Advanced Integration techniques

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Chapter 7 Advanced Integration Techniques Before introducing the more advanced techniques, we will look at a shortcut for the easier of the substitution-type integrals. Advanced integration techniques then follow: integration by parts, trigonometric integrals, trigonometric substitution, and partial fraction decompositions. 7.1 Substitution-Type Integration by Inspection In this section we will consider integrals which we would have done earlier by substitution, but which are simple enough that we can guess the approximate form of the antiderivatives, and then insert any factors needed to correct for discrepancies detected by (mentally) computing the derivative of the approximate form and comparing it to the original integrand. Some general forms will be mentioned as formulas, but the idea is to be able to compute many such integrals without resorting to writing the usual u-substitution steps. Z Example 7.1.1 Compute cos5xdx. 1 Solution: We can anticipate that the approximate form of the answer is sin5x, but then d d sin5x= cos5x· (5x) = cos5x·5 = 5cos5x. dx dx Since wearelookingfor a function whosederivativeis cos5x, andwefound onewhosederivative is 5cos5x, we see that our candidate antiderivative sin5x gives a derivative with an extra factor of 5, compared with the desired outcome. Our candidate antiderivative’s derivative is 5 times too large, so this candidate antiderivative, sin5x must be 5 times too large. To compensate and 1 arriveat a function with the proper derivative, we multiply our candidate sin5x by . This give 5 1 1 us a new candidate antiderivative sin5x, whose derivative is of course cos5x·5 = cos5x, as 5 5 desired. Thus we have Z 1 cos5xdx= sin5x+C. 5 It may seem that we wrote more in the example above than with the usual u-substitution method, butwhatwewrotecouldbeperformedmentallywithoutresortingtowritingthedetails. In future sections, an integral such as the above may occur as a relatively small step in the execution of a more advanced and more complicated method (perhaps for computing a much moredifficultintegral). Thissection’spurposeistopointouthowsuchanintegralcanbequickly dispatched, to avoid it becoming a needless distraction in the more advanced methods. 1 In this section, by approximate form we mean a form which is correct except for multiplicative constants. 591592 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES 7.1.1 The Method The method used in all the examples here can be summarized as follows: 1. Anticipate the form of the antiderivative by an approximate form (correct up to a multi- plicative constant). 2. Differentiate this approximate form and compare to the original integrand function; 3. IfStep1iscorrect,i.e.,theapproximateform’sderivativediffersfromtheoriginalintegrand function by a multiplicative constant, insert a compensating, reciprocalmultiplicative con- stant into the approximate form to arrive at the actual antiderivative; 4. For verification, differentiate the answer to see if the original integrand function emerges. For instance, some general formulas which should be quickly verifiable by inspection (that is, by reading and mental computation rather than with paper and pencil, for instance) follow: Z 1 kx kx e dx = e +C, (7.1) k Z 1 coskxdx = sinkx+C, (7.2) k Z 1 sinkxdx =− coskx+C, (7.3) k Z 1 2 sec kxdx = tankx+C, (7.4) k Z 1 2 csc kxdx =− cotkx+C, (7.5) k Z 1 seckxtankxdx = seckx+C, (7.6) k Z 1 csckxcotkxdx =− csckx+C, (7.7) k Z 1 1 dx = lnax+b+C. (7.8) ax+b a Example 7.1.2 Thefollowingintegralscanbecomputedwithu-substitution, butalsoarecom- putable by inspection: Z Z 1 x x 7x 7x e dx = e +C, cos dx = 2sin +C, 7 2 2 Z Z 1 1 1 2 dx= ln5x−9+C, sec πxdx = tanπx+C, 5x−9 5 π Z Z 1 1 sin5xdx=− cos5x+C, csc6xcot6xdx=− csc6x+C. 5 6 While it is true that we can call upon the formulas (7.1)–(7.8), the more flexible strategy is to anticipate the form of the antiderivative and adjust accordingly. For instance, we have the following antiderivative form, written two ways: Z 1 du= lnu+C, u Z ′ f (x) dx= lnf(x)+C. f(x)7.1. SUBSTITUTION-TYPE INTEGRATION BY INSPECTION 593 (As usual, the second form is the same as the first where u = f(x).) So when we see an integrand which is a fraction with the numeratorbeing the derivative of the denominator except for multiplicative constants, we know the antiderivative will be, approximately, the natural log of the absolute value of that denominator. Z x Example 7.1.3 Consider dx 2 x +1 Here we see that the derivative of the denominator of the integrand is present—up to a multiplicative constant—in the numerator. Our candidate approximate form can then be given 2 2 by lnx +1 = ln(x +1). Now we differentiate to see what constant factor we need to insert to get the correct derivative: d 1 2x 2 ln(x +1) = ·2x= . 