How would you do th following intgral? d 7 Wll of cours today you would us on of th modrn computr packags, such as Mathmatica or Mapl or Wolfram, that do calculus and algbraic manipulations for you Som of ths packags ar quit astonishing in what thy can do, and thy usually do giv a corrct answr (I hav nvr known thm not to do so!) I say "a" corrct answr, bcaus somtims th prssions thy giv, whil formally corrct, ar rathr complicatd and unrducd, or not always in th most convnint form For ampl I onc saw on of ths mathmatics packags giv, as th answr to a problm, th prssion cosh( ) This happnd to b formally corrct, but if you hav not bn handling hyprbolic functions rcntly (and most of us don't s thm all that oftn), it may b incomprhnsibl Quick calculat cosh( ) B honst now did you find th answr (559) in lss than fiv minuts? Dos your tabl of hyprbolic cosins (you havn t got on) hav an ntry for th squar root of minus two? (No it dosn t) Actually cosh( ) is th sam thing as cos, so that cosh( ) cos( ) 559 Ths on-lin intgration programs hav, in rcnt yars, improvd normously, and th problm of unduly complicatd or unrducd or incomprhnsibl solutions appars only occasionally - though you do somtims gt on On could argu that thr is no nd, ths days, to larn th art of intgration, or to wast on s tim on it W ar in th st cntury and w must us th tools availabl To insist on working out th intgrals onslf rathr than by using a computr is rathr lik insisting on doing numrical calculations by long multiplication and division with pncil and papr rathr than by just prssing th or sign on on s calculator What, thn, is th us of prparing a fil lik this on that shows us how to carry out intgrals ourslvs? I shan t argu with that If w com across an intgral that w nd in th cours of our scintific work, by all mans look it up in Wolfram and gt on with it Howvr, many of us njoy, for rlaation, doing th littl puzzls that appar in th daily nwspaprs, such as sudokus or crossword puzzls An intgral such as th on abov is probably slightly mor difficult than a nwspapr sudoku (although som sudoku dvots hav concoctd som vry difficult ons) Its difficulty lvl may b comparabl to a killr sudoku (for thos who havn t hard of a killr sudoku, it s a spcial varity with slightly diffrnt ruls), and not narly as difficult as a cryptic crossword in a British nwspapr I somtims spculat, wouldn t it b nic if our daily nwspaprs wr to giv, in addition to a sudoku and a crossword puzzl, an intgral for us to do? I don t know if that is vr likly to happn, but I suspct that many of us njoy or gt a bit of prsonal satisfactions from doing littl puzzls, and this fil may hlp with doing intgrations
d By th way, I hav just workd out myslf th old way, and I gt 7 ln [ k ( 8( 7) ) ] Now lt m try Wolfram and s what it givs m Hr it is, Wolfram s solution: sinh 7 Thy don t look at all alik, do thy? I ll lt you dcid if thy ar th sam And, if thy ar, which do you prfr? Wolfram is crtainly shortr and mor compact - but do you hav a sinh button on your calculator, or do you know how to program it on a computr? And by th way, if you hav concludd that thy ar not th sam and on of thm must b wrong, think again On way of tsting would b to choos som numbr for and s if both prssions giv th sam numrical answr This won t work - bcaus both prssions includ an arbitrary const of intgration I now giv a short list, for rfrnc, of th intgrals of th commonst simpl mathmatical functions, but, byond that I am not giving a long tabl of intgrals Rathr, I am giving a fw hints as to how to start Indd it is usually starting that is th most difficult part On oftn has to sk a Brilliant Substitution, and, onc on hits upon a suitabl substitution, th rst is straightforward Of cours, finding th bst Brilliant Substitution is somthing that coms partly with princ But is also coms in part from ralizing that not all succssful substitutions ar ncssarily "brilliant" thr ar som that should b routin Aftr th tabl of common intgrals, I ll giv a numbr of ampls, with hints on how to start I am assuming that th viwr dos know th basics of intgration, such as making Brilliant Substitutions, and how to intgrat by parts If not, you ar probably not quit rady for this fil, which is not for absolut bginnrs Do lt m know (jtatum at uvic dot ca) if you find any mistaks anywhr To rr is human, but on of th advantags of Wb publishing is that mistaks can b corrctd I m daling only with analytical intgration in ths nots For numrical intgration, s Sction of http://orcaphysuvicca/~tatum/clmchs/clmpdf
