Math 1B, lecture 8: Integration by parts

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1 Math B, lecture 8: Integration by parts Nathan Pflueger 23 September 2 Introuction Integration by parts, similarly to integration by substitution, reverses a well-known technique of ifferentiation an explores what it can o in computing integrals. Like integration by substitution, it shoul really be viewe not necessarily as a means of solving an integral, but as a means of transforming an integral. Whether this transformation prouces a more tractable integral integral or so will epen on how it is applie, an eventually you will evelop an intuition for what is likely to work an what is not. For our purposes, this technique is mainly inclue to expan the collection of antierivatives that will be familiar to you upon completing the course. As a seconary objective, a goo qualitative unerstaning of the technique shes a great eal of light on other parts of the subject. Integration by parts is a critical technique in unerstaning various phenomena in electricity an magnetism, for example, precisely because of the sort of transformation it inuces. In the option appenix, section 5, I inicate a geometric way to view integration by parts, which has been far more important for me, both in mathematical work an in stuying other subjects, than any calculations of any particular integrals. The reaing for toay is Gottlieb 29.. The homework is problem set 6 (which inclues weekly problems 5 an 6) an a topic outline. You shoul begin working on weekly problems 8 an 9. 2 The technique in general Integration by parts is simply the prouct rule in reverse. Suppose that u(x), v(x) are two ifferent functions of x. Then the prouct rule states that x (u(x)v(x)) = u (x)v(x) + u(x)v (x). To obtain the integration version of this, integrate both sies with respect to x to obtain the following, which is usually written with the arguments (x) suppresse, in ifferential notation. u(x)v(x) = uv = v(x)u (x)x + vu + uv u(x)v (x)x Here, we are regaring u an v as functions of x espite suppressing the argument, an u, v are the usual shorthan, enoting respectively u (x)x an v (x)x. The usual way of expressing integration by parts is the following. uv = uv vu

2 Integration by parts may be unerstoo as follows: it transforms an integral by ifferentiating one part an integrating another part. This explains the transition from uv to vu. The minus sign an the term uv are by-proucts of this transition. The reason to perform this transformation is usually that the integral vu is somehow easier than the original, but there may be other reasons, which are illustrate in the examples. Integration by parts is most likely to help in situations where u becomes much simpler upon ifferentiation. This is especially true if it is a polynomial. Other goo caniates are exponential functions, sines, an cosines. Often, it is necessary to perform integration by parts multiple times. This is especially the case when u is a polynomial (such as x 2 ). There is a systematic way to o this an keep track of all the signs, which is escribe in section Definite integrals To evaluate efinite integrals, you can either compute the inefinite integral an then plug in limits, or you can track the limits of integration as you go. For the secon option, you will en up with an expression like the following. b a b u(x)v (x)x = [u(x)v(x)] b a v(x)u (x)x Alternative, if you wish to evaluate the integral vu with respect to the variable u, rather than x, the following is also vali. Since I am now suppressing the arguments (x), it is necessary to specify which variable is going from a go b explicitly in the subscripts an superscripts. a x=b x=b uv = [uv] x=b vu u(b) = [uv] x=b vu In the secon line, I have converte the limits of integration from values of x to values of u. This will only be a useful thing to o if v can be written as a function of u, an thus the secon integral written as an integral with respect to u. For an example of where this can be one, see examples 3.5 an 3.9. This is also a useful viewpoint in the optional appenices 5 an 6. u(a) 3 Examples The following examples consist of the exempts on the worksheet that I use in class, with the exception of 3.3, which I have ae after the fact, an the ninth example from the hanout, which now comprises the optional section 6. Example 3.. xe x x This is a typical case: the e x is easy to integrate, an the x will become simpler if it is ifferentiate. xe x x u = x, v = e x x u = x, v = e x = xe x e x x = xe x e x + C 2

