Chpter 3 Symbolic Integrtion This chpter contins the bsic tricks of the symbolic integrtion trde. The gol of this chpter is not to mke you slow inncurte integrtion softwre, but rther to help you understnd the bsics well enough to use modern integrtion softwre effectively. The bsic methods of integrtion re importnt, but there re mny more tricks tht re useful in specil situtions. The computer knows most of the specil tricks but sometimes bsic preliminry hnd computtion llows the computer to clculte very complicted integrl tht it cnnot otherwise do. A chnge of vribles often clrifies the mening of n integrl. Integrtion by prts is theoreticlly importnt in both mth nd physics. We will encounter few of the specil tricks lter when they rise in importnt contets. The cble of suspension bridge cn be described by differentil eqution, which cn be ntidifferentited with the hyperbolic cosine. If you decide to work on tht project, you will wnt to lern tht trick. Prtil frctions is nother integrtion trick tht rises in the logistic growth model, the S-I-S disese model, nd the liner ir resistnce model in bsic form. We will tke tht method up when we need it. The Fundmentl Theorem of Integrl Clculus 2.3 gives us n indirect wy to ectly compute the limit of pproimtions by sums of the form f[] + f[ + ] + f[ +2 ] + + f[b ] = b X = step [f [] ] We hve b b X f[] d = lim [f [] ] = step b δ X = step δ [f [] δ] but the limit cn be computed without forming the sum. The Fundmentl Theorem sys: If we cn find F [] so tht df [] =f[] d, for ll b, then b b X f[] d = lim [f [] ] =F [b] F [] = step Finding n ntiderivtive lets us skip from the left side of the bove equtions to the right side without going through the limit in the middle. This indirect computtion of the integrl works ny time we cn find trick to figure out n ntiderivtive. There re mny such tricks, or techniques, nd the computer knows them ll. Your min tsk in this chpter is to understnd the fundmentl techniques nd their limittions, rther thn to develop skill t very elborte integrl computtions. 38
Chpter 3 - SYMBOLIC INTEGATION 39 The rules of integrtion re more difficult thn the rules of differentition becuse they mount to trying to use the rules of differentition in reverse. You must lern ll the bsic techniques to understnd this, but we will not wllow very deep into the swmp of esoteric techniques. Tht is left for the computer. 3. Indefinite Integrls The first hlf of the Fundmentl Theorem mens tht we cn often find n integrl in two steps: ) Find n ntiderivtive, 2) compute the difference in vlues of the ntiderivtive. It mkes no difference which ntiderivtive we use. We formlize nottion tht breks up these two steps. One ntiderivtive of 3 2 is 3,since d3 d =32, but nother ntiderivtive of 3 2 is 3 + 273, since the derivtive of the constnt 273 (or ny other constnt) is zero. The integrl b 32 d my be computed with either ntiderivtive. b 3 2 d = 3 b = b 3 3 =[ 3 + 273] b =[b 3 + 273] [ 3 + 273] = b 3 3 Our net result is the converse of the result tht sys the derivtive of constnt is zero. It sys tht if the derivtive is zero, the function must be constnt. This is geometriclly obvious - drw the grph of function with zero derivtive! Theorem 3. The ero-derivtive Theorem Suppose F [] nd G [] re both ntiderivtives for the sme function f [] on the intervl [, b]; tht is, df dg d [] = d [] =f [] for ll in [, b]. ThenF [] nd G [] differ by constnt for ll in [, b]. Proof: The function H [] =F [] G [] hs zero derivtive on [, b]. Wehve H [X] H [] = X d = by the first hlf of the Fundmentl Theorem nd direct computtion of the integrl of zero. If X is ny rel vlue in [, b], H [X] H [] =. This mens H [X] =H [], constnt. In turn, this tells us tht F [X] G [X] =F [] G [], constntforllx in [, b].
Chpter 3 - SYMBOLIC INTEGATION 3 Definition 3. Nottion for the Indefinite Integrl The indefinite integrl of function f [] denoted f [] d is equl to the collection of ll functions F [] with differentil or derivtive df d [] =f []. df [] =f [] d We write +c fter n nswer to indicte ll possible ntiderivtives. For emple, Cos[θ] dθ =Sin[θ]+c Eercise Set 3. Verify tht both 2 nd 2 + π equl 2d.
