TECHNIQUES OF INTEGRATION
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1 7 TECHNIQUES OF INTEGRATION Simpso s Rule estimates itegrals b approimatig graphs with parabolas. Because of the Fudametal Theorem of Calculus, we ca itegrate a fuctio if we kow a atiderivative, that is, a idefiite itegral. We summarize here the most importat itegrals that we have leared so far. d C e d e C d l C a d a l a C si d cos C sec d ta C sec ta d sec C sih d cosh C ta d l sec C cos d si C csc d cot C csc cot d csc C cosh d sih C cot d l si C a d a ta a C sa d si a C I this chapter we develop techiques for usig these basic itegratio formulas to obtai idefiite itegrals of more complicated fuctios. We leared the most importat method of itegratio, the Substitutio Rule, i Sectio 5.5. The other geeral techique, itegratio b parts, is preseted i Sectio 7.. we lear methods that are special to particular classes of fuctios, such as trigoometric fuctios ad ratioal fuctios. Itegratio is ot as straightforward as differetiatio; there are o rules that absolutel guaratee obtaiig a idefiite itegral of a fuctio. Therefore we discuss a strateg for itegratio i Sectio
2 7. INTEGRATION BY PARTS Ever differetiatio rule has a correspodig itegratio rule. For istace, the Substitutio Rule for itegratio correspods to the Chai Rule for differetiatio. The rule that correspods to the Product Rule for differetiatio is called the rule for itegratio b parts. The Product Rule states that if f ad t are differetiable fuctios, the d f t f t tf d I the otatio for idefiite itegrals this equatio becomes f t tf d f t or f t d tf d f t We ca rearrage this equatio as f t d f t tf d Formula is called the formula for itegratio b parts. It is perhaps easier to remember i the followig otatio. Let u f ad v t. the differetials are du f d ad dv t d, so, b the Substitutio Rule, the formula for itegratio b parts becomes udv uv v du EXAPLE Fid si d. SOLUTION USING FORULA Suppose we choose f ad t si. f ad t cos. (For t we ca choose a atiderivative of t.) Thus, usig Formula, we have si d f t tf d cos cos d cos cos d cos si C It s wise to check the aswer b differetiatig it. If we do so, we get si, as epected. 53
3 5 CHAPTER 7 TECHNIQUES OF INTEGRATION SOLUTION USING FORULA Let N It is helpful to use the patter: u dv du v u du d dv si d v cos ad so u d u du si d si d cos cos d cos cos d cos si C NOTE Our aim i usig itegratio b parts is to obtai a simpler itegral tha the oe we started with. Thus i Eample we started with si dad epressed it i terms of the simpler itegral cos d. If we had istead chose u si ad dv d, the du cos dad v, so itegratio b parts gives si d si cos d Although this is true, cos dis a more difficult itegral tha the oe we started with. I geeral, whe decidig o a choice for u ad dv, we usuall tr to choose u f to be a fuctio that becomes simpler whe differetiated (or at least ot more complicated) as log as dv t d ca be readil itegrated to give v. V EXAPLE Evaluate l d. SOLUTION Here we do t have much choice for u ad dv. Let u l dv d du d v Itegratig b parts, we get l d l d N It s customar to write d as d. N Check the aswer b differetiatig it. l d l C Itegratio b parts is effective i this eample because the derivative of the fuctio f l is simpler tha f.
