Tabular Integration –MATH 2411
Kawai
(#1) You have already seen the basic recipe for Integration by Parts (IBP). From the Product Rule (di¤erentiation), we have: d [uv] = u dv + v du
d [uv] = u dv + v du uv = u dv + v du ) Z Z Z Z Z u dv = uv v du Z Z This assumes that u and v are both functions of x: We use this recipe because:
(a) We do NOT know the antiderivative: u dv; but... (b) We DO know the antiderivative: v du;R and... (c) It is assumed that if we choose theR dv part correctly, then dv = v is easy to evaluate. (#2) Here’sa peculiar IBP: R xex dx (x + 1)2 Z This a contrived exception to L-I-A-T-E: x 2 We should choose u = xe and dv = (x + 1) dx: x x x x WHY? We see that [xe ]0 = xe + e = (x + 1) e and 2 1 (x + 1) dx = (x + 1) + C: 1 R u = xex dv = dx (x + 1)2
1 du = (x + 1) exdx v = x + 1 The undesirable factor CANCELS! xex xex 1 dx = (x + 1) exdx (x + 1)2 x + 1 x + 1 Z Z xex xex = + exdx = + ex + C x + 1 x + 1 Z If we factored this, we would obtain: x 1 ex ex + 1 + C = ex + C = + C: x + 1 x + 1 x + 1 Thus, the IBP was really a result of a messy Quotient Rule problem. In general, Quotient Rule problems often disguise their true nature. We must often rely on trial-and-error before a clear solution arises.
1 (#3) Tabular Integration allows us to evaluate certain families of IBP’s. Anything of this form can be evaluated using Tabular IBP:
p (x) ekxdx OR p (x) sin (kx) dx OR p (x) cos (kx) dx Z Z Z or possibly something like x3 (1 + x)1=2 dx because we can easily write down enough anti- derivatives of (1 + x)1=2 : R Tabular IBP is simply a way to perform SUCCESSIVE IBP steps with very easy bookkeeping. Suppose we have: 2 3x x e dx =??? Z We should choose: 2 3x u = x dv = e dx
& 1 du = 2x dx v = e 3x 3 Clearly, this simpli…ed our lives, somewhat. We can always …nd the uv portion of IBP by looking at the product down the diagonal.
2 3x 2 1 3x 1 3x x e dx = x e e (2x dx) 3 3 Z Z The last integral has one less degree in the polynomial part. 2 + xe 3xdx =??? 3 Z Another round of IBP, and we will have the complete answer. Instead, what if we leave the coe¢ cients where they are, and then use IBP directly on the two parts: 1 u = 2x dv = e 3xdx 3 & 1 du = 2dx v = e 3x 9 Spreading the coe¢ cients around cannot possibly change the …nal outcome, but it will cer- tainly give us a nice pattern to follow when we do Tabular IBP. From above, the new uv product is:
1 1 (2x) e 3x and the new integral is e 3x (2dx) : 9 9 Z Success! We know how to do this last one:
2 3x 2 3x e dx = e + C: 9 27 Z
2 Again, consider what happens if we phrase this last integral as an IBP: 1 u = 2 dv = e 3xdx 9 & 1 du = 0dx v = e 3x 27 2 The uv product is the desired term: e 3x and the v du = 0 dx = C: 27 It’sperfect. R R The diagonal product gives us the result for the …nal antiderivative.
Thus, if start out with u1 as a third-degree polynomial and v1 as the function which can be integrated many times, we have:
u1 v1 & u2 v2 && u3 v3 u1v2 ch && u4 v4 u2v3 && 0 v5 u3v4 ch & u4v5 In the left column, we di¤erentiate as many times as needed so that the result is zero. In the right column, we integrate the same number of times.
If u1 is a third-degree polynomial, then FOUR di¤erentiations are necessary to result in zero.
[u2 = u10 ; u3 = u20 ; etc.] We discard the constants of integration for the right column. In the end, we can tack on the “+C”for the inde…nite integral.
v2 = v1dx; v3 = v2dx; etc. (#4) WHYR do we need toR change the sign of the 2nd, 4th, 6th, etc. diagonal products? Remember, by IBP, we will have:
I = u1v1 u2v2dx; Z but by the next line, we have:
u2v2dx = u2v3 u3v3dx: Z Z By substitution, we obtain
I = u1v1 u2v3 u3v3dx = u1v1 u2v3 + u3v3dx: Z Z We see that the sign of the diagonal products alternate because of the IBP formula!
3 (#5) Ridiculous example: x5 cos (2x) dx:
Rx5 cos (2x)
& 1 5x4 sin (2x) 2 &&1 1 20x3 cos (2x) x5 sin (2x) 4 2 ch &&1 5 60x2 sin (2x) x4 cos (2x) 8 4 &&1 5 120x cos (2x) x3 sin (2x) 16 2 ch &&1 15 120 sin (2x) x2 cos (2x) 32 4 &&1 15 0 cos (2x) x sin (2x) 64 4 ch & 15 cos (2x) 8
Thus, the antiderivative is: 1 5 5 15 15 15 x5 sin (2x) + x4 cos (2x) x3 sin (2x) x2 cos (2x) + x sin (2x) + cos (2x) + C 2 4 2 4 4 8
This exactly matches the result from wolframalpha.com. If we evaluate at all the proper limits of integration, we have:
=4 1 5 15 15 x5 cos (2x) dx = 5 3 + : 2048 128 16 8 Z0
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