The Natural Logarithmic Function: Integration. ix u = 9-x 2,du =-2xdx. ~dx= --~1(9-x~)- /~(-2x)dx ]..--~-g dx = lnx-5[ +C

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1 Section 5. The Natural Logarithmic Function: Integration 48 Section 5. The Natural Logarithmic Function: Integration. u = + l, du = d 4. u = - 5, du = d C i ni +C X hl(4) + C. u = 9-,du =-d ~d= --~(9-~)- /~(-)d ]..--~-g d = ln-5[ +C 5. u = + 5, du = d 6. u = 4 -, du = - d I4@ : () d = _lln[4 - [+ C 7. u = -, du = d =+ln~- +C 8. U = 5 -- X,du = -X d ~ (_) + : l+ c 9. U = X 4 +, du = (4 + )d I 4X +d = IX4 (4X +)d : In 4 -[-- ] + C IO.U = X -- Z, du = ( - 6)d = ( - )d i~z_ - d= +I- (~ - 6) d _ X : --Nf~-- X +C. u = + + 9, du = ( + + ) d I"... ~ ++ d = r( ~_ + _+ ) d J + ~ +9 J + ~ +9 In = C 4. u = + -4, du =( + 6)d 5. I (+ +~_ ) d = I f ~ +6 d 4 -~J ~-~_-4 fx - + d = -4+ +l 6. [ X J d = +ll+ - f - X In I C d ni + + c d -- X + ll + 9In - + C = --+ 5ni - + c I +5 +5) X 5X + 9- ll5nl+ 5 + C I = ~ln -[+C 00 Brooks/Cole, Cengage Learning

2 484 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions.I -ft,- d= -+-T-~+ d 5. l(_l) X - +ln./ ~ + + C f_ dr=~ (-l) dr = [.(- ) d + l" l(- ) J(_ ) =! dr 7-i _ d +!( ) = ni II (-l) + C. u = ln, du = ld X. u = +l, du = d -- -~ lnl~ll]+ C ~ ~ = l(* + 0- /~ ~ lln( a+)+c = - + ~ + - = (+) / + C = -~/ ~ + + C 4. u = + /, du = ---:-=_._ d ~l~ ) ( - ) d -! dr _ [.(,, = I -~-_--{_ l dl.!... d - ) = ln] - + ( -) + C 7. u = + "~, du = --.~ dx (u ~ ) - du = dr where C = C~ +. I /( +,/) d : + V ~ d 00 Brooks/Cole, Cengage Learning

3 Section 5. The Natural Logarithmic Function: Integration 485 (u + )du dr 9. u = w/-~-, du = -~ dr ~ = = Iu +6U+9dU=u I(u+6+~du = [~ + 6u + 9n = u + u + 8n ul+ C = (~-)+ (/-~- )+ 8~ ~- + C~ = +6~+8n]~- +C -- where C = C u = ~/ -, du = ~ dr :=> dr (u + )d~t ~/~ f ~ dr= J~tu+ ~~ _,u +. )d~ = _---~.u fu + (+ u + ) du =. + +u+lnu +C - I (l/~ -) + (l/-) +(l/-)+ln l/- = In / - + /---~ + ~/ + + C. l cot(-~) do = I cot(-~)(~l do = nsin~ +C. I tan 50 do = ~ 5 sin 5. ~0 do cos 50 _- _! lc s + c 5 =-~ln csc+cot + C 4, sec = sec & =~see-+tan +C I 0() 5. J{cos0- do : Jcos do- JdO = lsin C 6, - tan do = odo - 4 jtan do 7. u = l+sint, du = costdt i cos t ~dt = lnll + sint + C 8. u = cot t, du = -csc ~ t dt i csc~tdt = -~ cott I+C cot t =0+4nlcos~+C 00 Brooks/Cole, Cengage Learning

