INTEGRATION TECHNIQUES (TRIG, LOG, EXP FUNCTIONS)

Transcription

1 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of MK HOME TUITION Mthemtics Revisio Guides Level: AS / A Level AQA : C Edecel: C OCR: C OCR MEI: C INTEGRATION TECHNIQUES (TRIG, LOG, EXP FUNCTIONS) Versio : Dte: 9--

2 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of INTEGRATION TECHNIQUES Polyomil Itegrls recp The list below reclls the geerl results for differetitio of polyomil fuctios Those highlighted re results obtied by the chi rule (We shll iclude frctiol d egtive powers of i this sectio, though they re ot polyomils i the true sese of the word) (Geerl rules) f() g() f ()g() kf() kf () (Polyomil) - ( + b) ( + b) - (f()) f ()(f()) - From this tble, we c produce similr list of rules for itegrtio (Geerl rules) f ) g( ) d ( ( ) d kf ( ) d k f ( ) d y = f() (provided ) f g( ) d f ( ) d + c ( + b) (provided ) b ( ) + c f ()(f()) (provided ) f ( ) + c Note the ptters i the results obtied by the chi rule Whe we differetite power of lier epressio i, we hve to reduce the power by, multiply by the origil power, d the multiply by the coefficiet of Whe itegrtig power of lier epressio, we dd to the power, divide by the updted power, d filly divide by the coefficiet of

3 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of Emples (): Itegrte with respect to : i) ; ii) +; iii) 8 ; iv) + 6 ; v) ; vi) i) d c ; ii) d c ; iii) 8 d c iv) vi) 6 d c ; v) = = -, so d Emple (): Fid the vlue of d Emple (): Fid the vlue of c or d c d, so d c or c d = = (9-8 + ) () = 6

4 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of Reversig the Chi Rule Itegrtio by ispectio The et emples ll mke use of itegrtio by ispectio, by usig the chi rule i reverse This method c be used if we c spot product of fuctio d some multiple of its derivtive Emple (): Differetite (-) d hece fid ( ) d Usig the chi rule, we hve dy du dy d u ( ) dy d ( ) d hece ( dy du du where u = (-), du d d ) d = ) c ( Our required itegrd, ( ) d, is - times smller th the result obtied by differetitig (-), so we divide by - to obti ( ) d = ) c ( d Altertively, we could hve used the result to give ( ) () Emple (): Fid ( ) c ( ) d ( ( ) b) d ( b) + c ( ) c We c use the sme formul s i the fil prt of Emple () to obti: 6 = ( ) d = We could lso mke educted guess d test the result by differetitio usig the chi rule Sice the power of epressio is rised by by itegrtio, we c guess tht the itegrl will be some multiple of 6 Differetitig ( - ) 6 by the chi rule would give ( - ), which is too lrge by fctor of We therefore djust the guess to 6

5 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of Emple (6): Fid ) d Here we c use the formul ( f ( )( f ( )) d ( f ( )) sice we c spot the cube of the fuctio f () = -, multiplied by, which is oe-sith of its derivtive, f ()6 Applyig the formul s it stds gives c, but becuse we hve d ot 6 i this itegrd, we must divide the result by 6 to get the true swer, c Altertively, we could hve mde guess of ( - ) d checked it by differetitio The derivtive works out s ( - ) result times higher th the required itegrd Dividig the guess by will give ) d ( = c Emple (7): Fid d Agi we c use the formul f ( )( f ( )) d ( f ( )) sice we c spot the reciprocl of the squre of the fuctio f () = + +, multiplied by the epressio +, which is oe-qurter of its derivtive, f () + Applyig the formul s it stds gives + i this itegrd, we must divide the result by to get the true swer, =, but becuse we hve + d ot = Altertively, we could hve mde guess of derivtive works out s d checked it by differetitio The, which is - times s lrge s the required itegrd Adjustig the guess d ddig the limits gives the correct itegrl of There is forml method correspodig to the lst four emples, mely itegrtio by substitutio This will be illustrted i the relevt documet

6 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge 6 of Trigoometric Itegrls - itroductio Suppose we were to estimte the re uder the grph of cos i the itervl < < 9, usig the trpezium rule with 9 strips y = cos b- = 9 h = = 9 The width of ech strip is therefore y (first & lst) y (other vlues) Sub-totls (): ) Sub-totls (): TOTAL Multiply by ½h (here ½ ) 7 The vlue of 9 cos d is therefore estimted t 7 to three sigifict figures However, we lered erlier tht the derivtive of f () = si is f () = cos The vlue of bout 7 seems t odds with the result si 9 or -, or The reso for the discrepcy is gi due to the use of degrees isted of rdis Usig rdis, h = d ot, d ½h =, or pproimtely The true itegrd is therefore cos d Its estimted vlue from the previous result is 87, or pproimtely 998, which is close to the true vlue of IMPORTANT Rdis re the defult uits of gle mesuremet i trig clculus