INTEGRALS OF THE OMMON SIMPLE FUNTIONS f ( ) f ( ) d n ( n ) n n / ln ln (ln ) a a lna sin cos cos sin ln ln( ) csc ln(csc cot ) cot ln sin sin sin cos cos ln( ) ln( ) csc csc ln( ) cot cot ln( ) sinh cosh
cosh sinh h ln(cosh ) h csch ln k coth ln( k sinh ) sinh sinh cosh cosh h h ln( ) h h ( ) csch csc h ln coth coth ln( ) I includd th invrs hyprbolic functions, for "compltnss" rathr than for thir imporc Thy won't man vry much unlss you ar awar of th following idntitis sinh cosh ln ln h ln h csch ( ) ( ) ( ) ( ) ln ln coth ln
5 d is of som intrst and nds som car Many of us might, in a momnt of d hast, writ ln( ) What about? W might, with similar hast, writ k ln( ), which can also b writtn ln or ln No problm so far But suppos w writ as and thn intgrat it W would find d d [ ln( )] ln( ) So, what is d? Is it ln( ), or is it ln( )? It sms to dpnd on whthr > or < You can t tak th logarithm of a ngativ numbr (Wll, you can t if all you hav hard of is ral numbrs Thos of you who ar familiar with compl numbrs will know that, for ampl, ln( ) 69 i and you ll know whr to find it on th Argand diagram Thos who ar not familiar with thm should prtnd you v nvr sn this paragraph) d Th truth is, of cours, that ln, and it dosn t mattr what th valu of is It might b intrsting to convinc yourslf that 6 d 5 d 69 and and similarly with othr limits, following 5 thm by looking at th corrsponding aras undr a graph of y
6 I now hav a look at svral sorts of intgrals that you might ncountr, with som suggstions as to how to dal with thm Following that, som intgrals for yourslf to try d (a) (b) b c d b c (c) b cd Thr ar thr cass to considr: i b c ii b > c iii b < c as i b c In this cas, th prssion b c can b writtn as a prfct squar, of th form ( α), aftr which th intgrals ar asy as ii b > c In this cas, th prssion b c can b writtn as th product of two ral linar trms, in th form ( α)( β ) For th intgral (a) split th intgrand into partial fractions:, ( )( ) ( ) ( ) aftr which th intgral is asy α β β α α β For th intgrals (b) and (c), lt y th form α, and w thn hav to dal with intgrals of (b) dy y( y h) or (c) y( y h) dy, whr h β α Thn lt y h Th rst should b straightforward as iii b < c In this cas, add and subtract b ("half th cofficint of squard") to th prssion b c, which bcoms b b c b ( b) h, whr h c b Thn lt b h From this, d h d and ( b) h h
7 Th thr intgrals thn bcom: (a) d h (b) d (c) h d S ampl 5 blow for this on d Try ltting d Try ltting d cos Writ cos cos 5 d n (a) If n is vn d d ( ) d d d 6 d ( ) d d d and so on for highr vn powrs ( ) d
8 (b) If n is odd d ln( ) I d d d d I d ( ) d and so on for highr odd powrs m n n / 6 sin cos d and ( ) d m Th ond intgral is th sam as th first if you lt sin, so w dal only with th first If on or both of m and n ar odd: For ampl: 5 sin cos d sin cos cosd Lt s sin, cos s, ds cos d, and so w hav s ( s ) ds If both m and n ar vn it's not quit so simpl For ampl sin cos d (cos cos 6 ) d Lt's just dal with cos 6 d, bcaus, if you can dal with that, you can probably also dal with cos d Th most straightforward way is to us th idntity cos 6 (cos6 6cos 5cos ) In th vry unlikly vnt that you did not know that cos 6 (cos6 6cos 5cos ), you'd nd to b abl to find it quickly I can think of two quick mthods With practic, it might b possibl to driv th idntity in your had, though I havn't trid it myslf All you nd is d Moivr's Thorm, which is th only thorm you nd in trigonomtry, bcaus all trigonomtric idntitis can b drivd quickly from it D Moivr's Thorm is: i m mi