3 Example 3.2. π/2 x cos xx This is very similar to the example 3.. π/2 x cos xx u = x, v = cos xx u = x, v = sin x = [x sin x] π/2 π/2 sin xx Example 3.3. e x sin xx = π 2 = π 2 [ cos x]π/2 This is a canonical example of a situation where integration by parts makes the integral no simpler, but accomplishes a curious swinle: it creates an equation from which the integral can be foun. It orer to achieve this swinle, integration by parts must be performe twice in a row. e x sin xx u = sin x, v = e x x u = cos xx, v = e x = e x sin x e x cos xx u = cos x, v = e x x u = sin x, v = e x = e x sin x e x cos x + = e x sin x e x cos x e x ( sin x)x e x sin xx Now, since the same inefinite integral appears on both sies, we can move it all to the left an solve for it. Notice that because all inefinite integrals have an implicit + C, once we move all inefinite integrals from the right sie, we must inclue this term on the right. 2 e x sin xx = e x sin x e x cos x + C e x sin xx = 2 (ex sin x e x cos x + C) Since 2C an C coul equally well be the arbitrary constants, this can be written also as e x sin xx = 2 ex (sin x cos x) + C Example 3.4. sin 2 xx We evaluate this integral in the previous class using the ouble-angle formula for cosine. However, just to illustrate the fact that the same integral can often be one multiple ways, here is a way to o it using integration by parts (an also a version of the swinle use in the previous example). 3

4 sin 2 xx u = sin x, v = sin xx 2 u = cos xx, v = cos x = sin x cos x ( cos x)(cos x)x = sin x cos x + cos 2 xx = sin x cos x + ( sin 2 x)x = sin x cos x + x sin 2 xx = sin x cos x + x sin 2 xx sin 2 xx = sin x cos x + x + C sin 2 xx = 2 sin x cos x + 2 x + C Have we obtaine the same answer using this metho as we i last time? At first blush, it appears that we have not. However, in fact this is the same answer; upon applying the ientity sin(2x) = 2 sin x cos x, this answer is equal to 4 sin(2x) + 2x + C, which is the answer obtaine by the other metho. Example 3.5. e ln xx This integral is one of the classics of integration by parts. The iea behin it is the eliminate the logarithm by taking its erivative an proucing the much simpler x. The price pai for this is the integral of x which is simply x, an in fact this is quite a small price. e ln xx u = ln x, v = x u = x x, v = x e = [x ln x] e x x x e = [x ln x] e x = [x ln x x] e = (e e) ( ) = Note that this is the same integral that we compute in an earlier class by means of re-slicing, an fining the area uner the curve y = ln x by instea fining the area between the curves x = e an x = e y, as an integral with respect to y. To see that what we were really oing in that example was integration by parts, consier this following alternative computation by integration by parts, where we write the secon integral as an integral with respect to y. 4

5 Let y = ln x ln xx = yx = xy xy = x ln x e y y = x ln x e y + C = x ln x x + C Notice that the only ifference here is that the new integral create by the integration by parts is now evaluate as an integral with respect to y, rather than with respect to x. What integration by parts has one, essentially, is reverse the way that the integral is slice, just as was one in the earlier computation. Example x 2 e x x x 2 e x x u = x 2, v = e x x u = 2xx, v = e x = x 2 e x (2x)e x x = x 2 e x 2 xe x x = x 2 e x 2xe x + 2e x 2. At the last stage, we have use the result of the first exercise. x 3 e x x x 3 e x x u = x 3, v = e x x u = 3x 2 x, v = e x = x 3 e x 3x 2 e x x = x 3 e x 3(x 2 e x 2xe x + 2e x ) + C = = e x (x 3 3x 2 + 6x 6) + C 3. Here, again, we have use the result of the previous part. Note that in each case, the integral is transforme, by integration by parts, into the same integral, with a smaller exponent for x. x n e x x By using u = x n, v = e x x, the following reuction formula is obtaine. x n e x x = x n e x n x n e x x 5

