TECHNIQUES OF INTEGRATION T2 _ _ R)2> 43. We use cylindrical shells and assume that R > r. m2 2 r2 — (y ~ R)2 => 3: = : so g(y) : 2 r2 — (y e R)2 and V : [5: 27w - 2 x/T’z — (y —‘R>2 dy : [mm + R)x/"—r2 — u2 du [men = y 4 R1 _ -7‘ 2 2 7‘ 2 2 Whereu=rsin9,du=rcosr9d0 —47rj‘Tu\/r —u du+47er_r\/r —u du in the second integral 1‘ : 471' [—ﬂrz — u2)3/2] + 47erj7/32 T2 cos2 6d9 = —4?7'(0 ~ 0) + 47rRr2 cos2 0d9 7r : 27rch2 + cos 26) d6 = 27rRr2 [6 + %sin 26] 713/2 2 2W2RT2 Another method: Use washers instead of shells, s0 V : 87rR for x / T2 ~ y2 dy as in Exercise 6.2.63(a), but evaluate the integral using y = r sin 6‘ 8.4 Integration of Rational Functions by Partial Fractions '\ . 2m _ A B 1/01)(a:+3)(3x+1)_a:+3+3m+1 k/ 1 1 1 A B C (b)m3+2m2+m~x(m2+2m+l)_\$(1‘+1)2_;+\$+1+(\$+1)2 , \ f 2(a)_\$___2£____ A + B I\\// \$2+x—2_(m+2)(m—1)~\$+2 22—1 2 2 (b ac _(w +m+2)—(m+2)_1_ m+2 )\$2+;I:+2_ m2+\$+2 _ m2+x+2 Notice that 2:2 + a: + 2 can’t be factored because its discriminant is b2 — 4ac = —7 < 0. 4 4 a; +1 9: +1 _A B C Dac—I—E 3'(a) m5+4x3—m3(m2+4)- \$+m2+m3+ w2+4 (b) 1 _ 1 _ 1 _ A + B + C + D (w2—9)2-[(x+3)(m—3)]2—(x+3)2(w—3)2_\$+3 (33+3)2 513—3 (313—3)2 / 3 / :c 13\$+12 13x+12 A B 1 . —‘_: —4 —"”"_: —4 ——: —4 I‘ya)m2+4w+3 m +x2+4cc+3 w +(a:+1)(:c+3) m +z+1+x+3 2m+1 A B C Dm+E Fat-PG (b) (m+1)3(m2+4)2 _m+1+(\$+1)2+(m+1)3+ m2+4 +(z2+4)2 I, 2:4 (m4—1)+1 1 1 i! 5. (21);]34ﬂ1 2—10? =1~I-\$4_1 [oruselongd1v1510n] =1+m " _1+(a:—1)(m+1)(m2+1)_1+m—1+x+1+m2+1 ((b‘) t4+t2+1 _At+B+Ct+D+Et+F y Ky/’ (t2 + 1)(t2 —I— 4)2 t2 + 1 t2 + 4 (152 + 4)2 V: SECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 643 4 4 3 cc _ :1: av _Aa:—|’BI Cm—I—D 6'(a) (m3+x)(\$2#x+3)—m(m2+1)(m2—x+3) — (m2——1)(w2—\$+3)_ 1724—1 'm2—m—I—3 1 1 A B C D Ens-FF :__———————-:—»——+———+ m3(a:u1)(x2+m+1) m :62 x3 m—1+:c2+a:+1 de=fo+ 66>d\$2m+6ln|m—6|+C x- 2 r #16 16 16 : —4 d (r+4 +T+4) dr /<r +T+4> r [oruselongdmsmn] 2 2%7" #4r—I—161nIT-I—4I—I—C (magnifm 9 53—9 _ A + B '(m+5)(m#2)—\$+5 x—2 . Multiply both sides by (a? + 5)(m — 2) to get an r 9 = A(a: A 2) + B(;r + 5)(*), or equivalently, x v 9 = (A + B):v — 2A + 5B . Equating coeﬂicients of :v on each side of the equation gives us 1 = A + B (1) and