《微积分》课程教学课件（Calculus）07. Integrals and transcendental functions

CALCULUSTHOMAS'CALCULUSARLYTRANSCENChapter 7JAMESSTEWARTIntegralsandTranscendental FunctionsERMStCIHeAA7.11.Integrals ofthetangent,cotangent,secant,andcosecantIntegrals ofTrigonometricfunctionsFunctions广TABLE 3.1 Derivatives of the inverse trigonometric functionsIntegrals of the tangent, cotangent, secant, and cosecant functionsdu/drd(sinw)]1(4)da/Ve-7d(cse"lw)du/dx(μ| > 1dxIu/V-1A

2016/11/15 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 7 Integrals and Transcendental Functions Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 7.1 Integrals of Trigonometric Functions Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 1. Integrals of the tangent, cotangent, secant, and cosecant functions Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley cot ln csc u du u C    

2ydy&rdIn(4)In(y-25)+C5H-CExereses243sec2fsecytanyh(2+see y)+CIn(6+3tant)+C 5.6+3tan/h62 + see)7.2seexdtnI+)+Cs..2 /n(sec x+tan x)+C2VT+2VIn(seex+tanx)21.+tcwttanwtdSeparableDifferential Equationse+not+dP22.AsineJarnrVhln(n)Zre cosfe'ydr[sine'-sin!23,Ze"cose"dyIn(1+e')+C26=0r=1-n(I+e)+C25.teLogn3Logiax20g0X+433837.x+1AJ10-ogz+Cin2-(og,t=2in2[n10-(og,(x+)=In10Slide4-8differentialEXAMPLE1Solve the differential equationtgiatogdyg(x)dydxh(y)=(I+ye, y>-1its differential form allows us to collect all y terms with dly and all x terms with drSinceI + y is never zero for ySolution1, we can solve theequation by separating(y)ay=g(x)dxthe variables.Now we simply integrate both sides of this equation:89黑-（+岁)dyg(x) dr(4)Treataridrasaountientotdiffesemtials md multiplydy=(1 + yledxboth silesi hy adr.After completing the integrations we obtain the solution y defined implicitly as a functionofx.=edDnidehylty1+ :[,-[ealetegrale borth sidesthoCEDIn(1+y)-e+ccoautama of istegration.The last equation gives y as an implicit fiunction of x.Slide 4-9Slide 4- 10SeparableDifferential EquationsExercisesSolve the differential equation in Exercises 9-22.dy层=+1)EXAMPLE2Solve the equation y(x + 1) -dyd10.-PVy>09.2Vxy>0-1.dSolutionWe change to differential form, separate the variables, and integrate:dy12.411.de=32e7dx(x + 1)dy =x(y2 + 1)da14. VE-Iydyad13.-VycosV,+*12+1x+1dx[-(-)9.2-1+c10. 2y -*+cDivider by # + 1.12.2=x+C11.e=e+cn(1+y)=x-nx+1|+c.14.c13.2tan o=x+cThe last equation gives the solution y as an implicit function of xSlide 4-11Slide 4.12

2016/11/15 2 Exercises 2 ln( 25) y C   2 ln(4 5) r C   ln(6 3tan )   t C ln(2 sec )   y C ln(1 )   x C 2 ln(sec tan ) x x C   1 sec t e C    csc( )t e C     n( /2) ln( /6) 2[sin ] 1 v l e    2 n 0 [sin ] sin1 x l e    ln(1 )r   e C (1 ) ln(1 ) 1 x x x e dx x e C e         2 10 1 ln10 (log ) 2   x C 2 4 2 1 1 [ ln 2 (log ) ] 2ln 2 2   x 2 9 2 0 [ln10 (log ( 1)) ] ln10    x Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 8 7.2 Separable Differential Equations Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 9 differential equation Slide 4 - 10 Slide 4 - 11 Slide 4 - 12 Exercises 3 1 2 2 2 9. 3 y x C   1 2 1 3 10. 2 3 y x C   11. y x e e C   3 12. y e x C   13. 2tan y x C   3 1 2 2 3 2 14. 2 y x C  