2 2 dx x +1 x +1 1 To correct for the extra factor of 2 and thus get the correct derivative, we insert the factor : 2   d 1 1 1 x 2 ln(x +1) = · ·2x= , 2 2 dx 2 2 x +1 x +1 as desired. Thus Z x 1 2 dx = ln(x +1)+C. 2 x +1 2 To be sure, a quick mental check by differentiation verifies the answer. Of course there are many other forms. Recall we had many other integration formulas, as in Subsection 6.6.1, page 578. For instance it is not difficult to see, or check, that Z Z 1 1 1 −1 −1 du =tan u+C =⇒ dx = tan (ax)+C. 2 2 2 u +1 a x +1 a Example 7.1.4 For instance, we have the following integral computations, which can be seen by relatively easily taking derivatives of the presented antiderivatives. Z Z 1 1 1 −1 dx = tan 3x+C, sec3xdx = lnsec3x+tan3x+C, 2 9x +1 3 3 Z Z   √ 1 1 1 −1 √ √ dx = sin 3x +C, tan2xdx= lnsec2x+C. 2 2 1−3x 3 The method can be used for somewhat more complicated integralsas well, though there does come a point where it seems more natural to simply execute the full substitution method, which is more “constructive” than our method here. However, our “approximateand correct” (read as verbs) method here can be reasonably employed on still more complicated integrals. Z 1 Example 7.1.5 Consider √ dx. 5x−9 R −1/2 Of course this can be rewritten (5x− 9) dx. Now it is crucial that a complete sub- stitution, u = 5x− 9 =⇒ du = 5dx, etc., would show that du and dx agree except for a multiplicative constant, so we know that the integral—up to a multiplicative constant—is of R −1/2 approximate form u du, which calls for the power rule. 1/2 1/2 The approximate form of the antiderivative is thus u = (5x−9) , which we write in x and then differentiate, d 1 1/2 −1/2 (5x−9) = (5x−9) ·5, dx 2594 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES 5 which has extra factors (compared to our original integrand) of collectively . To cancel their 2 2 effects we include a factor in our actual, reported antiderivative. Thus 5 Z √ 1 2 2 1/2 √ dx = (5x−9) +C = 5x−9+C. 5 5 5x−9 Note that a quick derivative computation, albeit involving a (simple) chain rule, gives us the √ correct function 1/ 5x−9. Z 5 2 2 Example 7.1.6 Consider 7x sin x cosx dx. For suchan antiderivative,our ability to guess the form depends upon our expertisewith the R originalsubstitutionmethod. Eachofthesewereofaform f(u)Kdu,wherewecouldanticipate both u and f, with du accounting for remaining terms, and K∈R which we can initially ignore by taking our shortcut path described in this section. Looking ahead, the student well-versed R 2 5 in substitution will expect u = sinx , and the integral being of the approximate form u du 6 (timesaconstant). Thuswewillhaveanapproximateantiderivativeofu (timesaconstant),i.e., 6 2 the approximate form should be sin x . Now we differentiate this and see what compensating 2 factors must be included to reconcile with the original integrand: d 2 6 2 5 2 5 2 2 (sinx ) = 6(sinx ) ·cosx ·2x= 12x sin x cosx . dx 7 Of course we want 7 in the place of the 12 (or separately, 2·6), so we multiply by (or again, 12 1 7· ). With this we have 6·2 Z 7 5 2 2 6 2 7x sin x cosx dx = sin x +C. 12 It would be perfectly natural to forego this method of “guess and adjust” in favor of the old-fashioned substitution method for this problem. Indeed the full substitution method has some advantages (see the next subsection). For instance, it is more “constructive,” and thus less error-prone; one is less tempted to skip steps while employing substitution, while one might attempt a purely mental derivative computation of the answer here and thus easily be off by a factor. It is important that each student find the comfortable level of brevity for himself or 3 herself. But the the method of this section is still often worthwhile. Z 3 4 Example 7.1.7 Compute x sinx dx. R 4 Solution: This is of the approximate form sinudu, with u=x . The approximate form of 4 4 the solutionis thus cosx +C (or−cosx +C, but these differ by a multiplicativeconstant−1), 4 3 1 which has derivative−sinx ·4x . We introduce a factor of− to compensate for the extra 4 factor of−4: Z 1 3 4 4 x sinx dx =− cosx +C, 4 which can be quickly verified by differentiation. 2 6 2 Notice that we are assuming fluency in the chain rule as we compute the derivative of sin x , rather than writing out every step as we did in Chapter 4. Each student must gage personal ability to omit steps. 