Integration by Parts 6-3-207 If u an v are functions of, the Prouct Rule says that (uv) = uv +vu Integrate both sies: (uv) = uv = uv + u v + uv = uv vu, vu v u, I ve written u an v as shorthan for u an v This is the integration by parts formula The integral on the left correspons to the integral you re trying to o Parts replaces it with a term that oesn t nee integration (uv) an another integral ( vu) You hope that the new integral is easier to o than the ol one I m going to set up parts computations using tables; it is much easier to o repeate parts computations this way than to use the stanar u-v-v-u approach To see where the table comes from, start with the parts equation: uv = uv vu Apply parts to the integral on the right, ifferentiating u an integrating v This gives [( )( ) ( )( u 2 ) ] ( )( ) ( )( u u 2 ) u uv = uv v v 2 = uv v + v 2 If I apply parts yet again to the new integral on the right, I woul get uv = uv ( )( u ) v + ( 2 )( ( u 2 ) ) ( ( v ) )( 3 u v 3 ) There s a pattern here, an it s capture by the following table: + u v u v + 2 u 2 3 u 3 ( v ) v
To make the table, put alternating + s an s in the left-han column Take the original integral an break it into a u (secon column) an a v (thir column) (I ll iscuss how you choose u an v later) Differentiate repeately own the u-column, an integrate repeately own the v-column (You on t write own the ; it s kin of implicitly there in the thir column, since you re integrating) How o you get from the table to the messy equation above? Consier the first term on the right: uv You get that from the table by taking the + sign, taking the u net to it, an then moving southeast to grab the v If you compare the table with the equation, you ll see that you get the rest of the terms on the right sie by multiplying terms in the table accoring to the same pattern: (+ or ) (junk) (stuff) The table continues ownwar inefinitely, so how o you stop? If you look at the last messy equation above an compare it to the table, you can see how to stop: Just integrate all the terms the last row of the table A formal proof that the table represents the algebra can be given using mathematical inuction You ll see that in many eamples, the process will stop naturally when the erivative column entries become 0 Eample Compute 3 e 2 Integration by parts is often useful when you have a prouct of ifferent kins of functions in the same integral Here I have a power ( 3 ) an an eponential (e 2 ), an this suggests using parts I have to allocate 3 e 2 between u an v remember that implicitly goes into v I will use u = 3 an v = e 2 Here s the parts table: + 3 e 2 3 2 2 e2 + 6 4 e2 6 8 e2 + 0 6 e2 You can see the erivatives of 3 in one column an the integrals of e 2 in another Notice that when I get a 0, I cut off the computation Therefore, 3 e 2 = 2 3 e 2 3 4 2 e 2 + 6 8 e2 6 6 e2 + 0 But 0 is just 0 (up to an arbitrary constant), so I can write 3 e 2 = 2 3 e 2 3 4 2 e 2 + 6 8 e2 6 6 e2 +C 2
Before leaving this problem, it s worth thinking about why the 3 went into the erivative column an the e 2 went into the integral column Here s what woul happen if the two were reverse: + e 2 3 2e 2 4 4 + 4e 2 20 5 This is ba for two reasons First, I m not getting that nice 0 I got by repeately ifferentiating 3 Worse, the powers in the last column are getting bigger! This means that the problem is getting more complicate, rather than less Here s another attempt which oesn t work: + 3 e 2 0 3 e 2 I got a 0 this time, but how can I fin the integral in the secon row? it s the same as the original integral! Putting the entire integran into the integration column never works On the other han, you ll see in eamples to follow that sometimes putting the entire integran into the ifferentiation column oes work Here s a rule of thumb which reflects the preceing iscussion When you re trying to ecie which part of an integral to put into the ifferentiation column, the orer of preference is roughly Logs Inverse trigs Powers Trig Eponentials L-I-P-T-E For eample, suppose this rule is applie to (ln) 2 You try the Log (ln) 2 in the ifferentiation column ahea of the Power Or consier e 2 sin5 Here you try the Trig function sin 5 in the ifferentiation column, because it has preceence over the Eponential e 2 Eample (a) Compute ln 3
(b) Compute (ln) 2 (a) + ln ln = ln = ln +C (b) (I compute + (ln) 2 2ln (ln) 2 = (ln) 2 2 ln = (ln) 2 2(ln )+C ln in part (a)) Eample Compute (+4) 50 First, (+4) 50 = u 50 u = 5 u5 +C = 5 (+4)5 +C [u = +4, u = ] The same substitution shows that (+4) 5 = 52 (+4)52 +C Now o the original integral by parts: + (+4) 50 5 (+4)5 + 0 2652 (+4)52 (+4) 50 = 5 (+4)5 2652 (+4)52 +C You can also o this integral using the substitution u = +4 4
Eample Compute sin Parts is also useful when the integran is a single, unsimplifiable chunk In this case, there isn t an obvious way to change or simplify sin Integration by parts replaces one integral with another, so I try it here an hope I get something that is easier to integrate: + sin 2 Therefore, sin = sin 2 I can o the new integral by substitution: Let u = 2, so u = 2, an = u 2 : sin = 2 sin u u 2 = sin + u = sin + 2 u 2 2 u+c = sin + 2 +C Eample Compute π/2 0 sin If you o a efinite integral using parts, compute the antierivative using parts as usual, then apply the limits of integration at the en + sin cos + 0 sin Thus, π/2 0 sin = [ cos+sin] π/2 0 = In some cases, integration by parts prouces a copy of the original integral Depening on how it arises, this may not be a ba thing Eample Compute e sin2 5
+ e sin2 e 2 cos2 + e 4 sin2 e sin2 = 2 e cos2+ 4 e sin2 4 e sin2 What s this? All that work an you get the original integral again! Look at the equation as an equation to be solve for the original integral It looks like this: (original integral) = (some junk) (original integral) Move the copy of the original integral on the right back to the left an solve for it: e sin2 = 2 e cos2+ 4 e sin2 e sin2 4 5 e sin2 = 4 2 e cos2+ 4 e sin2 e sin2 = 2 5 e cos2+ 5 e sin2+c Eample Compute (csc) 3 (csc) 3 = (csc) 2 (csc) = [+(cot) 2 ](csc) = csc+ (cot) 2 csc = + cot csccot (csc) 2 csc csc csccot (csc) 3 = ln csc+cot csccot (csc) 3 There is a copy of the original integral on the right sie Thus, (csc) 3 = csccot ln csc+cot 2 (csc) 3 = csccot ln csc+cot (csc) 3 = 2 csccot 2 ln csc+cot +c (csc) 3 Eample Compute 2 ( ) 3 6
2 + 2 ( ) 3 2 2( ) 2 + 2 2( ) + 0 2 ln 2 ( ) 3 = 2( ) 2 ln +C You coul also o this integral using the substitution u = c 207 by Bruce Ikenaga 7