7.3I,DefinitionsandIdentitiesHyperbolic Functionslide 4-13Slide 4- 14The hyperbolic sine and hyperbolic cosine functions are defined by the equationsEvery function f that is defined on an interval centered at the origincan be written in a unique way as the sum ofone even function andsinhx-et-ecoshxteandone odd function, The decomposition is()=()+ (-)+ ()-(-)ftanhx ≥ sinhxe-ecothr =coshx+ccoshx"e+ee-ersinhx..2evenpartoddpartsechx =cschx =e+eeecoshxsinhxIf we write e* this way, we gete'+e"2-ee=2折The hyperbolic functions++bear many similaraies to theeven partoddparttrigonometricfunctions儿coshxsinhxThey describe the motions of waves in elastic solids, the shapes ofhanging electric power lines, and the temperature distributions inmetal cooling fins.Slide 4-15Slide 4- 18Th1.Exeptfor differencesinadTABLE7.4 Identitiesforsign, these resemble2lhyperbolic functionsidentitiesweknow forthetrigorniometric functionscosh° x sinb’ x = 1sinh 2r 2 sinh x coshx2. The identities are provedcosh 2x = cosh°x + sinh directly from the definitionsByetest angest:Hyperbolike sise:Bypertolc cesisecosh x - Cosh 2r + 12.sinhxcoshx=2()(t)onhstSinb° x = cosh 2r - 1llyperbelic eotaagat:r-1-5tanh’- -1 --sech x sinh 2x.coth?x=1 +cschxHyperbolic secant:Byperhodice esecantecht-ochr-1-17Slide 4-18

2016/11/15 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 13 7.3 Hyperbolic Functions Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 14 1. Definitions and Identities Slide 4 - 15 Every function ƒ that is defined on an interval centered at the origin can be written in a unique way as the sum of one even function and one odd function. The decomposition is ( ) ( ) ( ) ( ) ( ) 2 2 f x f x f x f x f x       even part odd part If we write this way, we get x e 2 2 x x x x x e e e e e       even part odd part coshx sinhx They describe the motions of waves in elastic solids, the shapes of hanging electric power lines, and the temperature distributions in metal cooling fins. Slide 4 - 16 The hyperbolic functions bear many similarities to the trigonometric functions. Slide 4 - 17 Slide 4 - 18 1. Except for differences in sign, these resemble identities we know for the trigonometric functions. 2. The identities are proved directly from the definitions

Exercises Values and IdentitiesEach of Exercises I-4 gives a valuc of sinh x or cosh x. Use the defi-nitions and the identity cosh? x sinh? x 1 to find the values of theremaining five hyperbolic functions.1. sinhx = -32.shx=g417133. coshx =x>04, coshx =x>02.Derivatives and Integrals of15'3.Hyperbolic FunctionsRewrite the expressions in Exercises 510 in terms of exponentialsand simplify theresults as much as you can5, 2 cosh (ln.x)6. sinh (2 In x)7. cosh 5x + sinh 5x8, cosh 3x sinh 3x9. (sinhx + coshx)t10, In (coshx + sinh x) + In (coshx sinhx)lide 4-19Slide 4-20Exercises Finding DertvativesTABLE7.5 Derivativesof(）(一)InExrcises 1324, find the derivative ofywithrespect to the approhyperbelic functiospriate variabledad+edad13. y = 6 sinh 号14. y = sinh (2r + 1) (nh) 0ohm15. y = 2Vitunh Vi16. y = unh-oI(cobh) -ihww17, y = In(sinh z)18. y = In (eoshz2)19. y = sech e(1 In sech 9) 20. y = csch o(1 n csch o)（amha） ch层(cho- ()21 y In coh Itanh u 22 y ln sinhw Iohb u 22. coih(cobh) sch w23: y=2x, y'=223. y = (r + 1) sech (lnz)=_cosha daCHanit: Before differentiating,express in tenms of exponentials (och) -sch nhwmsinh’ μ dxand simplify.)24, y=4x,y'=424. y (4r2 ~ 1) oxch (In 2k)Icosh u dr (cschn) cschu.cothw -sinh w sinh d16.2-tamh -s8eeF！lanhf+seef13,2cosh=14. cosh(2x+1)15. V--schu cohwuoSinhz17.csh=dx=coth=18.19.sec h9-tanh.-n(seche)Lanh=sinh2cosh:20. coshe-cothe-n(csche)21. tanbySlide 4- 21Slide 4- 22EXAMPLE1TABLE7.5 Derivatives ofTABLE7.6Integral formulas for(a) (anh V1+7)= sictp VI+7-(V1+7)=sech2 V1 + hyperbolic functionshyperbolic functionsVi+t(ao) -o/-gm+c-gnmsc sinh rdi = osh w + Ccoth 5x dx =(b)(coh ) sahwcoshir dr = sinh u + C['sin='(o2-)[](c)（tanth）=Sechsech' du tanhur + C sinh2 - 1（cohm)=-chach' w dr coth w.+. C/n2102Ph4-(22-2)d(d)4e'sinhxdx l(sch) -schwahw 2sech a tanh w ahr - 0sech r + C=[e- - 2)2= (e26z- 21n2) (1 - 0) (cch) -ch rco.-csch a coth n ahi = csch w + C=42ln2-1n：1.6137Slide 4- 23Slide 4-24