3 It is the author’s experience that students in engineering and physics programs are often more interested in arriving at the answer quickly, while mathematics and other science students usually prefer the presentation of the full substitution method. The latter are somewhat less likely to be wrong by a multiplicative constant, though the former tend to progress through the topics faster. There are, of course, spectacular exceptions, and each group benefits from camaraderie with the other.7.1. SUBSTITUTION-TYPE INTEGRATION BY INSPECTION 595 Z p 2 Example 7.1.8 Compute x 9−x dx. R 2 1/2 Solution: Itisadvantageoustoreadthisintegralas x(9−x ) dx,whichisofapproximate R 1/2 2 2 3/2 form u du (where u = 9−x ). These observations, and the approximate form (9−x ) of the integral, can be gotten by this mental observation we are developing in this section. The 3 2 1/2 approximate antiderivative’s derivative is (9−x ) ·(−2x), which has an extra factor of−3 2 (after cancellation). Thus Z p  1 3/2 2 2 x 9−x dx=− 9−x +C. 3 7.1.2 Limitations of the Method There are two very important points to be made about the limitations of the method. The first point is illustrated by an example, and the second by making several related points. (I) It is imperative that the derivative of the approximate form differs from the original function to be integrated by at most a multiplicative constant. In particular, an extra variable function cannot be compensated for. To illustrate this point, and simultaneously warn against a common mistake, consider Z 1 dx. 2 x +1 2 Themistaketoavoidhereistotakeerroneouslytheapproximatesolutiontobeln(x +1), which we then notice has derivative d 1 2 ln(x +1) = ·2x. 2 dx x +1 4 Unfortunately we cannot compensate by dividing by the extra factor 2x, because 2   dln(x +1) d(2x) 2x 2 2 2 2x· −ln(x +1)· 2x· −2ln(x +1) d ln(x +1) 2 x +1 dx dx = = , 2 2 dx 2x (2x) 4x which is guaranteed (by the presence of the non-cancelling logarithm in the result) to be 1 something other than our original function . The method does not work because 2 x +1 multiplicative functions do not “go along for the ride” in derivative (or antiderivative) problems the way multiplicative constants do. Of course we knew from before that Z 1 −1 dx= tan x+C, 2 x +1 so this integral is not really suitable for a substitution argument, but is rather a special form in its own right. 4 Alternatively, a product rule computation can be used: » – » – 2 d 1 1 dln(x +1) d 1 1 2x 1 2 2 2 ln(x +1) = · +ln(x +1)· = · − ln(x +1), 2 2 dx 2x 2x dx dx 2x 2x x +1 2x which eventually gives the original function for the first product, but the second part of the product rule is a complication we cannot rid ourselves of easily, though a partial solution to this problem of extra, function-type factors in the integrand is given in the next section.596 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES (II) This method can not totally replace the earlier substitution method. (a) The skills used in the substitution method will be needed for later methods. In particular, the idea which is crucial and recurring is that the entire integral in x is often replaced by one in u (for instance)—including the dx-term and possibly others replaced by du—and, if a definite integral, the interval of integration also represents u-values. (b) If an integral is difficult enough, the more “constructive” substitution method is less error-prone than is this shortcut style developed here. (c) Anyhow,theideaofthesubstitutionmethodisembeddedinthismethod;anticipating what to set equal to u is equivalent to guessing the approximate form of the integral in u, and thus the approximate form of the antiderivative. (d) When using numerical and other methods with definite integrals, a substitution can sometimes make for a much simpler integral to be approximated or otherwise ana- 2 lyzed, even if the antiderivative is never computed. For instance, with u =x , giving then du = 2xdx, we can write Z Z 2 4 4 1 2 x u xe dx = e du. 2 −1 1 None of our usual techniques will yield antiderivatives for either integrand (as the reader is invited to try), but numerical methods such Riemann sums, Trapezoidal and Simpson’s Rules can find approximations for the definite integrals. The latter form of the integral (in u) will yield accurate numerical results more easily than the former (in x). To summarize, the method here has us making an educated guess about the form of the antiderivative, perhaps writing down our guess as our tentative (or “candidate”) answer, taking its derivative, and, assuming it is the same as the integrand except for some multiplicative constant(s), inserting other multiplicative constants into our answer to adjust for discrepancies. It will only work if the tentative antiderivative has derivative equal to some constant times the original integrand. The method is not sophisticated, but will be useful for streamlining later, much longer inte- gration techniques introduced in the rest of this chapter.7.1. SUBSTITUTION-TYPE INTEGRATION BY INSPECTION 597 Exercises For each of the following, first attempt to compute the antiderivative by finding an approxi- mate form of the antiderivative, differentiating it, and inserting a constant factor to compensate for any extra or missing constants. If that method is too unwieldy, compute the integral by the substitution method, showing all details. Z Z 3 2 4 2 2 7 1. (x +1) ·2xdx 18. (x +x ) (3x +2x)dx Z Z 4 3 5 2. cosx ·4x dx 19. sec 3x·sec3xtan3xdx Z √ Z 2 2 3 sin x 3. 