Chpter 3 - SYMBOLIC INTEGATION 3 3.2 Specific IntegrlFormuls p d = p + p+ + c, p 6= d =Log[]+c, > e d = e + c Sin[] d = Cos[]+c Cos[] d =Sin[]+c Emple 3. Guess nd Correct 6 Suppose we wnt to find 3 5 d = F []? We know tht if we differentite power function, we reduce the eponent by, so we guess nd check our nswer, F [] = 6 F [] =65 The constnt is wrong but would be correct if we chose F [] = 2 6 F [] = 2 6 5 =3 5 Here is nother emple of guessing nd correcting the guess by djusting constnt. Emple 3.2 Guess nd Correct Sin[3 ] Find Begin with the first guess nd check 7Cos[3] d = G[] G [] =Sin[3] G [] =3Cos[3]
Chpter 3 - SYMBOLIC INTEGATION 32 Adjusting our guess gives G[] = 7 3 Sin[3 ] G [] = 7 3 3Cos[3] =7Cos[3] It is best to check your work in ny cse so you only need to remember the five specific bsic formuls bove nd use them to djust your guesses. We will lern generl rules bsed on ech of the rules for differentition but used in reverse. Here is some bsic drill work. Eercise Set 3.2. Bsic Drill on Guessing nd Correcting ) 7 d=? b) 5 3 d =? c) e) 3 2 d =? d) (5 ) 2 d =? f) 3 2 d =? Sin[3 ] d =? g) e 2 d =? h) 7 d =? You cnnot do nything you like with indefinite integrls nd epect to get the intended function. In prticulr, the d in the integrl tells you the vrible of differentition for the intended nswer. 2. Eplin wht is wrong with the following nonsense: 2 d = d= d = [ 2 2 + c] = 2 3 + c nd 2 d = d= = [ 2 2 ]= 2 d
Chpter 3 - SYMBOLIC INTEGATION 33 3.3 Superposition of Antiderivtives f[]+bg[] d = f[] d + b g[] d Emple 3.3 Superposition of Derivtives in everse We prove the superposition rule f[]+bg[] d = f[] d + b g[] d by letting F [] = f[] d, G[] = g[] d nd writing out wht the clim for indefinite integrls mens in terms of these functions: F [] =f[] G [] =g[] (F[]+bG[]) = F []+bg [] =f[]+bg[] ( f[] d + b g[] d) = f[]+bg[] nd ( f[]+bg[] d) = f[]+bg[] Do the rbitrry constnts of integrtion mtter? No, s long s we interpret the sum of two rbitrry constnts s just nother rbitrry constnt. Emple 3.4 Superposition for Integrls (2 Sin[] 3 ) d =2 Sin[] d 3 d = 2Cos[] 3Log[]+c Now, use your rule to brek liner combintions of integrnds into simpler pieces.