equating constants gives us —9 = *2A + SE (2). Adding two times (1) to (2) gives us ‘7 : 7B <:> B : —1 and hence, A = 2. [Altemative1y, to ﬁnd the coeﬂicients A and B , we may use substitution as follows: substitute 2 for ac in to get —7 2 7B <:> B = —1, then substitute —5 for a: in to get —14 = —7A <=> A = 2.] Thus, /—(m—+%)_T§_—23dm:/(\$:5+x:12)da::2ln|m+5l~lnlx—2|+C. 1 A .—————:—— —— let—l Bt 4. 1° (t+4)(t—1) t+4+t—1 :> ( H (+) 75—1 4 1-53 4 3-515- 4 1 5A A %.Thus, /———1——-dt~/ :ME‘I—i/i dtevlln|t+4l+lln|t—1|+C or 11115;1 +0 (t+4)(t—1) _ t+4 t—l ‘ 5 5 5 t+4 5’11I1—— 1 — A + B Multil bothsidesb (a:—I—1)(:z:—1)to et1#A(\$—1)+B(m+1) K. sz2—1_(m+1)(a:r1)_—m+1 sc—l' py y g _ ' Substituting 1 for ac gives 1 = 23 ¢> B = Substituting ——1 for m gives 1 = —2A 42> A : —%. Thus, 3 3 1 _ —1/2 1/2 _ 1 1 3 /2mzﬁldw—A<\$+1+\$_1)dw—[/2lnlx—I—1I—I—2lnlm 1H2 = (—%ln4+%ln2)—(—;ln3+%ln1)2%(ln2—I—ln3—ln4) [orllng] 2 7 k }m2+3x+2 _ m+1 m+2 / . Multiply both sides by (a: + 1)(m —I— 2) to get an # 1 = A(w + 2) + B(m + 1). Substituting —2 for ac gives —3 = —B <:> B = 3. Substituting —1 for :1: gives —2 = A. Thus, 1 1 33*1 —2 3 1 __———— = d = —l 1 31 2 /0w2+3\$+2dm /(;<1‘+1+a:+2)m [2n|m+ H“ n|\$+l10 = (—2ln2+31n3) — (—21:11 +3ln2) = 3ln3 — 5ln2 [or In SECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 645 i 1 A B C + ','(a:+5)(ac—1) sc+5 (x+5)2 ct—l Z> 1=A(x+5)(m_1)+B(m_1)+C(x+5)2‘ Setting m = —5 gives 1 2 —6B, so B = —%. Setting m = 1 gives 1 = 360, so 0 = 31—6. Setting as = —2 gives 1,A(3)( 3)+B( 3) : C(32)——9A—3B+9C=—9A+§+§:—9A+§,sogA=—§andA:—%.Now _1 ~1/36 1/6 1/36 1 1 1 d .2: ————~ d :——1 5 —__ _1 _1 . /(a:+5)2(x~1) m /I:c+5 (\$+5)2+\$_1 3” 3611I93+ I+6(m+5)+36 nlm |+C / “‘_x2—5a:+16 A B C I .1. —__ 2- : _ 2 _ kW—2\$+l \$_2+(x_2)2 => \$ 51114-16 A(:L‘ 2) +B(;II 2)(2\$+1)+C(2IE+1). Setting :0 = 2 gives 10 : 50, so C = 2. Setting a: = —% gives 775 2 3453A, so A = 3. Equating coefﬁcients of 332, we get 1 = A+2B, so -2 : 2B andB = -—1. Thus, x2~5x+16 3 1 2 3 2 —————d= — -—d=—12 1* —2— (2m+1)(w—2)2 3” /(2x+1 m—2+(m—2)2) "3 2n|m+I In” I m—2+C 3 _ 21. or By long division, :2 = a: + Thus, w2+4|x3 + 01:2 + 0:0 + 4 \$3 + 43: —4x +4 x3+4 —4m+4 4x 4 [m2+4dm_/(m+ m2+4)d\$_/(\$_x2+4+x2+22>d\$ _l2_ 1 2 l —1f __12_ 2 —1 33) —2m 4 2lnlw +4I+4 2tan (2)+C—2av 2ln(ac +4)+2tan (2 +0 I’ 1 A B C D . =—— 1=A—ZB—2 2— 2. (E 82(3_1)2 8+52+s_1+(s_1)2 :> s(s 1)+ (s 1) +Cs(s 1)+Ds Sets = 0, giving B = 1. Then sets = 1 to get D = 1. Equate the coefﬁcients of s3 to get 0=A+C 0r A:—C,andﬁnallysets=2toget1=2A+1-—4A+4 or A:2.Now ds 2 1 2 1 1 1 ————= — —— d :21 ———21 — — . /sz(s—1)2 /Is+s2 s—l+ (5—1)2] 3 HM s nls 1| 3—1+C 5932+3at—2 _ 5w2+3x—2 _A B 23. —— —— — x3+2m2 m2 (m+2) :v m2 x+2 . Multiply by 3:2 (z + 2) to get5w2+3x—2 =Am(x+2)+B(m+2)+Cm2.Setcc= —2togetC=3,andtake a: = 0 to get B = —1. Equating the coeﬂicients of \$2 gives 5 = A + C => A = 2. So 2 _ /M—2dat=/<—2-—i+ 32)dw=2ln|ml+i+3ln|m+2|+0 m3+2\$2 a: 302 33+ ‘L, 646 CHAPTER8 TECHNIQUES OF INTEGRATION I \' \$2—x+6_\$2—x+6_é Bx+C . # ——— — .M ‘ l 2 t 2 — = 2 . 24 333+?” \$(m2+3) m + \$2+3 ultipybym(x +3)toge x m+6 A(a: +3)+(Bx+C)x Substituting 0 for 35 gives 6 2 3A <=> A : 2. The coefﬁcients of the xz-terms must be equal, so 1 = A + B :> B : 1 — 2 : A1. The coefﬁcients of the x—terms must be equal, so —1 = C. Thus, m2—\$+6 2 —:cA1 2 :c 1 ———d = — d = — ——— /\$3+3ac 3” /(m+\$2+3) m m2+3 x2+3>dx 1 1 m =2lnx——lnx2+3 ———tan‘1——+C 10 A B\$+C . . 2 5. —————— = .M 1t 1 th d i 2 (m_1)(x2+9) \$_1+ x2+9 u1pyb0 s1esby(m 1)(:c +9)toget 10 2 A(\$2 + 9) + (Bx + C)(a: — 1) (Jr) Substituting 1 for :3 gives 10 = 10A ¢> A z 1. Substituting 0 for x gives 10 = 9A — C => 0 : 9(1) — 10 : —1. The coefﬁcients ofthe x2-terms in (*) must be equal, so 0 = A + B 2 B = —1. Thus, 10 dwg 1 +Hm—1 dm_ 1 _ cc _ 1 d (:c—l)(:c2+9) _ m—l x2+9 — m—l w2+9 3024—9 3: = ln|x — 1| — éln(a:2 + 9) — %tan‘1(§) + C In the second term we used the substitution u = x2 + 9 and in the last term we used Formula 10. \$2 + m + 1 IE2 + 1 m 1 1 1 2 I __ — I d + _ _ — 3 _ 26 ( 2 1)2 d\$ /( 2 1)2 dzv /( 2 1)2 (13 / 2 1 dill 2/ 2 du [u \$ +1 du 21! 