15x sec 5x dx 20. √ dx x Z √ √ Z sec xtan x 4. √ dx 2 3 5 3 21. x sinx cos x dx 2 x  Z 1 2   Z csc 3 1 1 x sin cos 5. dx x x 2 22. dx x 2 x Z Z 7 2 6. tan x sec xdx x x 23. e cose dx Z x Z 7. dx 2 x 2 3 (x +1) 24. xe dx Z Z 9 8. (2x−11) dx 2x 2x 25. e sine dx Z Z 9. cos5xdx −x 2 −x 26. e sec e dx Z Z 10. csc9xcot9xdx 5x 27. e dx Z Z x 11. cosxsinxdx (See 13) e 28. dx 2x Z e +1 3 2 Z 12. tan 5xsec 5xdx 3x e 29. √ dx Z 6x 1−e 13. sinxcosxdx (See 11) Z dx √ 30. (Hint: multiply the inte- Z 2x e −1 3 x x 14. sin xcosxdx grand by e /e .) Z Z 4x 4x 10 5 2 31. e (9+e ) dx 15. tan xsec xdx Z Z 2 −2x 2 32. xe dx 16. xsinx dx Z Z 1/x p e 3 4 17. x · x −2dx 33. dx 2 x598 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES √ Z Z x sin(lnx) e 46. dx 34. √ dx x x Z Z lnx 3cos2x 47. dx 35. e sin2xdx x Z Z 1 cosx 48. p dx 36. dx 2 x 1−(lnx) sinx+1 Z Z 1 cosx 49. dx 37. dx 2 sinx x(1+(lnx) ) Z Z sinx 1 p 38. dx 50. dx 2 cosx xlnx (lnx) −1 Z Z 2 2x+1 sec (lnx) 39. dx 2 51. dx x +x x Z Z x 6 (9+lnx) 40. dx 2 52. dx x +1 x Z Z 1 1 41. dx 2 53. dx x +1 2 x(lnx) Z Z 1 42. dx 54. cot2xdx xlnx Z 2x Z e tan(lnx) 43. dx 55. dx 2x 1+e x Z Z 2x e x 44. dx 56. csc dx 4x 1+e 9 Z Z 2 sec x 2 3 45. dx 57. x sec5x dx 1+tanx7.2. INTEGRATION BY PARTS 599 7.2 Integration By Parts While integration by substitution in its elementary form takes advantage of the chain rule, by contrast integration by parts exploits the product rule. In applications it is a bit more complicated than substitution, and there are perhaps more variations on the theme than with substitution, at least at the college calculus level. For these reasons, being fluent in this method usually requires seeing more steps ahead than substitution required. Howeverit can be similarly mastered with practice. 7.2.1 The Idea by Examples 2 Suppose that we need to find an antiderivative of the function f(x) =xsec x. It is not hard to see that normal substitution in not going to easily yield a formula for our desired antiderivative F(x): Z 2 xsec xdx =F(x)+C. 2 However, a clever student might notice that xsec x contains terms that could have arisen from a product rule derivative such as: d dtanx dx xtanx =x· +tanx· dx dx dx 2 =xsec x+tanx. If we rearrange the terms above, we can rewrite this product rule derivative computation as follows: d 2 xsec x = xtanx−tanx. (7.9) dx In fact (7.9) above is perhaps where the spirit of the method is most on display: that the given function to be integrated is indeed one part of a product rule derivative. If we are fortunate, the other part of the product rule formula is easier to integrate, because the derivative term, namely d xtanx is trivial to integrate (see below). Indeed, if we take antiderivatives of both sides of dx R 2 (7.9), we then get the desired formula for the antiderivative xsec xdx: Z Z    dxtanx 2 xsec xdx = −tanx dx dx Z =xtanx− tanxdx =xtanx−lnsecx+C. 2 From such asthe aboveemerges a method wherebywe identify our given function (herexsec x) d as a part of a product rule computation ( xtanx), and integrate our original function by dx d instead (trivially) integrating the product rule derivative term (again xtanx), and then dx integrating the other part (tanx) of the product rule output. Often the other, hidden part of the underlying product rule is easier to integrate than the original function, and therein lies much of the usefulness of the method. While we will have a formal procedure to implement the method, one more example from first principles can further illustrate its spirit.600 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES Z Example 7.2.1 Compute xcosxdx. d Solution: The integrand is one part of a product rule computation for xsinx, namely dx d xsinx =xcosx+sinx, so we can write dx Z Z    d xcosxdx = xsinx −sinx dx dx =xsinx+cosx+C. It is interesting to compute the derivative of our answer, and see how the term we desire (xcosx) emerges, and how other terms which naturally emerge cancel each other. This is left to the reader. 7.2.2 The Technique in Its Simpler Applications Recall that when we completely developed the substitution method, the underlying principle— the chainrule—wasnot written out in complete derivativeform, but ratherin the morecompact ′ differential form. Having supposed that F was an antiderivative of f, i.e., F =f, we eventually settled on writing the argument below without the first two integrals: Z Z Z du(x) ′ f(u(x))u (x)dx = f(u(x))· dx = f(u)du=F(u)+C =F(u(x))+C. dx d At first we did write the first steps because the proof was in the chain rule: F(u(x)) = dx ′ ′ ′ F (u(x))u (x) = f(u(x))u (x). However, we eventually opted for the differential form, though for most it takes some practice for it to seem natural. We will adopt differential notation in integration by parts as well. For instance, recall