Chpter 3 - SYMBOLIC INTEGATION 34 Eercise Set 3.3. Superposition of Antiderivtives Drill ) b) c) 5 3 2 d =5 3 d 2 d =? 5 2 d=5 d 2 d=? 3 3 3 3 + 2 d =? d) 5Sin[] e 2 d =? e) Sin[5 ] 5Sin[] d =? f) Cos[5 ] 5 d =? emember tht the computer cn be used to check your work on bsic skills. 2. un the computer progrm SymbolicIntegr, nd use the computer to check your work from the previous eercise. 3.4 Substitution for Integrls One wy to find n indefinite integrl is to chnge the problem into simpler one. Of course, you wnt to chnge it into n equivlent problem. Chnge of vribles cn be done legitimtely s follows. First, let u = prt of the integrnd. Net, clculte du =. If the remining prt of the integrnd is du, mke the substitution nd; if it is not du, trydifferent substitution. The point is tht we must look for both n epression nd its differentil. Here is very simple emple. Emple 3.5 A Chnge of Vrible nd Differentil Find 2 p + 2 d = p[ + 2 ]{2 d} Begin with u =[+ 2 ] du = {2 d}
Chpter 3 - SYMBOLIC INTEGATION 35 We replce the epression for u nd du, thereby obtining the simpler problem: Find p[u] {du} = u 2 du = + u + 2 + c 2 = 2 3 u 3 2 + c The epression 2 3 u 3 2 + c is not n cceptble nswer to the question, Wht functions of hve derivtive 2 + 2? However, if we remember tht u =+ 2, we cn epress the nswer s 2 3 u 3 2 2 + c = 3 [ + 2 ] 3 2 + c Checking the nswer will show why this method works. We use the Chin ule: nd or y = 2 3 u 3 2 u =+ 2 so dy du = 2 3 3 2 u 3 du 2 d =2 dy d = dy du du d = u p 2 2 =2 + 2 dy =2 p + 2 d Emple 3.6 Another Chnge You must substitute for the differentil du ssocited with your chnge of vribles u =. Sometimes this is little complicted. For emple, suppose we try to compute 2 p + 2 d with the chnge of vrible nd differentil v = + 2 dv = + 2 d Our integrl becomes 2v[ rest of bove]
Chpter 3 - SYMBOLIC INTEGATION 36 with the rest of the bove equl to d.wecnfind dv = + d by multiplying numertor nd 2 denomintor by + 2,soourintegrlbecomes 2 p + + 2 2 ³p 2 d = 2 + 2 [ d] + 2 + 2 =2 v 2 dv = 2 3 v3 + c = 2 + 2 3 2 3 This is the sme nswer s before, but the substitution ws more difficult on the dv piece. Emple 3.7 AFiledAttempt Sometimes n ttempt to simplify n integrnd by chnge of vribles will led you to either more complicted integrl or sitution in which you cnnot mke the substitution for the differentil. In these cses, scrtch off your work nd try nother chnge. Suppose we try grnd simplifiction of 2 p + 2 d tking w = p + 2 dw = µ p+ 2 + 2 d + 2 We might substitute w for the whole integrnd, but there is nothing left to substitute for dw nd we cnnot complete the substitution. We simply hve to try different method. Emple 3.8 A Less Obvious Substitution with u = We my be slipping into the symbol swmp, but little wllowing cn be fun. Here is chnge
Chpter 3 - SYMBOLIC INTEGATION 37 of vrible nd differentil with twist: + d u = u 2 = ( >) u +u 2 2 udu=2 du = 2 d 2 udu= d u 2 µ du =2 +u2 =2 du 2 + d =2 2ArcTn[ ]+c µ +u 2 du du =2u 2ArcTn[u]+c +u2 Now, you try it. Eercise Set 3.4 Chnge of Vrible nd Differentil Drill ) d =? b) (3 2) 2 c) (3 + 7 2 ) 3 d =? d) 2t t 2 dt =? 3y (2+2y 2 ) 2 dy =? e) g) (Cos[]) 3 Sin[] d =? f) e Cos[θ] Sin[θ] dθ =? h) Cos[Log[]] d =? Sin[+ b] d =? If you hve checked severl indefinite integrtion problems tht you computed with chnge of vrible nd differentil, you probbly relize tht the Chin ule lies behind the method. Perhps you cn formlize your ide. Problem 3. The Chin ule in everse Use the Chin ule for differentition to prove the indefinite integrl form of Integrtion by Substitution. Once you hve this indefinite rule, use the Fundmentl Theorem to prove the definite rule.
