INTEGRATION BY PARTS (TABLE METHOD) Suppose you wat to evaluate cos d usig itegratio by parts. Usig the u dv otatio, we get So, u dv d cos du d v si cos d si si d or si si d We see that it is ecessary to perform itegratio by parts a d time. If we do that without multiplyig together ad simplifyig or factorig out ay costats, we get u dv si d du d v cos So, cos d si cos cos d At this poit, the remaiig itegral is simple eough to fiish without itegratio by parts ( ). But suppose we perform itegratio by parts a rd time, agai without chagig the costats, the we get u dv cos d du 0 d v si 7 So, cos d si cos si 0 si d 7 7 Sice the itegrad i the fial itegral is ow 0, its ati-derivative is simply a costat ( C ). We ca let C 0 for ow, ad add a costat of itegratio later. So, cos d si cos si 7 d cos si cos + si 7 cos d si + cos si 7 cos d (( )si + 6 cos) 7 ( ) If we had skipped the fial uecessary itegratio by parts, we would have doe these last lies ayway. Now, i the traditioal method, you would probably wat to simplify ad factor all the costats before each ew itegratio by parts, ad skip the last itegratio by parts. However, by ot doig so, we ca actually perform the etire multi-step itegratio by parts iside a sigle table.
To uderstad how the table method of itegratio by parts works, first otice that we did t really chage from the result of the st itegratio by parts, du d ad v si ad just reused them as u ad dv si d for the d itegratio by parts. That meas that we could elimiate writig that repeated lie. This also applies to from the d itegratio by parts du d ad v cos beig reused as u ad dv cos d for the rd itegratio by parts. Ad so o. So the etire itegratio by parts could have bee compressed ito the followig table. The dowward diagoal products (oe row i the u colum the et row i the dv colum) are the results of the uv portio of the itegratio by parts formula. However, because of the egative i the v du part of the formula, which distributes through each successive itegratio by parts, the products must be alterately added ad subtracted. I other words, to read the atiderivative from the table above: cos d si cos + (( si )si + 6 cos ) 7 7 The same result with much less writig. Notice that the fourth row of the u colum ( 0 ) is igored. We could have viewed it as a factor of the itegrad i the th itegratio by parts, which meas its atiderivative would have bee 0 (or C ). Or we could have viewed it as a factor of the et dowward diagoal product, which also meas its correspodig atiderivative would have bee 0. I either case, it cotributes othig to the atiderivative. However, because of that 0 i the u colum, all followig derivatives i the u colum would also have bee 0. So we ca stop takig ay more derivatives or atiderivatives. I fact, the 0 i the u colum tells us to stop fillig out the table, ad start collectig the aswer.
The orgaizatioal structure makes it a much faster ad less error-proe method which requires a lot less writig. It also makes it easier to fid mistakes later o. This method ca be used for atiderivatives of the types k ( a + a +... + a ) e d where is a positive iteger, ad k is a costat ( a + a +... + a ) si k d where is a positive iteger, ad k is a costat ( a + a +... + a ) cos k d where is a positive iteger, ad k is a costat ( a + a +... + a ) sih k d where is a positive iteger, ad k is a costat ( a + a +... + a ) k d where is a positive iteger, ad k is a costat k ( a + a +... + a )( m + b) d where is a positive iteger, ad m ad b are costats, ad k is a costat such that k (ad ever becomes durig the repeated atidifferetiatios) If you put the polyomial factor ito the u colum, the by repeatedly differetiatig it, you evetually get 0 i the u colum. I the meatime, the dv colum ca be atidifferetiated repeatedly usig basic techiques.