that the product rule for derivatives, d(uv) dv du =u· +v· , (7.10) dx dx dx can be rewritten in differential form d(uv) =udv+vdu. (7.11) This came from multiplying both sides of (7.10) by dx, all along assuming u and v are in fact functions of x. Now we rearrange (7.11) as follows: udv =d(uv)−vdu. (7.12) Equation (7.12) is perhaps the best equation to visualize the principle behind the eventual integration formula, because it is an easystep from the product rule. The actual formula quoted in most textbooks is still two steps away. First we integrate both sides: Z Z Z udv = d(uv)− vdu. (7.13) R Next we notice that the first integral on the right hand side of (7.13) is simply d(uv) = uv +Constant. Since there is also a constant present in the second integral on the right hand side of (7.13), we can omit mentioning the first constant and arrive at our final working formula for our integration by parts technique: Z Z udv =uv− vdu. (7.14)7.2. INTEGRATION BY PARTS 601 Most textbooks and instructors use the formula above in exactly this form (7.14). It should be memorized, though its derivation—particularly from (7.12)—should also not be forgotten. Furthermore, anytime it is used it is good practice to write the (boxed) formula (7.14) within the problem at the point at which it is used. Next we look at an example of the actual application of (7.14). In the example below, the arrangement of terms is as one would work the problem with pencil and paper, except for the implication arrows which we will omit in subsequent problems, and possibly the under-braces which become less and less necessary with practice. Z x Example 7.2.2 Compute xe dx. Solution: First recall our boxed formula (7.14). For this problem, we write x u=x dv =e dx ⇓ ⇑ x du =dx v =e Z Z x x e dx =uv− vdu z z u dv Z x x = (x)(e )− (e )dx x x =xe −e +C. It is interesting to note that we choose u and dv, and then compute du and v, with one qualifi- cation. That is that v is not unique; the computation from dv to v is of an antidifferentiation x nature and so we really only knowv up to an additive constant. In fact any v so thatdv =e dx x x (we took v = e =⇒ dv = e dx) will work in (7.14). Any additive constant, while legitimate, will eventually cancel in the final computation, and so we usually omit it. For instance, if we x had chosen v =e +100, we would have had Z Z x x uv− vdu=x(e +100)− (e +100)dx x x =xe +100x−e −100x+C x x =xe −e +C, with the same final answer as before. For most cases, we will just assume that the additive constant is zero and we will use the simplest antiderivative for v. We will also not continue to write the implication arrows as they are technical and perhaps confusing. Now we will revisit the first example we gave in Subsection 7.2.1, using what will be our basic style for this method. Z 2 Example 7.2.3 Compute xsec xdx. Solution: First, as is standard for these, we complete a chart like in the previous example. 2 u =x dv =sec xdx du =dx v =tanx602 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES Z Z 2 xsec xdx = udv Z =uv− vdu Z =xtanx− tanxdx =xtanx−lnsecx+C. Since this method is morecomplicated than substitution, there are morecomplicated consid- erations in how to apply it. First of course, one should attempt an earlier, simpler method. But 5 if those fail, and integration by parts is to be attempted, the following guidelines for choosing R R u and dv should be considered for our formula udv =uv− vdu: 1. u and dv must account for all factors of the original integral, and no more. 1.5. Of course, dv must contain the differential term (for example, dx) as a factor, but can contain more terms. R 2. v = dv should be computable with relative ease. ′ 3. du =u (x)dx (assuming the original integral was in x) should not be overly complicated. R R 6 4. The integral vdu should be simpler than the original integral udv. The next example illustrates the importance of consideration 2 above. Z 3 2 Example 7.2.4 Compute x sinx dx. 2 3 2 Solution: We donotwantto makeu =sinx , becausethendv =x dx, givingdu= 2xcosx R R 1 4 1 5 2 and v = x , and our vdu will be x cosx dx, which is worse than our original integral. 4 2 3 We will instead take u to be some power of x, but not all of x , else the terms remaining for 2 7 dv would be dv = sinx dx, which we cannot integrate with ordinary methods. 