2016/11/15 4 Slide 4 - 19 Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 20 2. Derivatives and Integrals of Hyperbolic Functions Slide 4 - 21 Slide 4 - 22 Exercises 13. 2cosh 3 x 14. cosh(2 1) x  1 2 15. tanh sec t h t t  1 12 16. 2 tanh sec t h t t   cosh 17. coth sinh z z z  sinh 18. tanh cosh z z z  19.sec tanh ln(sec ) h h      20. cosh coth ln(csc )      h 3 21. tanh v 3 22. coth v 23. 2 , ' 2 y x y   24. 4 , ' 4 y x y   Slide 4 - 23 Slide 4 - 24

Evaluating IntegralsExercisesEsercisesria4Evaluate the integrals in Exercises 4160,52.tanh 2r dt31coth.xdr42.41.sinh 2x dtsinhd53.20'cosh e d54,4esinhed)43. 6coh(-1n3)dr444 0osh (3x In 2) dr2 sinh (sin e) cos e d55.cosh (tan f) sec° e de5645;/tanhdx46.dcothcosh (In )so ViaV57.58,Vr47. / sap ( ~ )ad48.esch (5 x) dxs ()at4sin(6)60.59,csch (ln t) coth (In t) atsech Vf tanh Vr ar5052.1749.53.号+1n25.054.2hn2-3Vi283243. sinke号-ln3)+C42.5cosh=+C41.0sh2x+C57.355. e-!56. e+258. 8e+!e-e)-46.JEln(sinh45: 71n(cosh )+CJse44.sinb(3xIn2)+C99359.+1m2-2in1060821049:-2sechf+C47.tanh(r-)+(S0..-csehinn+C48,coth(5-x)+Cslide 4-25Slide 4- 283.InverseHyperbolicFunctions56The inverses ofthe sixbasic hyperbolicFIGURE7,5ThegraphsofntxNotsymmetries about thefunctions areveryusefulin integrationliney-.side 4- 27Slide 4- 28sinh+= In(r+ V2 + 1),00:00cosh-x.= In(x + V-1).x≥ 1474μ/！GURETpsothe iyolictng,coangt,andcSlide 4- 29Slide 4-30

2016/11/15 5 Slide 4 - 25 Exercises 1 41. cosh 2 2 x C 42. 5cosh 5 x C 43. sinh( ln 3) 2 x  C 4 44. sinh(3 ln 2) 3 x C   45. 7ln(cosh ) 7 x C 46. 3 ln(sinh ) 3 C   1 47. tanh( ) 2 x C   48. coth(5 )   x C 49. 2sec   h t C 50. csc (ln )   h t C Slide 4 - 26 Exercises 5 51. ln 4 1 17 52. ln 2 8 3 53. ln 2 32  3 54. 2ln 2 4  1 55. e e  1 56. 2 e e   3 57. 4 2 2 1 58. 8( ) e e e e    3 1 59. ln 2 8 2  99 60. 2ln10 10  Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 27 3. Inverse Hyperbolic Functions The inverses of the six basic hyperbolic functions are very useful in integration. Slide 4 - 28 Slide 4 - 29 Slide 4 - 30