Chpter 3 - SYMBOLIC INTEGATION 38 3.5 Chnge of Limits of Integrtion When we wnt n ntiderivtive such s 2 + 2 d, we hve no choice but to re-substitute the epressions for new vribles bck into our nswer. In the first emple of the previous section, 2 p + 2 d = 2 3 u 3 2 + c = 2 3 [ + 2 ] 3 2 However, when we wnt to compute definite integrl such s 7 5 2 p + 2 d we cn chnge the limits of integrtion long with the chnge of vribles. For emple, so the new problem is to find u =+ 2 du =2d u[5] = 26 u[7] = 5 5 which equls 2 3 u 3 2 5 26 = 2 3 [5 3 3 2 26 2 ] 235.72 88.383 = 47.39 We recommend tht you do your definite integrl chnges of vrible this wy: 26 u 2 du Procedure 3. To compute b f[] d. Set portion of your integrnd f[] equl to new vrible u = u[]. 2. Clculte du =?? d. 3. Clculte u[] =α nd u[b] =β. 4. Substitute both the function nd the differentil in f[] d for g[u] du. 5. Compute β α g[u] du
Chpter 3 - SYMBOLIC INTEGATION 39 Py Me Now or Py Me Lter: You cn compute the ntiderivtive in terms of nd then use the originl limits of integrtion, but there is dnger tht you will lose trck of the vrible with which you strted. If you re creful, both methods give the sme nswer; for emple, 7 5 2 p + 2 d = 2 3 [ + 2 ] 3 2 7 5 = 2 3 ³ [ + 7 2 ] 3 2 [ + 5 2 ] 3 2 47.39 Eercise Set 3.5. Chnge of Vribles with Limits Drill ) + 2 d =? b) π/6 Sin[3θ] dθ =? c) 3 2 2 ( 2 3) 2 d =? d) + bd=? e) 3 2 + d =becuse of the new limits. 3 f) e 2 d =? g) π/2 Cos[θ] Sin[θ] dθ =? 2. Use the chnge of vrible u = nd ssocited chnge of differentil to convert Cos[ ] d into multiple of u 2 Cos[u] du. 3.5. Integrtion with Prmeters Chnge of vribles cn mke integrls with prmeters, which re importnt in mny scientific nd mthemticl problems, into specific integrls. For emple, suppose tht ω nd re constnt. Then Sin[ωt] dt u = ωt du = ωdt du = dt ω Sin[ωt] dt = Sin[u] ω du = Sin[u] du ω
Chpter 3 - SYMBOLIC INTEGATION 32 nd 2 + 2 dt = 2 ( + (/) 2 ) d = 2 +(/) 2 d u = du = d du= d 2 + 2 dt = 2 +u 2 du= +u 2 du This type of chnge of vrible nd differentil my be the most importnt kind for you to think bout becuse the computer cn give you specific integrls, but you my wnt to see how n integrl depends on prmeter. Problem 3.2 Pull the Prmeter Out Show tht 3 p 9 2 d =3 3 p (/3) 2 d =9 p u 2 du Show tht the re of circle of rdius r is r 2 times the re of the unit circle. First, chnge vribles to obtin the integrls below nd then red on to interpret your computtion. r p p r 2 2 d = r 2 u 2 du 3.6 Trig Substitutions Study this section if you hve personl need to compute integrls with one of the following epressions (nd do not hve your computer): p p 2 2, 2 + 2 or 2 + 2 You should skim red this section even if you do not wish to develop this skill becuse the positive sign needed to go between (Cos[θ]) 2 = (Sin[θ]) 2 nd Cos[θ] = p (Sin[θ]) 2 cn cuse errors in the use of symbolic integrtion pckge. In other words, the computer my use the symbolic squre root when you intend for it to use the negtive.
Chpter 3 - SYMBOLIC INTEGATION 32 Emple 3.9 A Trigonometric Chnge of Vribles The following integrl comes from computing the re of circle using the integrl. A sine substitution mkes it one we cn ntidifferentite with two tricks from trig. p π/2 q u 2 du = Sin 2 [θ]cos[θ] dθ becuse we tke the substitution u =Sin[θ] du =Cos[θ] dθ u = θ = u = θ = π/2 q We know from high school trig tht Sin 2 [θ] =Cos 2 [θ], so, when the cosine is positive, Sin 2 [θ] =Cos[θ] nd π/2 q Sin 2 [θ]cos[θ] dθ = π/2 Cos 2 [θ] dθ We lso know from high school trig (or looking t the grph nd thinking little) tht Cos 2 [θ] = 2 [ + Cos (2θ)], so π/2 Cos 2 [θ] dθ = π/2 π/2 2 [ dθ + Cos (2θ) dθ] becuse we use nother chnge of vribles = π 4 + 4 π Cos [φ] dφ φ =2θ dφ =2dθ φ = θ = φ = π θ = π/2 Finlly, π Cos [φ] dφ = 4 4 [Sin [φ] π ]= swecouldhveesilyseenbysketchinggrphofcos [φ] from to π, with equl res bove nd below the φ-is.
