y udv uv y v du 7.1 INTEGRATION BY PARTS

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1 7. INTEGRATION BY PARTS Ever differetitio rule hs correspodig itegrtio rule. For istce, the Substitutio Rule for itegrtio correspods to the Chi Rule for differetitio. The rule tht correspods to the Product Rule for differetitio is clled the rule for itegrtio b prts. The Product Rule sttes tht if f d t re differetible fuctios, the d f t f t tf d I the ottio for idefiite itegrls this equtio becomes f t tf d f t or f t d tf d f t We c rerrge this equtio s f t d f t tf d Formul is clled the formul for itegrtio b prts. It is perhps esier to remember i the followig ottio. Let u f d v t. the differetils re du f d d dv t d,so,b the Substitutio Rule,the formul for itegrtio b prts becomes udv uv v du EXAPLE Fid si d. SOLUTION USING FORULA Suppose we choose f d t si. f d t cos. (For t we c choose tiderivtive of t.) Thus, usig Formul, we hve si d f t tf d cos cos d cos cos d cos si C It s wise to check the swer b differetitig it. If we do so, we get si, s epected. 53

2 5 CHAPTER 7 TECHNIQUES OF INTEGRATION SOLUTION USING FORULA Let N It is helpful to use the ptter: u dv du v u du d dv si d v cos d so u d u du si d si d cos cos d cos cos d cos si C NOTE Our im i usig itegrtio b prts is to obti simpler itegrl th the oe we strted with. Thus i Emple we strted with si d d epressed it i terms of the simpler itegrl cos d. If we hd isted chose u si d dv d, the du cos d d v, so itegrtio b prts gives si d si cos d Although this is true, cos d is more difficult itegrl th the oe we strted with. I geerl, whe decidig o choice for u d dv, we usull tr to choose u f to be fuctio tht becomes simpler whe differetited (or t lest ot more complicted) s log s dv t d c be redil itegrted to give v. V EXAPLE Evlute l d. SOLUTION Here we do t hve much choice for u d dv. Let u l dv d du d v Itegrtig b prts, we get l d l d N It s customr to write d s d. N Check the swer b differetitig it. l d l C Itegrtio b prts is effective i this emple becuse the derivtive of the fuctio f l is simpler th f.

3 SECTION 7. INTEGRATION BY PARTS 55 V EXAPLE 3 Fid t e t dt. SOLUTION Notice tht t becomes simpler whe differetited (wheres e t is uchged whe differetited or itegrted), so we choose u t dv e t dt Itegrtio b prts gives du tdt v e t 3 t e t dt t e t te t dt The itegrl tht we obtied, te t dt, is simpler th the origil itegrl but is still ot obvious. Therefore, we use itegrtio b prts secod time, this time with u t d dv e t dt. du dt, v e t, d te t dt te t e t dt te t e t C Puttig this i Equtio 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 esier method, usig comple umbers, is give i Eercise 5 i Appedi H. V EXAPLE Evlute e si d. SOLUTION Neither e or si becomes simpler whe differetited, but we tr choosig u e d dv si d w. du e d d v cos, so itegrtio b prts gives e si d e cos e cos d The itegrl tht we hve obtied, e cos d,is o simpler th the origil oe,but t lest it s o more difficult. Hvig hd success i the precedig emple itegrtig b prts twice, we persevere d itegrte b prts gi. This time we use u e d dv cos d. du e d, v si, d 5 e cos d e si e si d At first glce, it ppers s if we hve ccomplished othig becuse we hve rrived t e si d,which is where we strted. However,if we put the epressio for e cos d from Equtio 5 ito Equtio we get e si d e cos e si e si d

4 56 CHAPTER 7 TECHNIQUES OF INTEGRATION N Figure illustrtes Emple b showig the grphs of f e si d F e si cos. As visul check o our work, otice tht f whe F hs mimum or miimum. This c be regrded s equtio to be solved for the ukow itegrl. Addig e si d to both sides, we obti e si d e cos e si F Dividig b d ddig the costt of itegrtio, we get f e si d e si cos C _3 FIGURE _ 6 If we combie the formul for itegrtio b prts with Prt of the Fudmetl Theorem of Clculus, we c evlute defiite itegrls b prts. Evlutig both sides of Formul betwee d b, ssumig f d t re cotiuous, d usig the Fudmetl Theorem, we obti 6 b b f t d f t] b tf d EXAPLE 5 Clculte t d. SOLUTION Let u t du dv d d v So Formul 6 gives N Sice t for, the itegrl i Emple 5 c be iterpreted s the re of the regio show i Figure. t d t ] t t d d d =t! To evlute this itegrl we use the substitutio t (sice u hs other meig i this emple). dt d, so d dt. Whe, t ; whe, t ; so d dt l t t ] l l l FIGURE Therefore t d d l

