Integration by Parts

11 Key excerpts on "Integration by Parts"

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  Finite Mathematics and Applied Calculus

  • Stefan Waner, Steven Costenoble(Authors)
  • 2017(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  We also see how to extend the definition of the definite integral to include integrals over infinite intervals, and we show how such integrals can be used for long-term forecasting. Finally, we introduce the beau-tiful theory of differential equations and some of its numerous applications. 14.1 Integration by Parts Integration-by-Parts Formula Integration by Parts is an integration technique that comes from the product rule for derivatives. The tabular method we present here has been around for some time and makes Integration by Parts quite simple, particularly in problems where it has to be used several times. ✱ We start with a little notation to simplify things while we introduce Integration by Parts. (We use this notation only in the next few pages.) If u is a function, denote its derivative by D 1 u 2 and an antiderivative by I 1 u 2 . Thus, for example, if u 5 2 x 2 , then D 1 u 2 5 4 x and I 1 u 2 5 2 x 3 3 . [If we wished, we could instead take I 1 u 2 5 2 x 3 3 1 46, but we usually opt to take the simplest antiderivative.] Integration by Parts If u and v are continuous functions of x and u has a continuous derivative, then 3 u # v dx 5 u # I 1 v 2 2 3 D 1 u 2 I 1 v 2 dx . Quick Example (Discussed more fully in Example 1 below) 1. 3 x # e x dx 5 xI 1 e x 2 2 3 D 1 x 2 I 1 e x 2 dx 5 xe x 2 3 1 # e x dx I 1 e x 2 5 e x ; D 1 x 2 5 1 5 xe x 2 e x 1 C . 3 e x dx 5 e x 1 C As Quick Example 1 shows, although we could not immediately integrate u # v 5 x # e x , we could easily integrate D 1 u 2 I 1 v 2 5 1 # e x 5 e x . ✱ The version of the tabular method we use was developed and taught to us by Dan Rosen at Hofstra University. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 14.1 Integration by Parts 1061 Derivation of Integration-by-Parts Formula As we mentioned, the integration-by-parts formula comes from the product rule for derivatives.

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  Calculus for The Life Sciences

  • Sebastian J. Schreiber, Karl J. Smith, Wayne M. Getz(Authors)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  Integration by Parts Integration by Parts is a procedure based on inverting the product rule for differen- tiation. To derive a formula for this procedure, we begin with the product rule for differentiating functions f (x) and g(x), assuming these derivatives exist. d dx f (x)g(x) = f  (x)g(x) + f (x)g  (x) product rule  d dx f (x)g(x) dx =  [ f  (x)g(x) + f (x)g  (x)] dx antidifferentiate both sides  d dx f (x)g(x) dx =  f  (x)g(x) dx +  f (x)g  (x) dx properties of integrals f (x)g(x) =  f  (x)g(x) dx +  f (x)g  (x) dx FTC f (x)g(x) −  f  (x)g(x) dx =  f (x)g  (x) dx subtract  f  (x)g(x)dx from both sides If we let u = f (x) and v = g(x), then du = f  (x) dx, dv = g  (x) dx, and we ob- tain the following simplified formula. 5.6 Integration by Parts and Partial Fractions 395 Integration by Parts  u dv = uv −  v du To evaluate integrals using Integration by Parts, we want to choose u and dv so that the new integral is easier to integrate than the original integral. Example 1 Integration by Parts Find  xe x dx Solution For this example, there are two ways we can choose u and dv. Suppose we choose u = x and dv = e x dx. We differentiate u and integrate dv. Thus, du = dx, and v = e x . Now, substitute these values into the Integration by Parts formula.  u dv = uv −  v du  xe x dx = xe x −  e x dx = xe x − e x + C We noted that there were two possible choices for u and dv in Example 1. The other choice is to let u = e x and dv = x dx. If we make this choice, and substitute into the formula for Integration by Parts, then we obtain  xe x dx = 1 2 x 2 e x −  1 2 x 2 e x dx (Try this yourself to practice the technique.) In this case, instead of simplifying the problem and solving it, we just made it more complicated! Thus, it is important to try different substitutions to see which works best. Example 2 When the differentiable part is the entire integrand Find  ln x dx assuming x > 0.