7 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge 7 of Trigoometric Itegrls Stdrd Trig Derivtives (plus some chi rule emples) (Those for si, cos d t re the most importt) The gle must lso be mesured i rdis, ot degrees y = f () dy y d cos f () si si ( + b) cos ( + b) si si - cos cos -si cos ( + b) - si ( + b) cos - cos - si t sec t ( + b) sec ( + b) t t - sec cosec -cosec cot cosec ( + b) - cosec( + b) cot ( + b) cosec - cosec cot sec sec t sec ( + b) sec( + b) t ( + b) sec sec t cot -cosec cot ( + b) - cosec ( + b) cot - cot - cosec Note how the highlighted fuctios behve o differetitio by the chi rule The derivtive of si, for istce, is cos ; costt multiplier of hs ppered Similrly, the derivtive of cos is si, d tht of t ½ is ½ sec (½) Aother emple is tht of cos ; its derivtive is - cos si Here the power of cos hs bee reduced by, d multiplier of the origil power () d the derivtive of cos, ie si, hve lso ppered Similrly, differetitig si gives si cos The power of si hs bee reduced by, d multiplier of the origil power () d the derivtive of si, ie cos, hve lso ppered

8 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge 8 of From this tble, we c produce similr list of stdrd trig itegrls Agi, gles must be mesured i rdis! y = f() si si ( + b) f ( ) d -cos + c - cos( + b) + c si cos si + + c cos si + c cos ( + b) si ( + b) + c cos si - cos + + c sec t + c sec ( + b) t ( + b) + c t sec t + + c cosec cot -cosec + c cosec( + b) cot ( + b) - cosec ( + b) + c cosec cot - cosec + c sec t sec( + b) t ( + b) sec t cosec cosec ( + b) sec + c sec ( + b) + c sec + c -cot + c - cot ( + b) + c cot cosec - cot + + c Note how the highlighted fuctios behve o itegrtio, by reversig the process of differetitio by the chi rule The derivtive of si, for istce, ws cos ; multiplier of hd ppered Coversely, itegrtig cos would produce si ; this time we hd to divide by isted Aother emple is tht of cos si ; its itegrl is cos Here the power of cos hs bee icresed by to, d multiple of the reciprocl of tht ew power () hs ppered I dditio, we hve divided the result by the derivtive of cos, ie si To solve itegrls of these types, we c use the geerl formule bove, or use the reverse chi rule to mke educted guess, differetite the guess, d djust it if ecessry Emple (8): Fid si d We c either use the tbled result si( b) d cos( b) c to obti the swer cos + c - cos + c, or we c mke guess We kow tht the itegrl of si is cos + c, so we mke d iitil guess of -cos Differetitio of tht guess gives result of si, which is times too smll We must therefore djust the guess of si by multiplyig it by, givig cos + c Emple (9): Fid 6sec d

9 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge 9 of We c either use the tbled result sec ( b) d t( b) c to obti the swer 6 t + c t + c, or we c mke educted guess We kow tht the itegrl of sec is t + c, so we c guess t + c Differetitig t by the chi rule gives us result of sec which is of the right type, but too smll by fctor of We must therefore djust the guess of t by multiplyig it by Agi, this gives us t + c / Emple (): Fid si cos d / 6 (Remember rdi mesure must be used!) We c use the tbled result si cos d obti the swer si + c si + c si c Altertively, we c look t the itegrl d otice tht it icludes power of si (the fourth power) multiplied by its derivtive reverse chi rule result We c thus guess tht the itegrl will be somethig like si (compre itegrtig to get ) Differetitig si gives si cos, but our origil itegrl ws si cos The guess is too lrge by fctor of, so we eed to multiply it by to brig it to scle, gi givig si + c / This is defiite itegrl, so its vlue is (Remember: si (/) = ; si (/6) = ½) si ) / 6 ( 8 Emple (): Fid We c use the tbled result sec d obti the swer sec + c sec t d t sec c Altertively, we c rewrite the itegrd s sec sec t, thus showig the product of the cube of sec d its derivtive, sec t, more clerly This suggests swer of the form sec Differetitig sec gives sec sec t sec t This result is too lrge by fctor of, therefore the true itegrl is sec + c s bove Emple (): Fid cos 6 si d This type of itegrl is ot show i the tble, but lookig t it revels product of power of cos (cos 6 ) d its derivtive, - si The itegrl therefore looks s it if ws obtied by differetitig some multiple of cos 7 We will therefore mke first guess of cos 7 d differetite it Applyig the chi rule twice gives derivtive of - cos 6 si This guess is too smll by fctor of, d therefore the true itegrl is cos 7 + c Emples 8- bove c lso be evluted by substitutio (see Itegrtio by Substitutio)