Chpter 3 - SYMBOLIC INTEGATION 322 Putting ll these computtions together, we hve p u 2 du = π 4 One eplntion why the chnge of vribles in the previous emple works is tht sine nd cosine yield prmetric equtions for the unit circle. More techniclly speking, the (Pythgoren Theorem) identity (Sin[θ]) 2 +(Cos[θ]) 2 = (Cos[θ]) 2 = (Sin[θ]) 2 becomes the identity (Cos[θ]) 2 = u 2 when we let u =Sin[θ]. There is n importnt lgebric detil when we write Cos[θ] = p (Sin[θ]) 2 This is flse when cosine is negtive. For the definite integrl bove, we wnted u nd chose θ π/2 to put Sin[θ] in this rnge. It is lso true tht Cos[θ] for this rnge of θ, so tht Cos[θ] = p u 2 If the cosine were not positive in the rnge of interest, the integrtion would not be vlid. More informtion on this difficulty is contined in the Mthemticl Bckground Chpter on Differentition Drill. An epression such s 2 2 cn first be reduced to multiple of u 2 by tking u = / nd writing 2 2 = p (/) 2. This mens tht the epression 2 2 cn be converted to Cos[θ] with the substitutions u =(/) 2 nd u =Sin[θ] (provided cosine is positive on the intervl). Notice tht the substitution u =Cos[θ] converts u 2 into p (Cos[θ]) 2 =Sin[θ], provided sine is positive. Also note tht sine is positive over different rnge of ngles thn cosine. Another trig identity sys (Tn[θ]) 2 + = (Sec[θ]) 2 µ Sin[θ] 2 + Cos[θ] (Sin[θ]) 2 +(Cos[θ]) 2 = µ Cos[θ] 2 µ 2 = Cos[θ] Cos[θ] If we mke chnge of vrible u =Tn[θ], thenu 2 +becomes (Sec[θ]) 2 nd +u 2 = Sec[θ] provided secnt is positive. Emple 3. The Sine Substitution Without Endpoints The cost of not chnging limits of integrtion in Emple 3.9 is the following: The sme tricks s bove for indefinite integrls yield p u 2 du = 2 θ + 4 Sin[φ]
Chpter 3 - SYMBOLIC INTEGATION 323 where u =Sin[θ] nd φ =2θ. Ifwerellywntthentiderivtiveof u 2, then we must epress ll this in terms of u. The first term is esy, we just use the inverse trig function u =Sin[θ] θ =ArcSin[u] The term Sin[φ] = Sin[2 θ] must first be written in terms of functions of θ, (recll the ddition formul for sine), Sin[2 θ] = 2 Sin[θ] Cos[θ] Now, we use tringle tht contins the u =Sin[θ] ide. FromSOH-CAH-TOA,ifwetke right tringle with hypotenuse nd opposite side u, thenu = Sin[θ] (see Figure 3.6:). We lso know tht Cos[θ] is the djcent side of this tringle. Using the Pythgoren Theorem, we hve Cos[θ] = u 2 θ u Combining these fcts we hve, Figure 3.6:: Sin[θ] =u nd Cos[θ] =dj/hyp = u 2 Sin[2 θ] =2Sinθ] Cos[θ] =2u p u 2 nd p u 2 du = 2 θ + 4 Sin[φ] = 2 ArcSin[u]+u u 2 2 Eercise Set 3.6. Geometric Proof tht u 2 du = π 4 Sketch the grph y = 2 for. Wht geometricl shpe is shown in your grph? Wht is the re of one fourth of circle of unit rdius?