1 2(m2+1) +0 :tan‘laH- l +C=tan_1\$— u ’ 3 2 » ‘ +x +2m+1 Ax—f—B Cm—l—D . . : m2+1 + \$2+2 .Multiply both Sldes by (352+1) (x2+2) to get \$3+x2+2m+1=(Ax+B)(x2+2)+(Cm+D)(x2+1) ¢> 3:3——:v2+23:+1=(Ax3+Ba:2+2Ax+2B)+(Cm3+Dx2+Cx—l—D) e x3 —— m2 + 2x + 1 z (A + C)m3 + (B + D)m2 + (2A + C)w + (28 + D). Comparing coeﬂ‘icients gives us the following system of equations: A+C=1 (1) B+D=1 (2) 2A+C=2 (3) 2B+D=1 (4) Subtracting equation (1) from equation (3) gives us A = 1, so 0 = 0. Subtracting equation (2) from equation (4) gives us 3 2 1 3:0,50D=1_Thus,I=/H—m+—29de=/< w + 1 >dm.For/ x dm,letu=x2+1 (\$2+1)(a:2+2) 3324—1 m2+2 m2+1 I): l 1 1 1 2 1 sodu=2xdmandthen/w2+1dm=E/Eduzilnlu|+C=§ln(w +1)+C.For/\$2+2dm,use FormulalOwitha: So/ 1 dm:/———1——d\$: itan‘li+cl 202+ ﬂ ‘ SECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 647 I I I I I w2+2 «5 Thus I = lln(m2 +1) + —1— tan“)L 1 +0. ’ 2 «a v2 2— — B D i 28- m 22: 1 Z A + Ca3+ :> —— + 5 (m—1)2(m2+1) 37-1 (ac—U2 x2+1 332 — 2x — 1 : A(av * 1)(:102 + 1) + B(:c2 + 1) + (Cm + D)(m — 1)2. Setting a: = 1 gives B = —1. Equating the coefﬁcients of 2:3 gives A : ~C. Equating the constant terms gives ~1 = —A — 1 + D, so D : A, and settingm = 2gives ~1 = 5A — 5 ~ 2A+ A or A : 1. We have - w2—2x—1 1 1 17—1 1 1 2 _1 I m — ————— :1 — + ___1 ' ‘ /(a:~1)2(:c2+1)dm fIiE—l (\$—1)2 m2+1I dm nlm 1| x—l 2 nIm 1) +tan \$+C m+4 33+1 3 1 (2\$+2)d:c 311\$ . ——-——d : —-——— —_d =— .—._._. _— 29 /a:2+29:+5 m /m2+2\$+5d\$+/m2+2x+5 35 2/x2+2w+5+/(x+1)2+4 =élnim2+2m+5l+3/ 2 du Where z + l = 2u, 4(u2 +1) anddwz2du 1 =%1n(a:2+2x+5)+gtan‘lu+0:Eln(x2+2m+5)+gtantl<\$:l)+0 \$2+\$+4 3x2+x+4 _Aa:+B Cx+D 'x4+3m2+2_ (x2+1)(z2+2) _ \$2+1 + z2+2‘ .y._l.,.mw..ui..v.»‘..mWyn—“mgﬁ twaﬂwvww: Multiply both sides by (\$2 + 1)(:v2 + 2) to get ( 3m2+x+4:(Ax—I—B)(a:2+2)+(C\$+D)(ac2+1) c> 3\$2+\$+42(A333+Bm2+2Ax+2B)+(Cx3+Dac2+C'w+D) <:> 33:2 + as + 4 = (A + C'):v3 + (B + D):z:2 + (2A + C):c + (2B + D). Comparing coeﬁicients gives us the following system of equations: A+C=0 (1) B+D=3 (2) 2A+C=1 (3) 2B+D=4 (4) Subtracting equation (1) from equation (3) gives us A = 1, so 0 = —1. Subtracting equation (2) from equation (4) gives us B=1,soD=2.Thus, 3x2+m+4 :E—I—l —;z:—l—2 I /x4+3x2+2dm /m2+1 m+/a:2+2 a: 1 21‘ 1 1 2:1: 1 =— —- 2 —d 2/w2+1dx+/m2+1d\$ 2/x2+2dx+ /x2+(\/§)2 x =%ln|\$2+1l+tan_1x—§lnIx2+2|+2—%tan_1(—%>+C = %ln(as2 + 1) — %1n(ac2 + 2) + tan‘1 x + ﬂ tan‘1 + C , I I I ...
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