6 By repeate application of this formula, any such integral can be calculate. The result may be written as follows. x n e x x = e x (x n nx n + n(n )x n 2 n(n )(n 2)x n ( ) n (n!)) + C Example x ln xx If ln x is ifferentiate, then what remains will be only powers of x, so it seems most profitable to perform the substitution u = ln x. x ln xx u = ln x, v = xx 2. x n ln xx u = x x, v = 2 x2 = 2 x2 ln x 2 xx = 2 x2 ln x 4 x2 + C Try the same take as in the previous problem. x n ln xx u = ln x, v = x n x Example 3.8. (ln x) 2 x = = u = x x, v = n + xn+ n + xn+ ln x n + xn x n + xn+ ln x (n + ) 2 xn+ + C There are at least two approaches that will work here. One is to apply integration by parts immeiately, using u = (ln x) 2, v = x. This will reuce the problem to integrating ln xx, which is an integral that is alreay known. Another approach is to first perform a substation to eliminate the logarithm, as follows. (ln x) 2 x = = u = ln x, u = x x u 2 xu u 2 e u u Now this has been transforme to the integral from the first part of example 3.6. Hence using the result of that example, the result is (ln x) 2 x = e u (u 2 2u + 2) + C = x((ln x) 2 2 ln x + 2) + C 6

7 Example 3.9. tan xx This is another case where the erivative of the integran is of a somewhat simpler form than the original integran, so taking it for u will help. tan xx u = tan x, v = x u = + x 2 x, v = x = x tan x x x 2 + x This new integral now can be evaluate by means of a substitution for x 2 +, yieling the following result. tan x = x tan x 2 ln x2 + + C This integral, as well as that of ln x an a number of others, fits into a general scheme which can be use to evaluate integrals of many inverses of well-unerstoo functions. This is iscusses in more etail in section 6. 4 Tabular Integration This section shoul be consiere optional, although it may be useful in keeping track of notation while performing integrals assigne in this course. Tabular integration is a useful bookkeeping scheme for performing integration by parts multiple times in a row. The basic technique is to split the integran into to pieces, iteratively integrate one part an integrate the other, an arrange the results into a table. For example, consier x 2 e 2x. The integran x 2 e 2x can be split into the two parts x 2 an e 2x. Now ifferentiate x 2 until it reaches, an meanwhile integrate e 2x repeately (omitting the +C terms in this case; the only thing that matters is that the erivative of each entry of the secon column is equal to the function in the cell above it). The following table results. x 2 e 2x x x x 2x 2 x 2 e2x x 4 e2x x 8 e2x I have inicate with arrows the irection in which ifferentiation is taking place. So the lett column was create by successive ifferentiation, an the right column by successive integration. To obtain the inefinite integral from this table, raw slante lines with alternating signs, take the prouct of the terms at the en of each arrow, with the sign on the arrow, an sum. The process is shown below. 7

8 x 2 e 2x 2x + 2 e2x 2 4 e2x + 8 e2x x 2 e 2x = x 2 2 e2x 2x 4 e2x e2x + C = 2 x2 e 2x 2 xe2x + 4 e2x + C This technique will always work when the left column eventually becomes, that is, when the top of the left column is a polynomial. However, with a slight moification, the technique of tabular integration can be useful even when the left column oes not eventually become. The moification is to a a horizontal line on the bottom row, give it a sign (in the same alternating fashion), an inclue the integral of the prouct of these terms (with the sign) in the sum. For example, here is how you can use this more general type of tabular integration on e 2x sin(3x)x, which is example 29.5 in Gottlieb s textbook. e 2x sin(3x) 2e 2x + 3 cos(3x) 4e 2x + 9 sin(3x) e 2x sin(3)x = e 2x ( 3 ) cos(3x) 2e 2x ( 9 ) sin(3x) + 4e 2x ( 9 ) sin(3x) x = 3 e2x cos(3x) e2x sin(3x) 4 e 2x sin(3x)x 9 From here, both inefinite integrals can be move to one sie of the equation, an then compute by solving (as in the example in Gottlieb s book). Notice that the reason a horizontal line with an integral is not necessary when the left column becomes is quite simple: the integral is still there, but it is equal to. Notice also that the usual version of integration by parts in general can be shown by a tabular integration with only two rows. u(x) v (x) u (x) + v(x) 8