9 Thus: Lt i z cos i sin, so that / z cos i sin and hnc z z cos Th binomial cofficints for ( a b) (which you can gt from Pascal's pyramid) ar 6 6 5 5 6 6 so that z z 6 5 6 z z z z z z 6 6 That is cos cos6 cos cos, or cos 6 (cos6 6cos 5cos ) Th othr quick way to find this and similar idntitis is to look it up in th tabl that you will find in Sction 8 of http://orcaphysuvicca/~tatum/clmchs/clmpdf π / m n m Th dfinit intgrals sin cos d and ( ) a simpl formula, which is not difficult to driv, namly: n / d can b valuatd from / m n ( m )!!( n )!! sin cos d f ( m, n) ( m n)!! π Hr!! mans, for ampl, 9 7 5, and!!!! Th function f ( m, n) quals π/ if m and n ar both vn, and f ( m, n) othrwis For ampl: / 6 π sin cos d 8 5 7 5 π 8 6 5π 96 895 d 7 B D Th cubic prssion can always b prssd in th form ( α)( b c), and somtims vn in th form ( α)( β)( γ ) It may b asy to do so For ampl: ( ) ( )( ) Or it may b lss asy, for ampl 5 ( α )( b c),
whr α 7 6 8, b 7 6 8, c 89658 658 In any cas, you can split th intgrand into partial fractions: ( α)( b c) P α Q R b c, whr (I think but you'd bttr chck it) P c α( b, α) Q P and R c b α α( b α) Pd Rd and α b d will caus no difficulty, but what about? b c Try somthing lik this: b b b c b c b c b c ± a 8 n If n is odd, thr is no difficulty Lt y a and intgrat by parts 5 Thus d ( ) If n is vn, th intgral cannot b prssd in trms of th simpl lmntary functions If it has to b intgratd btwn finit dfinit limits, it has to b valuatd numrically t Howvr, th function (of t) d is calld th rror function rf t, and it is π supportd by many computr packags For mor on th rror function s Sction of http://orcaphysuvicca/~tatum/thrmodhtml Th dfinit intgral d has th valu π / This can b drivd as follows: ( y ) ( d) dyd ( t ) Lt y t, so that th innr intgral bcoms dt, in which, as far as this innr intgral is concrnd, is const
( t ) Thus ( d) dtd Rvrs th ordr of intgration (w can always do this with a wll-bhavd function whn calculating an ara, it dosn't mattr whthr w tak lmntal strips paralll to th y-ais and intgrat thm with rspct to, or do it th othr way round) ( t ) Thus ( d) ddt Th innr intgral now is ( t ) d, in which t is const It is asily found (g lt s ) that this intgral coms to / ( t ), and thrfor dt ( ) d This intgral is lmntary (g lt t ) and coms to ( t ) π/ Thrfor d π W can go furthr By substitution of a for, w asily s that a d π a Now ach sid of this quation is a function of a, not of If w now diffrntiat both sids again and again with rspct to a, w obtain progrssivly and so on a a d d 8 π a a π 5,, W can also do this with th odd powrs It is asy to obtain a d a, and by rpatd diffrntiation with rspct to a w obtain a 5 a d d, a, a
and so on Of cours, if n is vn, and zro if n is odd ( a b) b /(a) a( b /(a)) 9 d,tc a a b cos b d R d sin b a d Im a b d In cas th nwspaprs don t publish a daily intgral in addition to thir sudoku and crossword puzzl, hr ar a fw for fun, chosn at random As mntiond arlir, I stimat that on avrag ach is about as difficult as a killr sudoku puzzl, but not as difficult as a cryptic crossword from a British nwspapr From tim to tim as th spirit movs m, I may add a fw mor Following th list, I giv a fw hints And aftr th hints, in cas you ar absolutly stuck, I giv som workd solutions d d d d 5 d cos 6 ( ) d 7 ( ) d 8 d 9 d ( ) d 5 d 8 ( ) d ( ) d ( ) d 7
5 d 6 sin d 7 cos d 8 d d 9 ( ) d ( ) d d d d 5 ln d 6 sin d d 7 ( 9)( ) 8 cos sin d 9 d HINTS Hr ar som hints for th intgrals abov I giv workd solutions aftr th hints, but, whn you hav what you bliv to b th answr, you can always diffrntiat your answr to s if you arriv at th original intgrand Don t forgt to add a const to all of thm ( ) Split into partial fractions ( )( ) cos cos 5 csc csc cot cos cos sin