TALE7.7 Identities for inversehyperbolic functionssech'lx = cosh!The idersities are proveddirectly fiom thedefnisoncesch'lx = sinh-!4.UsefulIdentitiescoth"'x - tuanh-"Isech (cosh- ()) - -(cosh ()（)oSlide 4-31Slide 4-32wli5.Derivatives of InverseIf y = f(x) has a horizontal tangent line at (a, f(a) then the inverse function J-' has aHyperbolic Functionsvertical tangent line at (f(a); a), and this infinite slope implies that f'is not differentiableat f(a).The reciprocal relationship between the slopes of f and f-lholds for other functionsas well,but we must be carful to compareslopes atcoresponding points.Iftheslopeofy = f(x) at tbhe point (a, f(a)is y(a) and f(a) + 0, then the slope of y = (x) at thepoint (f(a), a) is the reciprocal 1/f'(a) (Figure 3.35). If we set b = f(a), then(-hy'(b)-r'(a)r((b)Slide 4- 33Slide 4- 34THEOREM3-The Derivative Rule for ImversesIf J has an interval Ias domainand()xistndievtzonthnsdiferentiableaeryointiEXAMPLE 1 The function f(x) = x*,x ≥ 0 and is inverse (x) = Vs have deriva-isdomn(theangeofThevalucoffatapointinthedomainoftives f'(x) = 2x and (/ly(r) 1/(2V).isthe reciprocal of the value of f'at the point a = f-l(b):Let's verify that Theorem 3 gives the sme fomula forthe derivutive of y"(x):(f=ly(x) =(f-"y(b) -(1)r(-(x)(s-(b))1. The conditions underowtich f-t is differentiable2(~l(x))df-!2. The formula for thedderivafive of f-t when it2(V)uy=exists.(y-(x)) xInverse finctiun eclintio%-() -1Differmtintmgbothside(-()-() =1Chrain Roulie()Solving fiorthedetrinaliSlide 4- 35Slide 4-38(())

2016/11/15 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 31 4. Useful Identities Slide 4 - 32 The identities are proved directly from the definitions. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 33 5. Derivatives of Inverse Hyperbolic Functions Slide 4 - 34 Slide 4 - 35 1. The conditions under which ƒ-1 is differentiable. 2. The formula for the derivative of ƒ-1 when it exists. 1 1 ( )' ' f f   Slide 4 - 36

TABLE 7.8Derivatives of inverse hyperbolic functionsd(sinh-la)didrVi+ead(eoshla)dui>1dxV-d6.Derivativesof Inversed(tanh""w)[el 1I-kdrd(sechl a)10≤≤1drVi-rard(cesch-la)dur.#*0uVi+Slide 4-37Slide 4-38EXAMPLE2Show that if a is a differentiable function of x whose values are greaterExercisesIn Exercises 2536, find the derivative of y with respect to the aprothan 1, thenpriate variable.d&(cohm)-?y-sinh-IViy-cosh-12V+1Vo-idr-(1-0)ah-828. y (o7 + 20)tanh- (0 + 1)First we find the derivuative of y. = cosh-' x for x > 1 by applying Theorem 3Solutiony-(1-ncath-Vi30. y = (1 P)coth",of Section 3.8 with f(x) = coshx and /'(x) = coshx. Theorem 3 can be applied be-cause the derivative of cosh x is positive for 00-SolutionThe indefinite integral is /-()+n>a>02dx/V5+40-Va+mAuh (a) +C.a= sinh-()+c.1/3J-()c0uTherefore,2dx（（3 J-d * 0 anda > 0)0=0.98665 sinhb-1slide 4- 42Slide 4- 41

2016/11/15 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 37 6. Derivatives of Inverse Hyperbolic Functions Slide 4 - 38 Slide 4 - 39 Slide 4 - 40 Exercises 1 1 25. 1 2 x x   1 26. (4 3)(1 ) x x   1 1 27. tanh 1      1 1 29. coth 2 t t    35. sec x 36. sec x Slide 4 - 41 Slide 4 - 42

