u x v x dx u x v x v x u x dx d u x v x u x v x dx u x v x dx Integration by Parts Formula

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1 7. Intgration by Parts Each drivativ formula givs ris to a corrsponding intgral formula, as w v sn many tims. Th drivativ product rul yilds a vry usful intgration tchniqu calld intgration by parts. Starting with th following form of th product rul, w intgrat both sids of th quation, and apply th fundamntal thorm of calculus: d u v u v u v d = ' ' d u ( ) v ( ) d = u '( ) v ( ) d u ( ) v '( ) d d = ' ' u v u v d u v d Subtracting u '( ) v ( ) d from both sids of th quation, w gt or in othr words, ' = ', u v u v d u v d ' = ' u v d u v u v d. This intgral formula is mor mmorabl in th abbrviatd form: Intgration by Parts Formula udv= uv vdu Mmorizing that formula is fin, but th procss of applying it (and whn to us it) is what you rally nd to mastr. It rally taks working through a fw ampls to gt th hang of this. Hr s th ovrall procss in a nutshll. You start with an intgral udv u v' ' ' = d, and thn st u = u dv= v d up th tmplat. Th product of th top row quals our du = u d v = v intgrand. Thn our original intgral quals th product of th diagonal lmnts u v, minus th intgral of th bottom row: ' = ' u v d u v v u d This tchniqu is usful whnvr you can t intgrat th intgral on th lft but can intgrat th intgral on th right.

2 Eampl. sin d=? ( This is a classic intgration by parts problm! ) u = dv = sin d To solv this, first st up th tmplat: du = d v = cos Thn, u dv = u v v du sin d = d ( cos ) ( cos ) Notic that w can intgrat th intgral on th right sid of th abov quation: = cos cos d= cos sin sin d= cos sin Notic that whn you st up th tmplat, you diffrntiat down th lft column, and you intgrat down th right column. Eampl. d=? u = dv= d To solv this, first st up th tmplat as follows: du = d v = Thn, d= d Notic that w can intgrat th intgral on th right sid of th abov quation: = d = d = Nt w ll look at a coupl of ampls involving dfinit intgrals.

3 Eampl 3. ln cosh d=? u = dv = cosh d To solv this, first st up th tmplat as follows: du = d v = sinh Thn, ln ln ln coshd= sinh sinh d ln ln ln ln = sinh sinh d= sinh cosh ( ln ) sinh ( ln ) sinh ( cosh ( ln ) cosh ) ( ln ) sinh ( ln ) ( cosh ( ln ) ) = = ( ln ) sinh ( ln ) cosh ( ln ) = But w can simplify this furthr, using th dfinitions of th hyprbolic sin and cosin: sinh ln cosh ln And so our intgral bcoms: ln ln ln ln ln( ) 3 = = = = = ln ln ln ln( ) 5 = = = = = 3 5 cosh d= ( ln ) sinh ( ln) cosh ( ln ) = ( ln) ln 3ln cosh d= Eampl. Thn, u = ln dv= d ln d=? To solv this, first st up th tmplat: du = d v = ln d = ln d ( ) = ln d= ln ln = ln d=

4 If you rpat th abov stps with indfinit intgrals, you find th intgral formula for th natural log: ln d = ln Tabular Intgration For any problm involving intgration by parts, thr s a short cut tchniqu calld tabular intgration. This tchniqu is spcially powrful in situations that would othrwis rquir two or mor applications of th intgration by parts tchniqu. So that you ll proprly apprciat this short-cut, w ll first do this problm th old-fashiond way, by using intgration by parts twic. Eampl 5. sinh d=? First st up th tmplat: u = dv = sinh d du = d v= cosh Our intgral bcoms: d cosh c osh = sinh = d cosh coh s d Th intgral on th far right rquirs anothr intgration by parts to compltly diffrntiat away that factor. So to intgrat cosh d, w st up anothr tmplat (onc w find this intgral, w hav to substitut it back into quation abov to solv our original intgral problm): u = dv = cosh d cosh d= sinh sinh d = sinh cosh du = d v = sin h ( W ll wait to th nd to put in th constant of intgration.) Now substitut this rsult into our quation : ( sh ) sinh d= cosh cosh d= cosh sinh co sinh d= cosh sinh cosh Som intgrals rquir svral applications of intgration by parts, which rally complicats things. It s vry asy to mak a mistak in solving such an intgral. But no far tabular intgration to th rscu!! Tabular intgration is actually asir than intgration by parts. So now w ll rdo th prvious ampl, sinh d, but this tim using tabular intgration. For this problm w want to diffrntiat away th. Anytim thr s a factor lik this in your intgrand that is making solving th intgral difficult, and that factor can b diffrntiatd to zro (with nough