Or you could try somthing that I oftn us as a last rsort whn daling with trigonomtric functions, lt It s oftn usful 6 t Th dnominator is a quadratic prssion You must look to s whthr b > < ac In this cas b < ac You must complt th squar in th dnominator by writing it as ) ( Thn, a substitution u may hlp 7 Th dnominator is ( )( ) Split th intgrand into partial fractions 8 If you had so try you d probably want to try cos But you hav, cosh instad This may rsult in somthing awful such as sinh d sinh cosh cosh ( But thn rmmbr that sinh ( ) and ), and you ll probably gt somthing that you can handl 9 Try You may gt an answr with and in it, and you won t know what to do with Pythagoras might hlp you out ( )( ) Split th intgrand into partial fractions Th dnominator is ( )( ) Split th intgrand into partial fractions ( )( ) Split th intgrand into partial fractions Lt u I had to work a littl with this on Th first thing I did was to try to mak th numrator look a bit lik th drivativ of th dnominator Thus ( ) W now hav two intgrals:
5 ( ) d 7 ( ) d I I 7 7 d I is asy Writ I as In th dnominator, I d 98 7 7 b < ac, so w complt th squar by writing th dnominator as 8 8 ( ), followd by a substitution such as lt I mak th 96 final answr ( ) d 7 ln(7 ) 7 8 8 5 6 Intgrat by parts 7 Writ ithr cos (cos cos ) or cos cos ( sin ) 8 ( ) 9 It s narly always a good ida, whn you s a, to lt a ( ), and do th ond two by intgration by parts Not particularly asy Lt Thn d d d Th intgral of n d is dalt with on pag 6 of this fil Th substitution cosh will also work, though I pct many will find that is asir
6 I mak th answr ln[ ( )] u Lt sin Aftr som manipulation of trigonomtric idntitis, you should arriv at ( sin ) d 5 Intgrat by parts Eithr way will do 6 Thr s probably mor than on way, but you might try Im d I mak it ( sin cos This looks unlikly, but try diffrntiating ) it and s what you gt You nvr know - it might b right 7 This looks lik nothing mor than hard work If my algbra is right, 9 9 ( 9)( ) 5 9 ( i ) Thn a fw dft substitutions, such as and should hlp 8 Try any of y cos, y cos or y cos Surly on of thm will put it in a form that you can cop with 9 ( )( ) Split th intgrand into partial fractions
7 SOLUTIONS d d d ln ln k( ) ln ( )( ) d d ( )( ) d d ln ln ln k ln d d d d d d ln 5 d ( cos ) d ( cos ) d (csc cos cos sin csc cot ) d cot csc ( / ) 6 ( ) d Th dnominator is a quadratic prssion W start by looking at b ac If it is, it will factoriz into two ral linar trms In this cas, it is lss than zro, so it won t factoriz W complt th squar by writing it as ( ) And if w now lt u, th intgral bcoms (or, rathr, th intgrals bcom)
8 u udu ln( 5 u du ) 5 ln( u ) 5 u 7 ( ) d ( ) d d ln k( ) ( )( ) ln ln ln If >, you d b saf in writing this as ln k ( ), but if < I wouldn t risk it 8 You nd to b prtty familiar with hyprbolic functions for this on d I Lt cosh Thn: I sinh d sinh cosh ( ) d ( ) [ cosh () ] 9 d I Try Thn I d ln k( ) ln k ( ) d It is always usful is ths situations to s if th dnominator factorizs; and, if it dos, split th prssions into partial fractions In this cas th dnominator factorizs into ( )( ) and th prssion to b intgratd th splits into
9 Thus: ( ) d ln[ k( ) ( )] d ( ) d ln[ k( ) ( ) This last form is OK whthr is lss than or gratr than [ ln ln( ) ln k] d d d ln 5 8 ( )( ) ( ) ( ) d ( ) ( )( d ) d k( ) ln ln ln( ) ( ) d ( ) d Lt u ( u u / u) du u u u ln u const ln[ k( ) ] ( ) d Th dnominator dos not factoriz into ral factors I m going to 7 manipulat th numrator to try to mak it look lik th drivativ of th dnominator So now w hav ( ) ( ) ( ) ( ) d 7 ( ) d I I 7 7 d I is asy It is just ln(7 ) const
I d 98 98 8 98 7 7 7 96 96 d d ( ) 8 96 du u Lt u Thn I plus 98 8 7 8 7 8 u 8 8 a const Hnc ( ) d 7 5 d d d 96 ln(7 ) 7 8 8 6 (a) sin d sin (b) dcos cos cos d d sin cos sin sin d cos sin d sin d sin d sin d sin d sin d sin cos cos 7 (a) cos d (cos cos ) d sin sin cos (b) d cos d cos sin d sin sin Th two solutions, (a) and (b), ar trigonomtric idntitis 8 d d d ln Wolfram givs ln cos Is that th sam?