ExercisesEvaluate the integrals in Exercises 6774 in terms ofa, imverse hyperbolic functions.74. [sinh"(n x) = ln/2+1)bh. natural logarithms,r2V5Co6dridr7.467.68.V4+2V1+921/2tdx69.70Jsa1 -x21-x3y3adRelativeRatesof Growth72.AVrV1- 16rJ1/ixV4+r50sx dra73,74.Vi+sin'xJ V1+ (lnx)67. [sih(=5+2)68. 2[sinh-(3) =2In(2 +1)69. [cob-x/, ---ln3 70. tanh- x-1m3371. -(sech"(4)=-2ln32-jgr--73. [sinh-(sinh ) =05+2lide 4- 43Slide 4- 44咖-y"r10020140fo金市直宝尚市商6BA1.GrowthRatesofFunctions146的古Youmay have noticed that exponential functionslike 2'and eseem togrow more rapidlyas x gets large than do polynomials and rational functions, These exponentials certainlygowmorepidy thanandyuansee2’outgowingineasesiFigure7Intout2ngowfthyw(Exercise 19). In contrast, logarithmic fiunctions like y log: and y = In x grow moreslowly as x→ oo than any positive power of x (Exercise 21).These important comparisons of exponential, polynomial, and logarithmic functionscan be made precise by defining what it means for a function f(x) to grow faster than an-otherfunction g(x)asX→00slide 4- 45Slide 4- 48EXAMPLE1Letscompare the growth rates of severalcommon functionsEXAMPLE2Show that V+Sand (2Vx-1 grow at the same rate as xco)(a) e growsfaster than xas x oo becauseSolutionWe show that the functions grow.at the same.rate by showing that they bohgrow athe same rate as the function g(x) = :=00Uhing IHopital Rule wice=1lim0/00(b) 3° grows faster than 2° as x → o0 becaust)(2)-V(c) ' grows faster than In x as x→ oo because28Tipitars Rule(d) Inx.groms slowerthan g/ as x co for any positive integer n because1/x白银Tafigitals Ral (1/n) /4/4-1.0.w la coutantSlide 4- 47Slide 4.48

2016/11/15 8 Slide 4 - 43 Exercises 1 2 3 0 67. [sinh ( )] ln( 3 2) 2  x   1 1 2 0 68. 2[sinh (3 )] 2ln( 2 1) x    1 2 5/4 1 69. [coth ] ln 3 2 x    1 1/2 0 1 70. [tanh ] ln 3 2 x   1 3/13 1/5 71. [sech (4 )] 2ln3 x     1 2 1 1 1 2 1 72. [csch ( )] ln 2 2 2 5 2  x      1 0 73. [sinh (sinh )] 0 x    1 1 74. [sinh (ln )] ln( 2 1) e x    Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 44 7.4 Relative Rates of Growth Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 45 1. Growth Rates of Functions Slide 4 - 46 Slide 4 - 47 Slide 4 - 48

The~litle-oh" and"big-oh" notation was invented by number theorists a hundred yearsago and is now commomplace in mathematical analysis and computer science.DEFINTIONA function J is ofsmaller order than g as xoo iffo. Weindicatethis by witing og)ilitleog)lamo gta)2.OrderandOh-Notationfgrows slowerthangasx-coEXAMPLE3Here we use little-oh notation,limlnr-0(a) In x = o(x) as x → 00 becausex(b) x2 = 0(r + 1) asx→00lim0because*x+1lide 4- 49Slide 4- 50Exercises10.True,or false? As r-→co,DEFINITIONLet fix) and g(x) be positive for x sufficiently large, Then f is市()b+++-0()ofatmost theerder ofg asx-+co ifthere is a positive integer Mfor whichf(x)SM1-+-0()d. 2 + cosx = 0(2)forxsufficiemtly large, We indicatethisby writing= O(g) ('fisbig-ohofg)1. xln o(r)e. e+x=Oe)h. In(x) = o(In (x° + 1)g In (Inx) = O(Inx)EXAMPLE4Here we use big-oh notation.fosuficenly lage(a)×+ sinx0(x)asx00becausee+rbecause(b) e +x=0(e)asx→00+1a8x.006408X00(c) = 0(e)asr→0because1.f=o(g) implies f=O(g)for functions that are positivefor xsufficiently large.2. f ard e grow at the same rate,then f=O(g) arnd g=O(f)Slide 4- 51580e 4- 82

2016/11/15 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 49 2. Order and Oh-Notation Slide 4 - 50 ƒ grows slower than ǥ as x→∞ Slide 4 - 51 1. ƒ=ο(g) implies ƒ=О(g) for functions that are positive for x sufficiently large. 2. ƒ and g grow at the same rate, then ƒ=О(g) and g=О(ƒ). Slide 4 - 52 Exercises