Chpter 3 - SYMBOLIC INTEGATION 324 2. Compute the integrl 9 9 2 d. (Check your symbolic computtion geometriclly). Use the chnge of vrible =Sin[θ] with n pproprite differentil to show tht v d = dθ = θ... = ArcSin[v] 2 How lrge cn we tke v? Use the chnge of vrible =Tn[θ] with n pproprite differentil to show tht v + 2 d = dθ = θ... =ArcTn[v] How lrge cn we tke v? 3. Working Bck from Trig Substitutions Suppose we mke the chnge of vrible u =Sin[θ] (s in the integrtion bove). Epress Tn[θ] in terms of u by using Figure 3.6: nd TOA. A tringle is shown in Figure 3.6:2 for chnge of vrible v =Cos[θ]. EpressSin[θ] nd Tn[θ] in terms of v by using the figure nd the Pythgoren Theorem. A tringle is shown in Figure 3.6:2 for chnge of vrible w =Tn[θ]. EpressSin[θ] nd Cos[θ] in terms of w by using the figure nd the Pythgoren Theorem. θ u θ w v Figure 3.6:2: Tringles for v =Cos[θ] =CAHndw =Tn[θ] =TOA We re beginning to wllow little too deeply in trig. The point of the previous emple could simply be: chnge the limits of integrtion when you chnge vrible nd differentil. However, it is possible to chnge bck to u, nd the previous eercise gives you strt on the trig skills needed to do this. Problem 3.3 A Constnt Use the chnge of vrible =Cos[θ] with n pproprite differentil to show tht d = dθ = θ... = ArcCos[]+c 2
Chpter 3 - SYMBOLIC INTEGATION 325 Also, use the chnge of vrible =Sin[θ] with n pproprite differentil to show tht d = dθ = θ... = ArcSin[]+c 2 Is ArcCos[] = ArcSin[]? Ask the computer to Plot ArcSin[] nd ArcCos[]. Why do they look like? How do they differ? Do the grphs of Cos[θ] nd Sin[θ] look like? How do they differ? 3.7 Integrtion by Prts Integrtion by Prts is importnt theoreticlly, but it ppers to be just nother trick. It is more thn tht, but first you should lern the trick. The formuls for the technique re b u [] dv [] =u [] v [] b b v [] du [] or, suppressing the dependence, b = udv= uv b = b = vdu The ide is to brek up n integrnd into function u[] nd differentil dv[] where you cn find the differentil du[] (usully esy), the ntiderivtive v[] (sometimes hrder), nd, finlly, where v[] du[] is n esier problem. Unfortuntely, often the only wy to find out if the new problem is esier is to go through ll the substitution steps for the terms. Helpful Nottion We encourge you to block off the four terms in this formul. First brek up your integrnd into u nd dv u = dv = nd then compute the differentil of u nd the ntiderivtive of dv du = v =
Chpter 3 - SYMBOLIC INTEGATION 326 Here is n emple of the use of Integrtion by Prts. Emple 3. Integrtion by Prts for b Log [] d =? Use the prts u =Log[] du = d nd dv = d v = 2 2 mking the integrls b udv = uv b b vdu b b b b Log [] d = b 2 2 Log [] b Log [] d = 2 2 Log [] b b Log [] d = 2 2 Log [] b 2 4 b 2 2 d 2 d Log [] d = 2 [b2 Log (b) 2 Log []] 4 [b2 2 ] provided tht both nd b re positive. (Otherwise, Log is undefined). Check: Notice tht the indefinite integrl of the clcultion bove is Log [] d = 2 2 Log[] 4 2 + c We check the correctness of this ntiderivtive by differentiting the right side of the eqution. First, we use the Product ule