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  How to Integrate It

  A Practical Guide to Finding Elementary Integrals

  • Seán M. Stewart(Author)
  • 2017(Publication Date)
  • Cambridge University Press

    (Publisher)

  7 Integration by Parts ‘Begin at the beginning,’ the King said, very gravely, ‘and go on till you come to the end: then stop.’ — Lewis Carroll, Alice in Wonderland After integration by substitution, the next most important technique is what is known as Integration by Parts. There is no counterpart for the product rule for the derivative when it comes to integration. Integration by Parts, however, is a consequence of the product rule for differentiation. We first state and give a proof for this very important result before giving a number of examples that make use of the rule. Theorem 7.1 (Integration by Parts). If f 0 .x/ and g 0 .x/ are continuous functions then Z f.x/g 0 .x/dx D f.x/ g.x/ Z f 0 .x/g.x/dx Proof From the product rule for the derivative we have .f.x/g.x// 0 D f 0 .x/g.x/ C f.x/g 0 .x/: Rewriting this as f.x/g 0 .x/ D .f.x/g.x// 0 f 0 .x/g.x/; one has Z f.x/g 0 .x/dx D Z .f.x/g.x// 0 dx Z f 0 .x/g.x/dx D f.x/ g.x/ Z f 0 .x/g.x/dx; which completes the proof. 79 80 How to Integrate It For a definite integral on the interval Œa;b the corresponding result will be Z b a f.x/g 0 .x/dx D f.x/ g.x/ ˇ ˇ ˇ b a Z b a f 0 .x/g.x/dx It is possible to write the rule more compactly. If u.x/ (for f.x/) and v.x/ (for g.x/) are two continuously differentiable functions of x, we can write du D u 0 .x/dx and d v D v 0 .x/dx, and the rule for Integration by Parts can be written more compactly as Z ud v D uv Z v du At first sight, using Integration by Parts may not seem very useful as it appears that one integral on the left side has simply been replaced with another integral on the right side. While this is true, the real power of the rule comes about from this exact change. In many instances the integral on the right side will be more amenable to integration than the integral one started out with. In many commonly encountered cases this is exactly what one finds. Also, to apply the rule, the function g 0 .x/ to be integrated needs to be relatively simple to begin with.

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  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

  • Adrian Banner(Author)
  • 2009(Publication Date)
  • Princeton University Press

    (Publisher)

  Section 18.2: Integration by Parts • 393 We are now supposed to work this out and try to move back to x -land. Well, by our nice equation, with x replaced by t , we see that the above integral is equal to F ( g ( t )) + C , where F is an antiderivative of f . This is just F ( x )+ C , which is exactly what we want. So this method works as well, and we have justified the method of substitution. 18.2 Integration by Parts We saw how to reverse the chain rule by using the method of substitution. There is also a way to reverse the product rule—it’s called Integration by Parts. Let’s recall the product rule from Section 6.2.3 of Chapter 6: if u and v depend on x , then d dx ( uv ) = v du dx + u dv dx . Let’s rearrange this equation and then integrate both sides with respect to x . We get Z u dv dx dx = Z d dx ( uv ) dx -Z v du dx dx. The first term on the right-hand side is the antiderivative of the derivative of uv , so it’s just equal to uv + C . The + C is unnecessary, though, because the second term on the right-hand side is already an indefinite integral: it includes a + C automatically. So we have shown that Z u dv dx dx = uv -Z v du dx dx. This is the formula for Integration by Parts. It’s perfectly usable in this form, but there’s an abbreviated form which is even more convenient. If we replace dv dx dx by dv , and replace du dx dx by du , we get the formula Z u dv = uv -Z v du. Again, this is just an abbreviation for the real formula, but it is pretty useful. Let’s see how it works in practice. Suppose we want to find Z xe x dx. Substitution seems useless (try it and see), so let’s try Integration by Parts. We’d love to get the integral in the form R u dv so we can apply the Integration by Parts formula. There are a number of ways to do this, but here’s one that works: set u = x and dv = e x dx . Then we certainly have R xe x dx = R u dv . Now, to apply the Integration by Parts formula, we need to be able to find du and v as well.