10 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of Other trigoometric itegrls c be evluted by usig idetities d compoud gle formule to simplify complicted itegrls ito forms which re esier to itegrte These will be discussed i seprte documet Epoetil d Logrithmic Itegrls Stdrd Epoetil d Logrithmic Derivtives (plus some chi rule emples) y = f() y dy d e e e + b e +b e f() f ( ) l l ( ) l (f()) f ( ) e l f ( ) f ( ) f () From this tble, we c produce similr list of stdrd epoetil d logrithmic itegrls y = f() f ( ) d e e +b f ( ) f ( ) e e f() + c e + c e + b + c + c l l + c Altertive form: l A l ( ) + c = l + c Altertive form: l (A ) f ( ) l (f()) + c f ( ) Altertive form: l A(f()) Note the ptters i the results obtied here Whe differetitig e rised to fuctio, the the result is the origil fuctio multiplied by its derivtive For emple, if e is rised to epressio ( + b), the the derivtive is the origil fuctio multiplied by Coversely, whe itegrtig e rised to epressio ( + b), the the itegrl is the origil fuctio divided by Whe differetitig we multiply by l : whe itegrtig we divide by l

11 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of The list lso shows how to itegrte, rememberig tht the stdrd rule d + c cot be used for = - The logrithm lws lso give rise to ltertive wys of deotig the costt of itegrtio, ie l + c = l (A) where A = e c Also, sice there c be o logrithm of egtive umber, the modulus fuctio f () must strictly be icluded i itegrds ledig to logrithmic fuctio, uless we re sure tht the fuctio whose logrithm is beig tke cot itself tke egtive or zero vlue Filly, ote the et result of the chi rule whe pplied to differetitig logrithmic epressio: the result is frctio whose top lie is the derivtive of the bottom lie Emples (): Fid i) 6 e d ii) e d I i) we tke the fctor of 6 outside the itegrl d use the result 6 e or e c e d b e b c to obti Altertively, we could hve mde first guess of e d differetited tht to give e This result is too smll by fctor of, so therefore the correct itegrl is e + c s before I ii) we recogise tht is oe-hlf the derivtive of, so we c tke ½ out s fctor d use the result f ( ) f ( ) f ( ) e e c to obti e ( ) e e ( e e ) Altertively, we could hve spotted reversed chi rule result d guessed t would give e e, which is too lrge by fctor of, so we would djust the itegrl to Differetitio (Both emples could lso hve bee worked out by substitutio (see documet Itegrtio by Substitutio ) e Emples (): Fid i) d ii) d iii) d iv) d For i) we use d c l, to give required itegrl of l = l l I ii), we use the stdrd result to obti l l l =l Becuse the bottom lie of the itegrd is positive withi the rge beig itegrted, there is o eed to iclude the modulus sig Prt iii) mkes use of the stdrd result gi to give l( ) ) l( l = (l l ) This time, we hd to use the modulus fuctio, sice the logrithmic fuctio is ot defied for egtive umbers The itegrl i prt iv) cot be evluted, becuse the fuctio is ot defied for ll vlues of withi the rge of the itegrl (- to ) The problem vlue here is = ; is udefied This is geerl coditio for ll defiite itegrls: if the fuctio is udefied for y vlue of betwee the limits, the itegrl cot be evluted (t lest usig techiques lered t A-level)

12 Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of Emples (): Fid i) d ii) 8 d I i) the top lie is the ect derivtive of the bottom lie, therefore the itegrl is l c or l A We do eed the modulus sig here, becuse the qudrtic withi the logrithmic epressio c tke egtive vlues, eg - whe = I ii) the top lie is the derivtive of the bottom lie multiplied by The itegrl is therefore l - (l 7 l ) There is o eed to iclude the modulus sig roud the logrithm, sice the qudrtic hs o rel roots (b c < ) d is thus > for ll This c be rewritte s sigle logrithm usig log lws: l 7 l l 7 ; 7 7 (l ) l l 9