9 u(x)v (x)x = u(x) v(x) uv = uv vu u (x)v(x)x It is not ifficult to see how iterating this process prouces the general technique of tabular integration as escribe above. 5 Appenix: geometric interpretation Integration by parts can be viewe as a generalization of the iea that an integral can be compute by slicing either horizontally or vertically. This appenix is meant to explain how this works. Suppose that u(x), v(x) are functions of x, usually abbreviate u, v, an use as usual the shorthan u = u (x)x an v = v (x)x. Suppose that we esire to compute the integral x=b uv. We can visualize the value of this integral by plotting a curve with coorinates (v(x), u(x))) from x = a to x = b; the esire integral is the area uner this curve, as shown in the picture. Now, we can attempt to integrate one one part of the integral uv, in effect assuming that u is constant, an guess that the inefinite integral is simply uv. What will be the error of this guess? The guess for the efinite integral woul be [uv] x=b = u(b)v(b) u(a)v(b), which can be seen geometrically as the L-shape regions shown. 9

10 The error of this guess is precisely the area to the left of the curve, which can be compute by means of horizontal slicing; it is precisely equal to x=b vu. Therefore the formula for integration by parts of a efinite integral is obtaine. x=b x=b uv = [uv] x=b vu The geometric effect of this transformation is to convert the effort of evaluating the original integral (slice vertically) to evaluating a new, horizontally slice integral, which may be easier. An example of an application which comes to min from this viewpoint is escribe in the following appenix. 6 Appenix: integrating inverse functions Recall from lecture 5 that the integral of an inverse function can often be more easily compute by slicing horizontally, that is, transforming an integral that was originally with respect to x to an integral with respect to y. This is the sort of transformation that integration by parts accomplishes, although integration

11 by parts is a much more general technique. The relationship is most easily seen when attempting to integrate inverse functions, such as ln x or tan x, as in the examples above. In fact, the same sort of technique work for f x, whenever f(x) is a function whose inverse can be integrate. To see this, suppose that f(x) is a function with known antierivative F (x). Then we can compute the integral of f (x) as follows. Note that I o not actually write own the erivative of f (x) because I o not care; in this case it is not actually neee in orer to finish evaluating the integral. I also use the letter y rather than u since I really am thinking of this as a y-coorinate in a graph. f (x)x y = f (x), v = x y = ( ), v = x = xf (x) xy = xf (x) f(y)y = xf (x) F (y) + C = xf (x) F (f (x)) + C Notice that this commutes a number of integrals that we have previously compute (an not surprisingly, it gives the same answer). Example 6.. Let f(x) = x 2, so that f (x) = x, an F (x) can be taken to be 3 x3. Then F (f (x)) = 3 ( x) 3 = 3 x3/2. Therefore xx = x x 3 x3/2 + C = 2 3 x3/2 + C Example 6.2. Let f(x) = e x. Then we can take F (x) to be e x as well, an this last result gives: ln xx = x ln x e ln x + C = x ln x x + C Example 6.3. Let f(x) = tan x. Then we can take F (x) to be ln sec x (see problem set ). Therefore F (f (x)) = ln sec(tan x) = ln + x 2 = 2 ln + x2. The penultimate step follows because if θ = tan x, then θ is the angle opposite x in a right triangle with legs of length x an, hence the hypotenuse is of length + x 2, an sec θ = + x 2. Therefore we obtain tan x = x tan x 2 ln + x2 + C