5 SECTION 7. INTEGRATION BY PARTS 57 EXAPLE 6 Prove the reductio formul N Equtio 7 is clled reductio formul becuse the epoet hs bee reduced to d. 7 si d cos si si d where is iteger. SOLUTION Let u si du si cos d dv si d v cos so itegrtio b prts gives si d cos si si cos d Sice cos si, we hve si d cos si si d si d As i Emple, we solve this equtio for the desired itegrl b tkig the lst term o the right side to the left side. Thus we hve si d cos si si d or si d cos si si d The reductio formul (7) is useful becuse b usig it repetedl we could evetull epress si d i terms of si d (if is odd) or si d d (if is eve). 7. EXERCISES Evlute the itegrl usig itegrtio b prts with the idicted choices of u d dv.. rct t dt. p 5 l p dp. l d; u l, dv d 3. t sec t dt. s s ds. cos d; u, dv cos d 5. l d 6. t sih mt dt 3 3 Evlute the itegrl. 7. e si 3 d 8. e cos d 3. cos 5 d 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 si 3t dt.. t cosh t dt. e d 3. l. 3 cos d d 9 l s d

6 58 CHAPTER 7 TECHNIQUES OF INTEGRATION 5. d 6. e 7. cos d cos lsi d 3. s3 rct d l 3 d r 3 s r dr (b) Use prt () to evlute si 3 d d si 5 d. (c) Use prt () to show tht, for odd powers of sie, si d 6. Prove tht, for eve powers of sie, l d 3. t e s sit s ds si d First mke substitutio d the use itegrtio b prts to evlute the itegrl. 33. cos s d 3. t 3 e t dt 35. s s 3 cos d ; 39 Evlute the idefiite itegrl. Illustrte, d check tht our swer is resoble, b grphig both the fuctio d its tiderivtive (tke C ). 3. () Use the reductio formul i Emple 6 to show tht l d 38. sil d 39. 3e d. 3 l d. 3 s d. si d (b) Use prt () d the reductio formul to evlute si d.. () Prove the reductio formul (b) Use prt () to evlute cos d. (c) Use prts () d (b) to evlute cos d. 5. () Use the reductio formul i Emple 6 to show tht si d cos d cos si si d where is iteger. si e cos t si t dt C si d cos d 7 5 Use itegrtio b prts to prove the reductio formul Use Eercise 7 to fid l 3 d. 5. Use Eercise 8 to fid e d Fid the re of the regio bouded b the give curves. 53. e.,, 5. l d l l d e d e e d t d t t d sec d t sec 5 l, ; Use grph to fid pproimte -coordites of the poits of itersectio of the give curves. fid (pproimtel) the re of the regio bouded b the curves. 55. si, 56. rct 3, l 5 sec d 57 6 Use the method of clidricl shells to fid the volume geerted b rottig the regio bouded b the give curves bout the specified is. 57. cos,, ; bout the -is 58. e, e, ; bout the -is 59. e,,, ; bout 6. e,, ; bout the -is

7 SECTION 7. INTEGRATION BY PARTS Fid the verge vlue of f l o the itervl, A rocket ccelertes b burig its obord fuel, so its mss decreses with time. Suppose the iitil mss of the rocket t liftoff (icludig its fuel) is m, the fuel is cosumed t rte r, d the ehust gses re ejected with costt velocit ve (reltive to the rocket). A model for the velocit of the rocket t time t is give b the equtio where t is the ccelertio due to grvit d t is ot too lrge. If t 9.8 ms, m 3, kg, r 6 kgs, d ve 3 ms,fid the height of the rocket oe miute fter liftoff. 63. A prticle tht moves log stright lie hs velocit vt t e t meters per secod fter t secods. How fr will it trvel durig the first t secods? 6. If f t d f d t re cotiuous, show tht 65. Suppose tht f, f 7, f 5, f 3, d f is cotiuous. Fid the vlue of f d. 66. () Use itegrtio b prts to show tht (b) If f d t re iverse fuctios d f is cotiuous, prove tht b vt tt ve l f d bfb f f b t d [Hit: Use prt () d mke the substitutio f.] (c) I the cse where f d t re positive fuctios d b,drw digrm to give geometric iterprettio of prt (b). (d) Use prt (b) to evlute l d. e m rt m f t d f t f t f t d f d f f d f prts o the resultig itegrl to prove tht d c b 68. Let I si d. () Show tht I I I. (b) Use Eercise 6 to show tht (c) Use prts () d (b) to show tht d deduce tht lim l I I. (d) Use prt (c) d Eercises 5 d 6 to show tht lim l This formul is usull writte s ifiite product: V b f d =g() = =ƒ I I I I =b d is clled the Wllis product. (e) We costruct rectgles s follows. Strt with squre of re d ttch rectgles of re ltertel beside or o top of the previous rectgle (see the figure). Fid the limit of the rtios of width to height of these rectgles. 67. We rrived t Formul 6.3., V b f d,b usig clidricl shells, but ow we c use itegrtio b prts to prove it usig the slicig method of Sectio 6., t lest for the cse where f is oe-to-oe d therefore hs iverse fuctio t. Use the figure to show tht V b d c d ke the substitutio c t d f d the use itegrtio b