4 SECTION 7. INTEGRATION BY PARTS 55 V EXAPLE 3 Fid t et dt. SOLUTION Notice that t becomes simpler whe differetiated (whereas e t is uchaged whe differetiated or itegrated), so we choose Itegratio b parts gives u t du tdt dv e t dt v e t 3 t e t dt t e t te t dt The itegral that we obtaied, te t dt, is simpler tha the origial itegral but is still ot obvious. Therefore, we use itegratio b parts a secod time, this time with u t ad dv e t dt. du dt, v e t, ad te t dt te t e t dt te t e t C Puttig this i Equatio 3, we get t e t dt t e t te t dt t e t te t e t C t e t te t e t C where C C N A easier method, usig comple umbers, is give i Eercise 5 i Appedi H. V EXAPLE Evaluate e si d. SOLUTION Neither e or si becomes simpler whe differetiated, but we tr choosig u e ad dv si dawa. du e d ad v cos, so itegratio b parts gives e si d e cos e cos d The itegral that we have obtaied, e cos d, is o simpler tha the origial oe, but at least it s o more difficult. Havig had success i the precedig eample itegratig b parts twice, we persevere ad itegrate b parts agai. This time we use u e ad dv cos d. du e d, v si, ad 5 e cos d e si e si d At first glace, it appears as if we have accomplished othig because we have arrived at e si d, which is where we started. However, if we put the epressio for e cos d from Equatio 5 ito Equatio we get e si d e cos e si e si d
5 56 CHAPTER 7 TECHNIQUES OF INTEGRATION N Figure illustrates Eample b showig the graphs of f e si ad F e si cos. As a visual check o our work, otice that f whe F has a maimum or miimum. This ca be regarded as a equatio to be solved for the ukow itegral. Addig e si dto both sides, we obtai e si d e cos e si F Dividig b ad addig the costat of itegratio, we get f e si d e si cos C _3 FIGURE _ 6 If we combie the formula for itegratio b parts with Part of the Fudametal Theorem of Calculus, we ca evaluate defiite itegrals b parts. Evaluatig both sides of Formula betwee a ad b, assumig f ad t are cotiuous, ad usig the Fudametal Theorem, we obtai 6 b a b f t d f t] a b tf d a EXAPLE 5 Calculate ta d. SOLUTION Let u ta du dv d d v So Formula 6 gives N Sice ta for, the itegral i Eample 5 ca be iterpreted as the area of the regio show i Figure. ta d ta ] ta ta d d d =ta! To evaluate this itegral we use the substitutio t (sice u has aother meaig i this eample). dt d, so d dt. Whe, t ; whe, t ; so d dt l t t ] l l l FIGURE Therefore ta d d l
6 SECTION 7. INTEGRATION BY PARTS 57 EXAPLE 6 Prove the reductio formula N Equatio 7 is called a reductio formula because the epoet has bee reduced to ad. 7 si d cos si si d where is a iteger. SOLUTION Let u si du si cos d dv si d v cos so itegratio b parts gives si d cos si si cos d Sice cos si, we have si d cos si si d si d As i Eample, we solve this equatio for the desired itegral b takig the last term o the right side to the left side. Thus we have si d cos si si d or si d cos si si d The reductio formula (7) is useful because b usig it repeatedl we could evetuall epress si di terms of si d(if is odd) or si d d (if is eve). 7. EXERCISES Evaluate the itegral usig itegratio b parts with the idicated choices of u ad dv.. arcta tdt. p 5 l pdp. l d; u l, dv d 3. t sec tdt. s s ds cos d. ; u, dv cos d 5. l d 6. t sih mt dt 3 3 Evaluate the itegral. 7. e si 3 d 8. e cos d 3. cos 5d 5. re r dr 6. t si t dt 7. si d 8. cos m d 9. l d. si d. e d 9.. t cosh tdt. 3. l d. 3 cos d t si 3tdt. e d 9 l s d
7 58 CHAPTER 7 TECHNIQUES OF INTEGRATION 5. d 6. e 7. cos d cos lsi d 3. s3 arcta d l d 3 r 3 s r dr (b) se part (a) to evaluate si 3 dad si 5 d. (c) se part (a) to show that, for odd powers of sie, si d 6. Prove that, for eve powers of sie, l d 3. t e s sit s ds si d irst make a substitutio ad the use itegratio b parts to evaluate the itegral. 33. cos s d 3. t 3 e t dt 35. s s 3 cos d ; 39 Evaluate the ide ite itegral. llustrate, ad check that our aswer is reasoable, b graphig both the fuctio ad its atiderivative (take ). 3. (a) se the reductio formula i Eample to show that l d 38. sil d 39. 3e d. 3 l d. 3 s d. si d (b) se part (a) ad the reductio formula to evaluate si d.. (a) Prove the reductio formula cos d cos si (b) se part (a) to evaluate cos d. (c) se parts (a) ad (b) to evaluate cos d. 5. (a) se the reductio formula i Eample to show that si d si d where is a iteger. si e cos t si t dt si d cos d 7 5 se itegratio b parts to prove the reductio formula se Eercise 7 to d l 3 d. 5. se Eercise to d e d id the area of the regio bouded b the give curves. 53. e.,, 5. l d l l d e d e e d ta d ta ta d sec d ta sec 5 l, ; se a graph to d approimate coordiates of the poits of itersectio of the give curves. d (approi matel) the area of the regio bouded b the curves. 55. si, 56. arcta 3, l 5 sec d 57 6 se the method of clidrical shells to d the volume geerated b rotatig the regio bouded b the give curves about the speci ed ais. 57. cos,, ; about the ais 58. e, e, ; about the ais 59. e,,, ; about 6. e,, ; about the ais
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