Chpter 3 - SYMBOLIC INTEGATION 327 d(f[] g[]) 3 = df d d g + f dg d d( 2 Log[]) d =2 Log[]+ 2 =2 Log[]+ f[] = 2 df d =2 Net, we differentite the whole epression d( 2 2 Log[] 4 2 + c) = Log[]+ d 2 2 += Log[] 4 This verifies tht the indefinite integrl is correct. g[] =Log[] dg d = Emple 3.2 e d To compute this integrl, use integrtion by prts with so u = du = d dv = e d v = e This gives udv= uv vdu = e d = e e d = e e =( ) e + c e d =( ) e + c Check: Differentite using the Product ule: d(f[] g[]) d d(( ) e ) d = df d g + f dg d = e +( ) e = e f[] =( ) df d = g[] =e dg d = e
Chpter 3 - SYMBOLIC INTEGATION 328 Emple 3.3 A eduction of One Integrl to Previous One The integrl 2 e d cn be done by prts in two steps. First, tke the prts so u = 2 du =2d dv = e d v= e This gives udv= uv vdu = 2 e d = 2 e 2 e d = 2 e 2( ) e + c Note tht the second integrl ws computed in the previous emple, so 2 e d =( 2 2 +2)e + c Check: Differentite using the Product ule: d(f[] g[]) d = df d g + f dg d f[] =( 2 2 +2) g[] =e df d =2 2 dg d = e d(( 2 2 +2)e ) d =(2 2) e +( 2 2 +2)e = 2 e
Chpter 3 - SYMBOLIC INTEGATION 329 Emple 3.4 Circulr Prts Still Gives n Answer We compute the integrl e 2 Sin[3 ] d by the prts so u = e 2 dv =Sin[3] d du =2e 2 d v = Cos[3 ] 3 This gives udv = uv vdu = e 2 Sin[3 ] d = 3 e2 Cos[3 ]+ 2 3 e 2 Cos[3 ] d Now, use the prts on the second integrl, so w = e 2 dz =Cos[3] d dw =2e 2 d z = Sin[3 ] 3 This gives wdz = wz zdw = e 2 Cos[3 ] d = 3 e2 Sin[3 ] 2 3 e 2 Sin[3 ] d Substituting this into the second integrl bove, we obtin e 2 Sin[3 ] d = 3 e2 Cos[3 ]+ 2 3 3 e2 Sin[3 ] 2 e 2 Sin[3 ] d 3 = 3 e2 Cos[3 ]+ 2 9 e2 Sin[3 ] 4 e 2 Sin[3 ] d 9 Bringing the like integrl to the left side, we obtin e 2 Sin[3 ] d + 9 e 2 Sin[3 ] d = 2 4 9 e2 Sin[3 ] 3 e2 Cos[3 ]+c 4+9 e 2 Sin[3 ] d = 2 4 9 e2 Sin[3 ] 3 e2 Cos[3 ]+c e 2 Sin[3 ] d = 2 3 e2 Sin[3 ] 3 3 e2 Cos[3 ]+c
Chpter 3 - SYMBOLIC INTEGATION 33 Emple 3.5 A Two-Step Computtion of (Cos[]) 2 d This integrl cn lso be computed without the trig identities in Section 3.6. Use the prts so u =Cos[] du = Sin[] d dv =Cos[] d v =Sin[] which yields the integrtion formul (Cos[]) 2 d =Sin[]Cos[]+ =Sin[]Cos[]+ =Sin[]Cos[]+ (Sin[]) 2 d [ (Cos[]) 2 ] d d (Cos[]) 2 d so 2 (Cos[]) 2 d = Sin[]Cos[]+c nd (Cos[]) 2 d = 2 2 Sin[]Cos[]+c Here re some prctice problems. (emember tht you cn check your work with the computer). Eercise Set 3.7. Drill on Integrtion by Prts ) θ Cos[θ] dθ =? b) θ 2 Sin[θ] dθ =? c) e) g) π/2 θ Sin[θ] dθ =? d) e 2 d =? f) e 5 Sin[3 ] d =? h) π/2 θ Cos[θ] dθ =? 2 e 2 d =? e 3 Cos[5 ] d =? 2. Compute Log[] d using integrtion by prts with u =Log[] nd dv = d. Check your nswer by differentition.