2 What we will settle on is dv =xsinx dx, because its integral is an easy substitution we can 2 short-cut as in Section 7.1. We leave the remaining terms, collectively x , for u: 2 2 u=x dv =xsinx dx 1 2 du = 2xdx v =− cosx 2 5 Of course with practice one can see ahead whether or not integration by parts is likely to achieve an answer for a particular integral. R 6 Later, in a twist on the method, we will see that the we do not require vdu be easier than the original, R udv, in all cases, but it is desirable in most cases. R 7 2 In fact, we cannot compute sinx dx using any kind of substitution or parts, or any other method of this text for that matter, and arrive at an antiderivative in simple terms of the functions we know so far such as powers, exponentials, logarithms, trigonometric or hyperbolic functions or their inverses. However, when we 2 study series we will find other expressions with which we can fashion an antiderivative of sinx .7.2. INTEGRATION BY PARTS 603 Z Z 3 2 2 2 x sinx dx= x ·xsinx dx z z u dv Z =uv− vdu     Z 1 1 2 2 2 = (x ) − cosx − − cosx 2xdx 2 2 Z 2 x 2 2 =− cosx + cosx ·xdx 2 2 x 1 2 2 =− cosx + sinx +C. 2 2 We omitted the details of computing v from dv, and computing the last integral, in the spirit of Section 7.1. R This last example shows how the requirement that v = dv (up to an additive constant) be computable helps to guide us to the proper choice of u and dv. It was lucky that the second integral was easily computable (which would not have been the case if the originalintegral were, R R 2 2 4 2 say, x sinx dx or x sinx dx), but anyhow we can not even get to the second integral if we can not compute v. Z 9 x Example 7.2.5 Compute √ dx. 5 1−x R Solution: This is similarto the previousexample, inthe sensethat computability ofv = dv dictates our choices of u and dv. 5 4 5 −1/2 u =x dv =x (1−x ) dx 2 4 5 1/2 du =5x dx v =− (1−x ) 5 Z Z 9 x √ dx = udv 5 1−x Z =uv− vdu Z 2 5 5 1/2 4 5 1/2 =− x (1−x ) +2 x (1−x ) dx 5 p 2 −1 2 5 5 3/2 5 =− x 1−x +2· · (1−x ) +C 5 5 3 p 2 4 5 5 3/2 5 =− x 1−x − (1−x ) +C 5 15 p   2 5 5 5 =− 1−x 3x +2(1−x ) +C 15 p 2 5 5 =− (x +2) 1−x +C. 15 As will often be the case in this section and subsequent sections, to check our answers it will usually be much easier to carefully check each step in our work of computing the integrals, than to compute the derivatives of our answers. (If we do wish to check answers by differentiation, it is often simpler to do so with a nonsimplified expression of the solution.)604 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES 7.2.3 Repeated Use of Integration by Parts The nextexamplesshowadifferentlesson: that it issometimesappropriatetointegratebyparts R more than once in a given problem. At each step, the hope is that the integral vdu is simpler R than the integral it came from ( udv) in the integration by parts formula. Sometimes the new R integral vdu is indeed simpler, but not so much so that it can be integrated readily. Indeed, at times that second integral also needs integration by parts, and so on, until we arrive at an integral that we can compute easily. Z 2 Example 7.2.6 Compute x cos3xdx. 2 Solution: The x term is complicating our integral, and so we reduce its effect somewhat by an integration by parts step. 2 u =x dv =cos3xdx 1 du = 2xdx v = sin3x 3 Z Z 2 x cos3xdx=uv− vdu Z (7.15) 1 2 2 = x sin3x− xsin3xdx. 3 3 While we still cannot compute this last integral directly with old methods, it is better than the original in the sense that our trigonometric function is multiplied by a first-degree polynomial, where in the original the polynomial was second-degree. One more application of integration by parts and there will be no polynomial factor at all in the new integral. A strict use of the language would force us to introduce two new variables other than u and 1 2 v, but since they have “disappeared” in the present form of our answer, namely x sin3x− R 3 2 xsin3xdx, it is not considered such bad form to “reset” (or “recycle”) u and v for another 3 R integration by parts step, this time involving the integral xsin3xdx: u =x dv =sin3xdx 1 du =dx v =− cos3x 3 Z Z xsin3xdx =uv− vdu Z x 1 =− cos3x+ cos3xdx 3 3 x 1 1 =− cos3x+ · sin3x+C . 1 3 3 3 Now we insert this last result into our original computation (7.15): Z   2 x 2 x 1 2 x cos3xdx = sin3x− − cos3x+ sin3x+C 1 3 3 3 9 2 x 2 2 = sin3x+ xcos3x− sin3x+C. 3 9 277.2. INTEGRATION BY PARTS 605 Several lessons can be gleaned from the example above. (1) it is very important for proper “bookkeeping,” as these problems can beget several “subproblems,” and the proper placement of resulting terms is crucial for getting the correct final answer; (2) it is not unknown to use integration by parts more than once in a problem; (3) if we can take as many antiderivatives of a function f(x) as we like (i.e., the antiderivative, the antiderivative of the antiderivative, etc.), R n n then for an integral x f(x)dx we can let u =x , and integration by parts will yield a second integral with a reduction in the power of x, namely Z Z n n n−1 x f(x)dx =x F(x)− nx F(x)dx (7.16) ′ where F =f. If this process is repeated often enough, the complicating polynomial factors will have their degrees diminished until there is no polynomial factor, and we can finally finish the integration. 