5 drivativs), us tabular intgration! As with intgration by parts, w st up a tablau with two columns. And as bfor w diffrntiat down th lft column as w intgrat down th right column. This tim howvr, w kp going until th factor has bn diffrntiatd down to. Eampl 5. sinh d=? ( using tabular intgration ) sinh cosh sinh cosh sinh cosh sinh cosh d= S how much asir that was?! Not th altrnating,,, to th lft of th main tablau. Th nt ampl is asy using tabular intgration, but would b a nightmar using intgration by parts svral tims. Eampl 6. 5 cos d=? d d 5 cos 5 sin 3 cos 6 sin cos sin cos cos = sin 5 cos sin 6 cos sin cos d

6 Th tabular intgration procss dosn t hav to kp going until on of th factors has bn diffrntiatd to dath (all th way to ). On can stop at any row. On writs out th diagonal products of th ntris in th tablau, as bfor. But whn thr s no in th last row, that last row contains th intgrand of a rmaining intgral. Our nt ampl shows how tabular intgration can b usd to do a normal singl intgration by parts problm. Hncforth, w will clusivly b using tabular intgration! Eampl 7. d=? If w stop thr, w v don intgration by parts, to rwrit our original intgral in trms of a nw, asy to solv intgral: d= d W could finish this on off, as th intgral that rmains can asily b solvd. But it s lss hassl had w simply kpt going with our tabular intgration. So lts tak it again, from th top : = d With dfinit intgrals, on valuats th antidrivativ at th uppr and lowr intgral bounds, and subtracts thos valus, as usual. Somtims a mystrious circumstanc unfolds whn applying tabular intgration at som point, th intgral of a row rsults in th sam intgral w startd with. And somtims this actually mans you v solvd th intgral (othr tims it simply mans you v gon in a circl). Whn this occurs, it fls a bit lik lifting onslf up by on s own bootstraps. Whn this occurs, it hlps to call th original intgral I; on nds up with a simpl quation on can solv for I. Eampl 8. = = I sin d? Notic that in row 3 of our tabular intgration tablau, th product of th ntris quals our original intgrand. W thn stop and rwrit our intgral:

7 sin cos sin sin I = cos sin I I = cos sin I = ( sin cos ) I = sin d= cos Always b sur to add in that constant of intgration; it ariss with vry indfinit intgral. Not: If th abov intgral involvd th sinh instad of th sin, th I can t b solvd for instad it cancls out (Try this out to s for yourslf!). In that cas this mthod dosn t work, and w would hav just gon in a circl, to no avail. Similarly, th intgral cosh d can t b solvd using this tchniqu (you only find that out by trying!). Strangly, if thr ar numrical cofficints in thr, as in cosh 3d, which mak th intgral look mor difficult, it thn can b solvd using tabular intgration: Eampl 9. I = cosh 3 d=? cosh 3 sinh 3 3 cosh 3 9 sinh 3 cosh sinh 3 cosh 3 I = I I = I = 3 I = 5 3 sinh3 cosh 5 3 sinh3 cosh3 ( 3sinh3 cosh3 ) I = d= cosh 3 5 Oftn it s tricky, knowing whn to apply intgration by parts (or it s dscndnt, tabular intgration). Oftn, it s a mattr of trial and rror. Succss is nvr guarantd, nor vn possibl.

8 In th following problm th rasoning is simply that w can tak th drivativ of th natural log trm. And w can intgrat th factor. In this nt ampl you ll s a tabular intgration nhancmnt discovrd by th author (as far as h knows; but it s vry likly that many othrs hav also indpndntly discovrd this littl tchniqu). Eampl. 3 I = ln d=? ( ln ) 3 3ln At this point, notic somthing intrsting w can do, that hasn t com up bfor. In this tabular intgration procss, you can plac undr a tablau row, a nw row with any ntris that rsult in th sam product, and thn continu as usual. This is bst sn by ampl, and in th following tablau w ll do this thr tims. It s just too tmpting to cancl thos s! W ll start again from th bginning: ( ln ) 3 3ln 3ln 6ln 6ln I = ln d= ln 3 ln 6ln 6 This nw procss is rally nothing othr than svral intgration by parts. But th bookkping is gratly simplifid by using this modification of th tabular intgration procss. Th nt problm could b don with but on intgration by parts. Nonthlss, w ll continu to us tabular intgration.

9 Eampl. ln? 5 d = ( Stop whn th intgral of th bottom row can b calculatd. ) 5 ln 5 5 ln ln ln ln ln d = d = = 6 ln ln ln 5 ln 5 = = = = I ln 5 ln = d= 5 56 Eampl. I = sin cos3 d=? ( Stop whn th bottom row contains intgral I ) sin cos3 sin 3 cos 3 cos3 sin 9 sin sin 3 cos cos3 I = sin cos3 d= I sin sin cos cos sin sin cos cos I = = I = = = I = I = sin cos3d=

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