Chpter 3 - SYMBOLIC INTEGATION 33 3. Check the previous indefinite integrl from Emple 3.4 by using the Product ule to differentite 3 2 e2 Sin[3 ] 3 3 e2 Cos[3 ]. Compute the integrl e 2 Sin[3 ] d by the prts so u =Sin[3] du =3Cos[3] d dv = e 2 d v = 2 e2 4. Clculte (Sin[]) 2 d. There is something indefinite bout these integrls. 5. Wht is wrong with the eqution d = d d = + 2 when you use integrtion by prts with u =, dv = d, du = d, ndv = 2? Subtrcting from both sides of the equtions bove yields d = The proof of the Integrtion by Prts formul is ctully esy. 6. The Product ule in everse Use the Product ule for differentition to prove the indefinite integrl form of Integrtion by Prts. Notice tht if H[] is ny function, dh[] =H[]+c, bydefinition. Let H = uv nd show tht dh = udv+ vdu.indefinitely integrte both sides of the dh eqution, u v = H[] = dh = udv+ vdu Once you hve this indefinite rule, use the Fundmentl Theorem to prove the definite rule. Here re some tougher problems in which you need to use more thn one method t time. 7. () Cos[]Sin[] d =? Notice tht (Cos[θ]) 2 dθ is done bove two wys. (b) θ (Cos[θ]) 2 dθ =? Use integrls from the previous drill problems. (c) 3 d =? 2 HINT: Use prts u = 2 nd dv = d 2 nd compute the dv integrl.
Chpter 3 - SYMBOLIC INTEGATION 332 (d) q 3 d =? q Use prts u = nd dv = 2. Compute the dv integrl. (e) ArcTn[] d =? Use prts u =ArcTn[] nd dv = d. (f) ArcTn[] d =? Use prts nd the previous eercise. (g) ArcTn[ ] d =? Use prts u =ArcTn[ ] nd chnge vribles in the resulting integrl. (h) 9 4 Sin[ ] d =? Use prts u =Sin[ ] nd dv = d. Then chnge vribles with w =. (i) (Log[]) 2 d =? Use the prts u =Log[] nd dv =Log[] d. Clculte the dv integrl. 3.8 Impossible Integrls There re importnt limittions to symbolic integrtion tht go beyond the prcticl difficulties of lerning ll the tricks. This section eplins why. Integrtion by prts nd the chnge of vrible nd differentil re importnt ides for the theoreticl trnsformtion of integrls. In this chpter, we tried to include just enough drill work for you to lern the bsic methods. Before the prcticl implementtion of generl ntidifferentition lgorithms on computers, development of humn integrtion skills ws n importnt prt of the trining of scientists nd engineers. Now, the computer cn mkes this skill esier to mster. The skill hs lwys hd limittions. Erly in the dys of clculus, it ws quite impressive tht integrtion could be used to lern mny mny new formuls such s the clssicl formuls for the re of circle or volume of sphere. We sw how esy it ws to generlize the integrtion pproch to the volume of cone. However, some simple-looking integrls hve no ntiderivtive wht so ever. This is not the result of peculir mthemticl emples. The rclength of n ellipse just mens the length mesured s you trvel long n ellipse. Problem 4.2 sks you to find integrl formuls for this rclength. Erly developers of clculus must hve tried very hrd to compute those integrls with symbolic ntiderivtives, but fter
Chpter 3 - SYMBOLIC INTEGATION 333 more thn century of trying, Liouville proved tht there is no nlyticl epression for tht ntiderivtive in terms of the clssicl functions. The fct tht the ntiderivtive hs no epression in terms of old functions does not men tht the integrl does not eist. If you find the following integrl with the computer, you will see peculir result: Cos[ 2 ] d The innocent-looking integrl Cos[ 2 ] d is not innocent t ll. The function Cos[ 2 ] is perfectly smooth nd well behved, but it does not hve n ntiderivtive tht cn be epressed in terms of known functions. The bottom line is this: Integrls re used to define nd numericlly compute importnt new functions in science nd mthemtics, even when they do not hve epressions in terms of elementry functions. Functions given by integrl formuls cn still be differentited just s you did in Eercise 2.8.. Eercise Set 3.8 un the SymbolicIntegr progrm.