7.2.4 Integrals of Certain Other Functions Herewelookatcaseswherefunctionsappearwhosederivativesareknown,butwhoseantideriva- tives might not be standard knowledge of the averagecalculus student. In these cases we choose that function to be u, and the other terms to constitute dv, assuming of course we can compute R v = dv. Z 2 Example 7.2.7 Compute (x +1)lnxdx. Solution: We cannot let dv = lnxdx, since as yet we do not know the antiderivative of lnx. (Even if we did, such a choice for dv would not be advantageous, as Example 7.2.8 will help to show later.) So we have little choice but to let lnx be u. 2 u =lnx dv =(x +1)dx 3 1 x du = dx v = +x x 3 Z Z 2 (x +1)lnxdx =uv− vdu   Z   3 3 x x 1 = (lnx) +x − +x dx 3 3 x Z   2 1 x 3 = (lnx)(x +3x)− +1 dx 3 3 1 1 3 3 = (x +3x)lnx− x −x+C. 3 9 Next we have an interesting case where we are forced to take dv =dx. Z Example 7.2.8 Compute lnxdx. R Solution: Here we can not let dv = lnxdx, for computing v = dv would be the same as computing the whole, original integral. As in Example 7.2.7, we also note that placing lnx in the u-term makes for a simple enough du term. Thus we write u =lnx dv =dx 1 du = dx v =x x606 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES so that Z Z lnxdx =uv− vdu Z   1 = (lnx)(x)− (x) dx x Z =xlnx− 1dx =xlnx−x+C. In fact the same type of computation will be used for finding antiderivatives of arctrigonometric functions(thougharcsecantandarccosecantneedlatertechniquesforfindingtheresultingsecond R integral vdu). Z −1 Example 7.2.9 Compute sin xdx. −1 Solution: Again we have no choice but to let u =sin x, and dv =dx. −1 u = sin x dv =dx 1 √ du = dx v =x 2 1−x R −1 For brevity, we label the desired integral (I), so here (I) = sin xdx. (The second integral below is computed “by inspection.”) Z Z x −1 (I) =uv− vdu =xsin x− √ dx 2 1−x  1/2 −1 2 =xsin x+ 1−x +C. 7.2.5 An Indirect Method The following method, summarized at the end, is useful in surprisingly many settings. Z 2x Example 7.2.10 Compute e cos3xdx =(I). Solution: Here we again name the desired integral (I) for brevity in later steps. It should be obvious (especially after a few attempts) that simple substitution methods will not work. So we attempt an integration by parts. Step 1. We will let the trigonometric function be part of dv: 2x u =e dv = cos3xdx 1 2x du =2e dx v = sin3x 3 So far, after some rearrangement and simplifying, we have Z (I) =uv− vdu Z 1 2 2x 2x = e sin3x− e sin3xdx. (7.17) 3 3 z (II)7.2. INTEGRATION BY PARTS 607 Thisdoesnotseemanyeasierthanthefirstintegral,soperhapswemightcontinue,butthis time let the trigonometric function be u and the exponential (along with dx) be contained in dv. R 2x Step 2–First Attempt. Compute (II) = e sin3xdx in light of the comments at the end of the first step. 2x u =sin3x dv =e dx 1 2x du =3cos3xdx v = e 2 Z Z 1 3 2x 2x (II) =uv− vdu = e sin3x− e cos3xdx. 2 2 Combining this with the conclusion (7.17) of Step 1 gives us:  Z  1 2 1 3 2x 2x 2x (I) = e sin3x− e sin3x− e cos3xdx 3 3 2 2 Z 1 1 2x 2x 2x = e sin3x− e sin3x+ e cos3xdx 3 3 Z 2x = e cos3xdx. Unfortunately that puts us right back where we started. However, a minor change in our effort above will eventually lead us to the solution. While keeping Step 1, our next step towards a solution is to replace Step 2 by the same strategy as used in Step 1, namely that we use the exponential function for u and the trigonometric function in the dv term. R 2x Step 2—Second Attempt. Again we attempt to compute (II) = e sin3xdx, though with different choices of u and dv. 2x u =e dv =sin3xdx 1 2x du = 2e dx v =− cos3x 3 Z (II) =uv− vdu Z 1 2 2x 2x =− e cos3x+ e cos3xdx. (7.18) 3 3 It may seem that (7.18) is also a dead end, since it contains the original integral. But this attempt is different. In fact, when we combine (7.18) with (7.17) we get 1 2 2x (I) = e sin3x− (II) 3 3  Z  1 2 1 2 2x 2x 2x = e sin3x− − e cos3x+ e cos3xdx 3 3 3 3 Z 1 2 4 2x 2x 2x = e sin3x+ e cos3x− e cos3xdx, 3 9 9 z (I)608 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES which we can summarize by the following equation: 1 2 4 2x 2x (I) = e sin3x+ e cos3x− (I). (7.19) 3 9 9 Nowwearereadytoderive(I),notbyanothercalculuscomputation,butinfactbysimple algebra: we solve for it. 4 Step 3. Solve (7.19) for (I). First we add (I) to both sides of (7.19): 9 13 1 2 2x 2x (I) = e sin3x+ e cos3x+C . (7.20) 1 9 3 9 Here we include C because in fact each (I) in (7.19) represents all antiderivatives, which 1 differ from each other by additive constants. Now (7.19) made sense because of the fact that there are (hidden) additive constants on both sides of that equation (though on the 4 8 right side they are multiplied by− , but that still yields additive constants). Solving 9 (7.20) for (I) we now have   9 1 2 2x 2x (I) = e sin3x+ e cos3x+C 1 13 3 9 3 2 2x 2x = e sin3x+ e cos3x+C, (7.21) 13 13 9 where C = C . 