4 486 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 9. u = sec-l, du = sectand sec tan i [ + C ~ -- = i d : ~ls~ I(sec + tan)d = ½I(sec + tan )() d = ½inlsec + tan - In cos + C 4. y = d = 4n[+C 0,): = 4n}ll+ c ~ c = y = 4inl + ~o 45. s : I tan(0)do : _~ I tan(0)( do) = -± lnicos 0 + c (0, ): = -} Inlcos (O)l+ C => C = s = -½ Inl cos oi + (0, )~ : (-,0):0 =-l-inl- +C=-I+C=> C=I y = - l.ll+ Y= - : 8 - =-n - +C (,0): 0 =-nl- + C ~ C = 0 y -- - ~l - ~ ~o -~0 (0,4): 4 = lnlo-9 + C =~ C : 4-n9 y = ln ~-9+4-n9 -a 46. r = tan ~ sec t = + t dt = ln ltant+l[+c (n, 4): 4 = In c ~ c = f"()= r = In tan t S (~<) = s O) = s () = S() = f() = S() = lo 75= -, >O - ~+C =-+C~ C= - ~+ -n + + C~ = -(0) + + C, :=> C = - -in Brooks/Cole, Cengage Learning

5 Section 5. The Natural Logarithmic Function." Integration = -4(- ) 4~. f"() (-l) z 4 S () : (_l) ~+c () = 0 = 4-4+c=> c = 0 4 f () - - f() = 4n(-)- +Cl f() = = 4(0)-4+C => Cl = 7 f() = 4n(-)- +7 ~9. dy _ (0, ) d + (a) (o, l) >l 5. (a) -~ ill///tt//tl/l l ~ t/i//tt!/tlf /I l/!lliflil!/ll //t//i/////it// --~, / (b) dy =l+d, (, 4) y=+in+c 4=+0+C~C= y=+ln+ - Ill I-~ (b) y= d = ln + + C 5. (a) ), y(o) = => = ln+c => C= -in So, y:ln++l-in:ln(f-~/+l. 50. dy d (a) - f ln (, -),i -i,, (b) dy = sec, (0, ) y = Insec+tan +C = Inl + oi + c ~ c = y = ln l see + tan I + " (b) Y =!ln-7 y() = - => - - (In ) (In ) So, y _~_ d = In + = - In = o l d = ~[ln + ]l_l 54.!~l X +--- ~ 55. u = l + In, du= -l d X = l[in5iinl]~ = lin5 ~ lle ( + in ) i~ d ( ]e= + In ) Brooks/Cole, Cengage Learning

6 488 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 56. u = ln, du = -d +l [l - ~ 0,69 ~ ~ I - e d = Ilnlln lq = ln,..t~ - lnl + = -In o = I- ni + ll~; = - ~ ~ ~ - cos~ do = [ln O - sin 0 ~ 0 - sin 0 I _ 60.!~i (csc 0- cot 0)d0 = I~] (csc 0- csc 0 cot 0+ cot = 0)dO Note: In Eercises 67-70, you can use the Second Fundamental Theorem of Calculus or integrate the function. 67. F() =!i ~ ~ dt F () = F() = I~: tan t dt F () = tan 69. F(): a, 7 dt F () = ~()=- (by Second Fundamental Theorem of Calculus) Alternate Solution: F() = o, -fdt = Et.ltl~, = lnl I F () = g( ) - = 70. F()= a, 7 dt F () = _ 7. = [-cot 0 + csc 0-0~:=~ = ~. I + ~ dv = -,/~ - In( + /~)+ C ] 6~. J l + ~ ~ = 4w/~ ln( + ~) + C A ~.5; Matches (d) 6. [~_~ d = +/7+C a- ~-,J~ + 7. X X ~. I75_± = ~l-~ +-++C /4 [ ~ /:~ (csc - sin )d = ln(/~ + ) - T ~ ~ a-~/4 cos A ~ ; Matches (a) 7. A = I~l~d= E6inl[~i =6n ~ Brooks/Cole, Cengage Learning