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  Calculus

  Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  Show that the two answers are equivalent. Integration by Parts for Definite Integrals For definite integrals the formula corresponding to (3) is  b a u dv = uv  b a −  b a v du (7) Remark It is important to keep in mind that the variables u and v in this formula are functions of x and that the limits of integration in (7) are limits on the variable x. Sometimes it is helpful to emphasize this by writing (7) as  b x=a u dv = uv  b x=a −  b x=a v du (8) The next example illustrates how Integration by Parts can be used to integrate the inverse trigonometric functions. Example 7 Evaluate  1 0 tan −1 x dx. Solution Let u = tan −1 x, dv = dx, du = 1 1 + x 2 dx, v = x Thus,  1 0 tan −1 x dx =  1 0 u dv = uv  1 0 −  1 0 v du The limits of integration refer to x; that is, x = 0 and x = 1. = x tan −1 x  1 0 −  1 0 x 1 + x 2 dx But  1 0 x 1 + x 2 dx = 1 2  1 0 2x 1 + x 2 dx = 1 2 ln (1 + x 2 )  1 0 = 1 2 ln 2 so  1 0 tan −1 x dx = x tan −1 x  1 0 − 1 2 ln 2 =  π 4 − 0  − 1 2 ln 2 = π 4 − ln √ 2 Reduction Formulas Integration by Parts can be used to derive reduction formulas for integrals. These are formulas that express an integral involving a power of a function in terms of an integral that involves a lower power of that function. For example, if n is a positive integer and n ≥ 2, then Integration by Parts can be used to obtain the reduction formulas 414 CHAPTER 7 Principles of Integral Evaluation  sin n x dx = − 1 n sin n−1 x cos x + n − 1 n  sin n−2 x dx (9)  cos n x dx = 1 n cos n−1 x sin x + n − 1 n  cos n−2 x dx (10) To illustrate how such formulas can be obtained, let us derive (10).

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  Single Variable Calculus

  Concepts and Contexts, Enhanced Edition

  • James Stewart(Author)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  SECTION 5.6 Integration by Parts 383 Every differentiation rule has a corresponding integration rule. For instance, the Substitu- tion Rule for integration corresponds to the Chain Rule for differentiation. The rule that corresponds to the Product Rule for differentiation is called the rule for Integration by Parts. The Product Rule states that if and are differentiable functions, then In the notation for indefinite integrals this equation becomes or We can rearrange this equation as Formula 1 is called the formula for Integration by Parts. It is perhaps easier to remem- ber in the following notation. Let and . Then the differentials are and , so, by the Substitution Rule, the formula for Integration by Parts becomes Integrating by parts Find . SOLUTION USING FORMULA 1 Suppose we choose and . Then and . (For we can choose any antiderivative of .) Thus, using Formula 1, we have It’s wise to check the answer by differentiating it. If we do so, we get , as expected. x sin x x cos x sin x C x cos x y cos x dx x cos x y cos x dx y x sin x dx f x t x y t x f x dx t t t x cos x f x 1 t x sin x f x x y x sin x dx EXAMPLE 1 y u dv uv y v du 2 dv t x dx du f x dx v t x u f x y f x t x dx f x t x y t x f x dx 1 y f x t x dx y t x f x dx f x t x y f x t x t x f x dx f x t x d dx f x t x f x t x t x f x t f 5.6 Integration by Parts SOLUTION USING FORMULA 2 Let Then and so Note: Our aim in using Integration by Parts is to obtain a simpler integral than the one we started with. Thus in Example 1 we started with and expressed it in terms of the simpler integral . If we had instead chosen and , then and , so Integration by Parts gives Although this is true, is a more difficult integral than the one we started with. In general, when deciding on a choice for and , we usually try to choose to be a function that becomes simpler when differentiated (or at least not more complicated) as long as can be readily integrated to give .

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  Calculus

  Early Transcendental Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  THE PRODUCT RULE AND Integration by Parts Our primary goal in this section is to develop a general method for attacking integrals of the form  f (x)g(x) dx As a first step, let G(x) be any antiderivative of g(x). In this case G  (x) = g(x), so the product rule for differentiating f (x)G(x) can be expressed as d dx [ f (x)G(x)] = f (x)G  (x) + f  (x)G(x) = f (x)g(x) + f  (x)G(x) (1) This implies that f (x)G(x) is an antiderivative of the function on the right side of (1), so we can express (1) in integral form as  [ f (x)g(x) + f  (x)G(x)] dx = f (x)G(x) or, equivalently, as  f (x)g(x) dx = f (x)G(x) −  f  (x)G(x) dx (2) This formula allows us to integrate f (x)g(x) by integrating f  (x)G(x) instead, and in many cases the net effect is to replace a difficult integration with an easier one. The application of this formula is called Integration by Parts. In practice, we usually rewrite (2) by letting u = f (x), du = f  (x) dx v = G(x), dv = G  (x) dx = g(x) dx This yields the following alternative form for (2):  u dv = uv −  v du (3) Example 1 Use Integration by Parts to evaluate  x cos x dx. Note that in Example 1 we omitted the constant of integration in calculat- ing v from dv. Had we included a constant of integration, it would have eventually dropped out. This is always the case in Integration by Parts [Exer- cise 76(b)], so it is common to omit the constant at this stage of the computa- tion. However, there are certain cases in which making a clever choice of a con- stant of integration to include with v can simplify the computation of  v du (Exercises 77–79). Solution. We will apply Formula (3). The first step is to make a choice for u and dv to put the given integral in the form  u dv. We will let u = x and dv = cos x dx (Other possibilities will be considered later.) The second step is to compute du from u and v from dv. This yields du = dx and v =  dv =  cos x dx = sin x