1 13 Of course with this our original problem is solved: Z 3 2 2x 2x 2x e cos3xdx = e sin3x+ e cos3x+C. 13 13 What is important to understand about the example above is that sometimes, though we cannot perhaps directly compute a particular integral, it may happen that an indirect method gives us the answer. Here we found an equation, namely (7.19), which our desired integral satisfies, and for which (I) could be solved algebraically. We must be open to the possibility— indeed, theopportunity—offindingadesiredquantitybysuchindirectmethods, aswellasdirect computations. It should be pointed out that we could have computed the integral in Example 7.2.10 by 2x instead letting u be the trigonometric function, and dv =e dx in both Steps 1 and 2. In fact it is usually best to pick similar choices for u and dv when an integration by parts will take more R n than one step. (Recall the discussion for x f(x)dx.) The method of Example 7.2.10, namely solving for (I) after an integration by parts step, is available perhaps more often than one would think, though it is not a method of first resort. Thenextexampleisalsooneinwhichwewilleventually“solve”fortheintegralalgebraically. Z 2 Example 7.2.11 Compute sin xdx =(I). 8 In fact, two simultaneousappearances of (I)donot have to have thesame additiveconstants, so (I)−(I) = C , not zero. 27.2. INTEGRATION BY PARTS 609 Solution: The only reasonable choice here seems to be to let u =sinx and dv = sin dx, if we 9 are to integrate this by parts. u =sinx dv =sinxdx du =cosxdx v =−cosx Z Z Z 2 (I) = uv− vdu =−sinxcosx+ cos xdx. We could perform the same integration by parts with the second integral, which might or might not yield an equation we can solve for (I) (as the reader is invited to explore), but instead we 2 2 will use the fact that cos x= 1−sin x: Z 2 (I) =−sinxcosx+ (1−sin x)dx Z 2 =−sinxcosx+x− sin xdx =x−sinxcosx−(I). 10 Adding (I) to both sides we get 2(I) =−sinxcosx+x+C 1 1 =⇒ (I) = (x−sinxcosx)+C. 2 7.2.6 Miscellaneous Considerations First we look at a definite integral arising from integration by parts. It should be pointed out 11 that the general formula will look like the following: Z Z x=b x=b x=b udv = uv − vdu. (7.22) x=a x=a x=a 9 The integral in Example 7.2.11 can also be computed directly if we first use the trigonometric identity 1 2 sin x = (1−cos2x), and then the identity sin2x =2sinxcosx: 2 Z Z 1 1 1 2 sin xdx= (1−cos2x)dx= x− sin2x+C 2 2 4 1 1 1 1 = x− ·2sinxcosx+C = x− sinxcosx+C, 2 4 2 2 which is the same as the answer in the text of Example 7.2.11. In Section 7.3 we will opt for this alternative method, and indeed will make quite an effort to exploit the algebraic properties of the trigonometric functions wherever possible, but some integrals there will still require integration by parts. 10 In fact many textbooks do not bother writing the C term, preferring to remind the student at the end that 1 an indefinite integral problem necessitates a “+C.” 11 Sometextsleaveoutthe“x=”parts,assumingtheyareunderstood,butwewillcontinuetousetheconvention that, unless otherwise stated, the “limits of integration” should correspond to values of the differential’s variable. Another popular way to write (7.22) avoids the issue: Z ˛ Z b b b ˛ ′ ′ ˛ u(x)v (x)dx = u(x)v(x) − v(x)u (x)dx. ˛ a a a610 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES Z π Example 7.2.12 Compute xsinxdx. −π Solution: The antiderivative is an easier case than many of our previous examples, but care has to be taken to keep track of all the signs (+/−) in computing the definite integral: u =x dv =sinxdx du =dx v =−cosx Z Z π π π xsinxdx = (−xcosx) + cosxdx −π −π −π π = −πcosπ−−(−π)cos(−π)+ sinx −π = (−π)(−1)−(π)(−1)+sinπ−sin(−π) =π+π+0−0 = 2π. R π In the example above, we could also have noticed that cosxdx is zero because we are −π integrating over a whole period −π,π of cosx, and both sinx and cosx have definite integral zero over any full period a,a+2π. (Think of their graphs, or their definite integrals over any such period.) b It is typical to compute that part u(x)v(x) separately, but one could instead separately a compute the entire antiderivative, and then evaluate at the two limits and take the difference: Z π π xsinxdx = (−xcosx+sinx) −π −π = (π+0)−(−π+0) = 2π. The choiceofmethodisamatterofbookkeepingpreferences, andperhapswhetherornotpartof the right-hand side of (7.22) is particularly simple. If not, it is reasonable to solve the indefinite R integral xsinxdx as a separate matter, and then write the definite integral with the formula R b for the antiderivative inserted, as in f(x)dx = F(x) and so on as above. a R The next example gives us several options for computing the new integral vdu along the way, though in each the original choices of u and dv are the same. Z −1 Example 7.2.13 Compute xtan xdx =(I). Solution: Again we have little choice on our selection of u and dv. −1 u = tan x dv =xdx 1 1 2 du = dx v = x 2 x +1 2 Z (I) =uv− vdu Z 2 1 1 x 2 −1 = x tan x− dx. 2 2 2 x +1

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