7 Section 5. The Natural Logarithmic Function: Integration 489 ~ [4 4 d = a~tdx = In In~ = [ln(in4)-in(ln)~ ln(n = / ~, = In) In 74. A =!~In ~r/4 i"] ~r/4 ~ ln 75. A= I0 tand =-lnlcosll =-ln~ +0 = In~ =~ q~r/4 =-ln(-~)+ln(l+~)in(+-) ( "f~ [,d4 sin~ d = -Inll + cos -,~/4 = ) = In A = Ja-/4 + cos - 4 [4+4d= Ii~(-~) X-I- 77. A = d= +4n = (8+4n4) 5 = +8n =.045 dl ~ 78. A = I~ ~+6 ~d = IlS(~) s =5+6n5-=4+6n5~" d= [ + 6 In ]~ X 8 a- (a /a" d = -7-a0 ~I I sec a~ rl 79. l~see 6 \ 6)6 6 T J0 d = In sec-+ tan = h sec--+ tan + = ~ h + ~ a tan(o.))d = + -- (0. = 6 + In cos(. - + In cos(0. ~ ~ 8. f() =, b_a = 5- = 4, n = 4 f()+ f()+ f(4) + f(5)] = -~[ ] = 0. f()+ 4f()+ f()+ 4f(4) + f(5)~ = ~[ ] a [5 Calculator: al 7 -d = 9. Eact: In 5 00 Brooks/Cole, Cengage Learning

8 490 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 8 8. f() = + 4 b - a = 4-0 = 4, n = 4 Trapezoid: ~Ef(O ) + f() + f() + f() + f(4)~ = ] ~ Simpson: ~Ef(O ) + 4f() + f() + 4f() + f(4)~ ~ Calculator: [4 8~ d ~ 6.48 ~0 +4 Eact: 4 In 5 8. f() = ln, b-a = 6- = 4, n = 4 Trapezoid: -~Ef( 4 ) + f() + f(4) + f(5) + f(6)~ = ] ~ Simpson: --~[f( ) + 4f() + f(4) + 4f(5) + f(6)~ ~ 5.6 Calculator: In d ~ y() = sec, b-a = -- Trapezoid: -~--j-kf~.-~ + f - + f(0)+ f + f ~ -i-~[ ] ~.780 Simpson: ~(4)[fl--~)+ 4f - + f(o)+ 4f + f ~.6595 Calculator: J-~/ sec d = Power Rule 86. Substitution: (u = + 4)and Power Rule 87. Substitution: (u = + 4)and Log Rule 88. Substitution: (u = tan )and Log Rule 89. X dt = /47dr I( n=~- (a) ln=ln5~=5 (b) ln = ~ = e assume > 0) fcos u 9. Icotu du = -sinu du = lnlsinul+ C Alternate solution: ~ulln d~ sinul,+c-]~ =--sinucosu+c = cotu+c ~=-~(~)=ln4 ~ =--ln4 = ln = 9. csc I i Icsc u du u.+ = csc cot u\-d-~scu ~) du = + _i cot ( csc u + cot u -csc u cot u - csc ) u du = -lnlcsc u + cot u[+ C Alternate solution: ~d ~-lnlcsc u + cot u I + C~ = CSC du~-, \ t-osc u cot u - = o~c u(oot u + csc u) csc u + cot u \! csc u + cot u = CSC U 00 Brooks/Cole, Cengage Learning

9 Section 5. The Natural Logarithmic Function: Integration 49 sec - tan sec tan +C = -ln lsec-tanl+c csc - cot 97. Average value = 4~- -~- d = 4~X- d 00. A _vera=e. value & z 0 Jo 6 6 In m tan z L h = ~[ln( + ~)- ~(+ = ~ l~ Average value = [44(X + ) 4 - J ~ d = n4-4 + d =n+ =ln Average value = [e In X _ I(ln)]e - ~ P(t) = I~ t dt = (000)(4)I + 0.5t 0.5 =,000 lnll C P(O) =,000 lnll + 0.5(0) + c = 000 C = 000 P(,) =,000 lnl! + o.5ti + looo = looo~ ln[ + o.5t + ~ P() = looo~(ln.75)+ ~ ~ t = ln~as 0 foo ot_0~ dt ~_~o~ lo = ln(t -00)~ s o = ~-~[ln 00 - ln50] - ~ 0.58 e- 0. _~ f 90,000 d = ~000 m400 + ~]:oo 50 40,/4o ~ $68.7 O 00 Brooks/Cole, Cengage Learning