d 9 It s narly always a good ida, whn you s ( ) a In this cas, a and w obtain 8 d d d cos (cos ) sin 8 6 6 a,to lt 6 6 (sin cos ) 6 d ( ) d d ln ( ) d d d d ( ) d Lt Thn d d d d I I
const ) ln( const ) ln( ) ( I I d d d d d d d d I const ) ln( I Thrfor ( ) [ ] ) ( ln const ] [ln( d Altrnativly: Lt sinh, sinh, cosh d d Thn [ ] ) ( ( ln cosh cosh sinh sinh ) (cosh ) ( ) ( sinh d d d d d d Lt u u u u u du u d ) 8 ( ) (8 ) ( 5 5 8 /
d Lt sin, thn d sin cos d sin Multiply top and bottom by sin d sin cos d sin sin cos d sin ( sin ) d cos sin Incidntally I trid d and sin cos d on Wolfram rcntly, and in sin both cass it rturnd impossibly complicatd answrs! I think thy wr actually corrct, though it was vry hard work to simplify thm Such cass ar mrcifully rar ths days, and usually th answr is rturnd in a convnint form 5 ln d On way: I ln d d[ (ln )] (ln ) (ln ) d (ln ) I d I (ln ) (ln ) I (ln ) Th othr way: d d d ln ln ( ) ln ln ln d ln 6 sin d i Im d i i ( ) d d i ( i) ( i) ( i) ( i)(cos i sin )
Th imaginary part is (sin cos ) You could also try intgrating by parts (ithr way) - but that s not so intrsting 7 d 9 9 ( 9)( ) 5 9 ( ) 5 9 5 ( ) 8 cos sin d Lt s try th first of my suggstions First, y cos, dy sin d cos cos y cos sin d (y ) dy Thn mayb try ltting y That is, y This looks promising, so lt s go back to th bginning and mak th truly Brilliant Substitution cos, sin d d, cos cos Thn: cos sin d W alrady know, from pags 6 and 7, how to intgrat and hnc writ th intgral as cos n d, so w ll us 5 sin d ( ) d
5 On carrying out th procdurs suggstd on pags 6 and 7, I obtain 5 ln( ) 8 ln( ) 8 ln( ) 5 Thus th intgral bcoms, aftr a littl tidying up, cos 6 sin d [ (5 ) ln( )] W hav to gt back to, rmmbring that cos cos Thus cos, from which cos sin d [ cos )cos (5 cos ) ln( cos cos )] 6 Th ond suggstion was lt y cos y This lads to cos sin d dy 8( y) Th third suggstion was lt y This lads to cos sin d dy ( y ) You could try ithr of ths and s whr thy lad For numrical intgration (s http://orcaphysuvicca/~tatum/clmchs/clmpdf ) ths ar far fastr than th original prssion 9 d d d ( )( )
6 d d ) ( ) ( ) ( 6 I I I ) ln( I ) ln( I ) ( ] ) ( ) [( d d I Hnc d ) ( ) ln( ) ln( 6