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  Calculus

  Single Variable

  • Deborah Hughes-Hallett, William G. McCallum, Andrew M. Gleason, Eric Connally, Daniel E. Flath, Selin Kalaycioglu, Brigitte Lahme, Patti Frazer Lock, David O. Lomen, David Lovelock, Guadalupe I. Lozano, Jerry Morris, Brad G. Osgood, Cody L. Patterson, Douglas Quinney, Karen R. Rhea, Ayse Arzu Sahin, Adam H. Spiegler(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  Problem 59 asks you to do this integral by another method. The previous example illustrates a useful technique: Use Integration by Parts to transform the integral into an expression containing another copy of the same integral, possibly multiplied by a coefficient, then solve for the original integral. Example 7 Use Integration by Parts twice to find   2 sin(3) . Solution Using Integration by Parts with  =  2 and   = sin(3) gives   = 2 2 ,  = − 1 3 cos(3), so we get   2 sin(3)  = − 1 3  2 cos(3) + 2 3   2 cos(3) . On the right side we have an integral similar to the original one, with the sine replaced by a cosine. Using Integration by Parts on that integral in the same way gives   2 cos(3)  = 1 3  2 sin(3) − 2 3   2 sin(3) . 7.2 Integration by Parts 357 Substituting this into the expression we obtained for the original integral gives   2 sin(3)  = − 1 3  2 cos(3) + 2 3 ( 1 3  2 sin(3) − 2 3   2 sin(3)  ) = − 1 3  2 cos(3) + 2 9  2 sin(3) − 4 9   2 sin(3) . The right side now has a copy of the original integral, multiplied by −4∕9. Moving it to the left, we get ( 1 + 4 9 )   2 sin(3)  = − 1 3  2 cos(3) + 2 9  2 sin(3). Dividing through by the coefficient on the left, (1+4∕9) = 13∕9 and adding a constant of integration  , we get   2 sin(3)  = 9 13 ( − 1 3  2 cos(3) + 2 9  2 sin(3) ) +  = 1 13  2 (2 sin(3) − 3 cos(3)) + . Example 8 Use a computer algebra system to investigate ∫ sin( 2 ) . Solution It can be shown that sin( 2 ) has no elementary antiderivative. A computer algebra system gives an antiderivative involving a non-elementary function, the Fresnel Integral, which you may not recog- nize. Exercises and Problems for Section 7.2 Online Resource: Additional Problems for Section 7.2 EXERCISES 1. Use Integration by Parts to express ∫  2    in terms of (a)   3    (b)     2. Write arctan  = 1 ⋅ arctan  to find ∫ arctan  .

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  Introduction to Integral Calculus

  Systematic Studies with Engineering Applications for Beginners

  • Ulrich L. Rohde, G. C. Jain, Ajay K. Poddar, A. K. Ghosh(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  For example, to evaluate the integral Ð x e x 2 dx, the method of Integration by Parts is not needed. (Why). (6) Example (20): Evaluate Ð x n log x dx ¼ I ðsayÞ Solution: I ¼ Ð ðlog xÞ ðx n Þdx d dx ðlog xÞ ¼ 1 x ð x n ¼ x nþ1 n þ 1 8 > > < > > : ) I ¼ ðlog xÞ  x nþ1 n þ 1  ð 1 x  x nþ1 n þ 1 dx ¼ x nþ1 n þ 1  log x  1 n þ 1 ð x n dx ¼ x nþ1 n þ 1  log x  1 n þ 1  x nþ1 n þ 1 þ c ¼ x nþ1 n þ 1  log x  x nþ1 n þ 1 ð Þ 2 þ c Ans: Example (21): Evaluate Ð 2x sin 4x  cos  2x dx Solution: Let I ¼ ð 2x sin 4x  cos 2x dx ¼ ð 2x  1 2 sin 6x þ sin2x ½ dx ) I ¼ ð x sin 6x dx þ ð x sin 2x dx Now, we can apply the method of Integration by Parts to each integral on right-hand side. (6) Check if this integral can be evaluated by simple substitution. 112 Integration by Parts Remark: If the integrand were 2xsin 2xcos 2x, then it could be written as x[2 sin 2x cos 2x] ¼ xsin 4x, which can be easily integrated by parts. Example (22): Evaluate Ð xcos 2 x dx ¼ I ðsayÞ Here again we must use trigonometric identities to convert the integrand into some convenient form. We have cos 2x ¼ 2 cos 2 x  1 ) cos 2 x ¼ 1 þ cos 2x 2 ) I ¼ ð x 1 þ cos 2x 2  ! dx ¼ x 2 4 þ 1 2 ð x cos 2x dx Note (7): Integrals like Ð sin 1 x  log x dx (where in both the functions cannot be integrated easily) cannot be evaluated by the method of Integration by Parts. 4a.3 HELPFUL PICTURES CONNECTING INVERSE TRIGONOMETRIC FUNCTIONS WITH ORDINARY TRIGONOMETRIC FUNCTIONS In practice, inverse trigonometric functions are often combined with ordinary functions. Trigonometric substitutions, if successful, help in converting a troublesome integral in x to a simpler integral in t. However, the problem of translating back to an expression in x always remains. Such difficulties are easily overcome by drawing a suitable triangle based on the equation like x ¼ sin t. In this case we can write sin t ¼ (x/1), and draw the required picture as done above.