The method caot be used directly for atiderivatives of the types e a si k d where a ad k are costats e a cos k d where a ad k are costats but it ca be used for part of the itegratio. Cosider e cos d. After differetiatig/atidifferetiatig twice, it becomes clear we will ever get 0 i the u colum. However, we have arrived at a row that is essetially the same as the origial row (ecept for costat factors). That is eough, because rememberig that the product across the rd row is a itegrad, we get e e cos d (cos) e ( si ) e + ( cos) e d cos d e cos + e si e cos d The origial itegral appears o both sides of the equatio, so we ca isolate it. e cos d + e cos d e cos + e e cos d e cos e si + e cos d e cos + e si e cos d e cos + e si e cos d e (cos + si ) This techique works wheever repeated itegratios by parts results i itegrads which are costat multiples of the origial itegral, which icludes atiderivatives of the types (where k ad m are o-zero costats ad ( ) where k m ) si si k si m d ( ) si k cos m d cos k cos m d si k sih m d si k m d cos k sih m d cos k m d ( )
The techique also works for atiderivatives of the types (where k ad m are o-zero costats ad ( ) where k m ) e a sih k d e a k d sih k sih m d ( ) sih k m d k m d However, these could all be doe much more easily by replacig the hyperbolic fuctios with their epoetial defiitios, simplifyig the itegrad, ad the atidifferetiatig without itegratio by parts. ( ) By the way, the origial itegral could also have bee doe by reversig the iitial choice of u ad dv. e cos d e si e ( cos) + e ( cos) d e cos d e si + e cos e cos d e cos d + e cos d e si + e cos e cos d e si e cos + e cos d e si + e cos e cos d e si + e cos e cos d e (si + cos) The result ad amout of work is the same as the origial choice. However, mistakes are more commo whe atidifferetiatig sies ad cosies tha whe differetiatig them. [Mistakes are much less commo whe atidifferetiatig or differetiatig epoetials.] You could also have differetiated u ad atidifferetiated dv, 6 or ay eve umber of times ad gotte back to a costat multiple of the origial itegrad. If you had doe so, ad completed the algebra i a similar way as show above, you would gotte the same aswer, but with a lot more (uecessary) work. Ad remember that doig uecessary work icreases the chaces of makig mistakes.
The method ca be used for atiderivatives of the type p (l ) d where p is a costat ad is a positive iteger with a ot-so-slight modificatio to the process. Cosider (l ) d First, let s cosider what does t work well. We caot put l ito the dv colum because we do t kow its atiderivative right ow. Eve if you kow the atiderivative of l, it is more complicated tha l itself, which goes agaist the basic rule of itegratio by parts, that each successive itegrad should ot be more complicated tha the previous. So we are forced to put l ito the u colum. This does ot look promisig, because we kow the et row i the u colum will ivolve the quotiet rule, which agai goes agaist the basic rule of itegratio of parts, that each successive itegrad should ot be more complicated tha the previous. So, perhaps like the previous eample, we eed to rewrite the origial itegrad usig what we ve doe so far. (l ) (l ) d (l ) d (l ) d (l ) l d The ew itegral is simpler tha the origial because the power of the l has bee reduced by. I fact, you might guess that if we repeat the itegratio by parts a secod time, we ca reduce the power of the l by more, to 0. We ca do that, but if we re goig to do a d itegratio by parts, should t there be a way to do it without goig ito ad out of the table method repeatedly? There is, ad it amouts to rebalacig the u ad dv colums i the middle of the repeated differetiatios ad atidifferetiatios.
The rebalacig of the table allows us to cotiue differetiatig the u colum without itroducig a complicated epressio ito the table. This last product ca be itegrated without itegratio by parts, but we ca complete the whole process i the table by rebalacig oce more.
Now, we have a 0 i the u colum, which meas we are doe with the table, ad ca collect the dowward diagoal products. But we must be careful which factors we multiply. Remember that the red lies above correspod to rewritig the same itegral, ot to a itegratio by parts. So our dowward diagoal products must ever ivolve a factor from above a red lie with a factor below the red lie. ( ) That meas, d (l ) (l ) (l ) + 7 6 6 7 (l ) d (l ) l + (l ) d ((l ) l + ) ( ) If you multiply a factor from above a red lie with a factor below the red lie, you are ot multiplyig u by v, but rather u by dv, which has othig to do with itegratio by parts. This techique is usually limited to specialized atiderivatives (like the origial problem), ad some comple situatios i which itegratio by parts must be combied with aother itegratio techique such as substitutio, trigoometric/hyperbolic substitutio, polyomial log divisio or partial fractios. A eample of such a situatio is tah : d tah d tah ( + tah ) tah d tah + tah tah d tah + tah d (( ) tah + )
A eample of a much more comple situatio is d : ) ( d + C d + + C d + C d + ) ) ((