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  Calculus

  Single and Multivariable

  • Deborah Hughes-Hallett, Andrew M. Gleason, William G. McCallum, Daniel E. Flath, Patti Frazer Lock, David O. Lomen, David Lovelock, Brad G. Osgood, Douglas Quinney, Karen R. Rhea, Jeff Tecosky-Feldman, Thomas W. Tucker, Otto K. Bretscher, Sheldon P. Gordon, Andrew Pasquale, Joseph Thrash(Authors)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  Problem 59 asks you to do this integral by another method. The previous example illustrates a useful technique: Use Integration by Parts to transform the integral into an expression containing another copy of the same integral, possibly multiplied by a coefficient, then solve for the original integral. Example 7 Use Integration by Parts twice to find   2 sin(3) . Solution Using Integration by Parts with  =  2 and   = sin(3) gives   = 2 2 ,  = − 1 3 cos(3), so we get   2 sin(3)  = − 1 3  2 cos(3) + 2 3   2 cos(3) . On the right side we have an integral similar to the original one, with the sine replaced by a cosine. Using Integration by Parts on that integral in the same way gives   2 cos(3)  = 1 3  2 sin(3) − 2 3   2 sin(3) . 7.2 Integration by Parts 357 Substituting this into the expression we obtained for the original integral gives   2 sin(3)  = − 1 3  2 cos(3) + 2 3 ( 1 3  2 sin(3) − 2 3   2 sin(3)  ) = − 1 3  2 cos(3) + 2 9  2 sin(3) − 4 9   2 sin(3) . The right side now has a copy of the original integral, multiplied by −4∕9. Moving it to the left, we get ( 1 + 4 9 )   2 sin(3)  = − 1 3  2 cos(3) + 2 9  2 sin(3). Dividing through by the coefficient on the left, (1+4∕9) = 13∕9 and adding a constant of integration  , we get   2 sin(3)  = 9 13 ( − 1 3  2 cos(3) + 2 9  2 sin(3) ) +  = 1 13  2 (2 sin(3) − 3 cos(3)) + . Example 8 Use a computer algebra system to investigate ∫ sin( 2 ) . Solution It can be shown that sin( 2 ) has no elementary antiderivative. A computer algebra system gives an antiderivative involving a non-elementary function, the Fresnel Integral, which you may not recog- nize. Exercises and Problems for Section 7.2 Online Resource: Additional Problems for Section 7.2 EXERCISES 1. Use Integration by Parts to express ∫  2    in terms of (a)   3    (b)     2. Write arctan  = 1 ⋅ arctan  to find ∫ arctan  .

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  Differentiation and Integration

  • W. Bolton(Author)
  • 2016(Publication Date)
  • Routledge

    (Publisher)

  4 6.3 Integration by Parts TECHNIQUES OF INTEGRATION 125 Review problems 16 Evaluate the following integrals: (a) r: ( 4 :lxl) dx using X= 2 tan a substitution, (b) J': x dx using u = J 1 + 2x 2 substitution, 0 b +2xl (C) I~ J 4 -X 2 dx using X = 2 sin a substitution The product rule for differentiation (see section 2.1.3) gives J!.(uv) = udv +v$. dx dx dx where u and v are functions of x. Integrating both sides of this equation with respect to x gives I !

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