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Techniques of Integration Because of the Fundamental Theorem of Calculus, We Can Integrate a Function If We Know an Antiderivative, That Is, an Indefinite Integral

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Techniques of Integration Because of the Fundamental Theorem of Calculus, We Can Integrate a Function If We Know an Antiderivative, That Is, an Indefinite Integral The techniques of this chapter enable us to find the height of a rocket a minute after liftoff and to compute the escape velocity of the rocket. Techniques of Integration Because of the Fundamental Theorem of Calculus, we can integrate a function if we know an antiderivative, that is, an indefinite integral. We summarize here the most impor- tant integrals that we have learned so far. x nϩ1 1 y x n dx ෇ ϩ C ͑n Ϫ1͒ y dx ෇ ln Խ x Խ ϩ C n ϩ 1 x a x y e x dx ෇ e x ϩ C y a x dx ෇ ϩ C ln a y sin x dx ෇ Ϫcos x ϩ C y cos x dx ෇ sin x ϩ C y sec2xdx෇ tan x ϩ C y csc2xdx෇ Ϫcot x ϩ C y sec x tan x dx ෇ sec x ϩ C y csc x cot x dx ෇ Ϫcsc x ϩ C y sinh x dx ෇ cosh x ϩ C y cosh x dx ෇ sinh x ϩ C y tan x dx ෇ ln Խ sec x Խ ϩ C y cot x dx ෇ ln Խ sin x Խ ϩ C 1 1 x 1 x y dx ෇ tanϪ1ͩ ͪ ϩ C y dx ෇ sinϪ1ͩ ͪ ϩ C x 2 ϩ a 2 a a sa 2 Ϫ x 2 a In this chapter we develop techniques for using these basic integration formulas to obtain indefinite integrals of more complicated functions. We learned the most important method of integration, the Substitution Rule, in Section 5.5. The other gen- eral technique, integration by parts, is presented in Section 7.1. Then we learn methods that are special to particular classes of functions such as trigonometric func- tions and rational functions. Integration is not as straightforward as differentiation; there are no rules that absolutely guarantee obtaining an indefinite integral of a function. Therefore, in Section 7.5 we discuss a strategy for integration. |||| 7.1 Integration by Parts Every differentiation rule has a corresponding integration rule. For instance, the Substi- tution 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 f and t are differentiable functions, then d ͓ f ͑x͒t͑x͔͒ ෇ f ͑x͒tЈ͑x͒ ϩ t͑x͒f Ј͑x͒ dx 476 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION In the notation for indefinite integrals this equation becomes y ͓ f ͑x͒tЈ͑x͒ ϩ t͑x͒f Ј͑x͔͒ dx ෇ f ͑x͒t͑x͒ or y f ͑x͒tЈ͑x͒ dx ϩ y t͑x͒f Ј͑x͒ dx ෇ f ͑x͒t͑x͒ We can rearrange this equation as 1 y f ͑x͒tЈ͑x͒ dx ෇ f ͑x͒t͑x͒ Ϫ y t͑x͒f Ј͑x͒ dx Formula 1 is called the formula for integration by parts. It is perhaps easier to remem- ber in the following notation. Let u ෇ f ͑x͒ and v ෇ t͑x͒. Then the differentials are du ෇ f Ј͑x͒ dx and dv ෇ tЈ͑x͒ dx, so, by the Substitution Rule, the formula for integration by parts becomes 2 y udv ෇ uv Ϫ y v du EXAMPLE 1 Find y x sin x dx. SOLUTION USING FORMULA 1 Suppose we choose f ͑x͒ ෇ x and tЈ͑x͒ ෇ sin x. Then f Ј͑x͒ ෇ 1 and t͑x͒ ෇ Ϫcos x. (For t we can choose any antiderivative of tЈ.) Thus, using Formula 1, we have y x sin x dx ෇ f ͑x͒t͑x͒ Ϫ y t͑x͒f Ј͑x͒ dx ෇ x͑Ϫcos x͒ Ϫ y ͑Ϫcos x͒ dx ෇ Ϫx cos x ϩ y cos x dx ෇ Ϫx cos x ϩ sin x ϩ C It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as expected. SOLUTION USING FORMULA 2 Let |||| It is helpful to use the pattern: u ෇ x dv ෇ sin x dx u ෇ ᮀ dv ෇ ᮀ du ෇ ᮀ v ෇ ᮀ Then du ෇ dx v ෇ Ϫcos x and so u d√ u √ √ du y x sin x dx ෇ y x sin x dx ෇ x ͑Ϫcos x͒ Ϫ y ͑Ϫcos x͒ dx ෇ Ϫx cos x ϩ y cos x dx ෇ Ϫx cos x ϩ sin x ϩ C SECTION 7.1 INTEGRATION BY PARTS ❙❙❙❙ 477 NOTE I 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 x x sin x dx and expressed it in terms of the simpler integral x cos x dx. If we had chosen u ෇ sin x and dv ෇ x dx, then du ෇ cos x dx and v ෇ x 2͞2, so integration by parts gives x 2 1 y x sin x dx ෇ ͑sin x͒ Ϫ y x 2 cos x dx 2 2 Although this is true,x x 2 cos x dx is a more difficult integral than the one we started with. In general, when deciding on a choice for u and dv, we usually try to choose u ෇ f ͑x͒ to be a function that becomes simpler when differentiated (or at least not more complicated) as long as dv ෇ tЈ͑x͒ dx can be readily integrated to give v. EXAMPLE 2 Evaluate y ln x dx. SOLUTION Here we don’t have much choice for u and dv. Let u ෇ ln x dv ෇ dx 1 Then du ෇ dx v ෇ x x Integrating by parts, we get dx y ln x dx ෇ x ln x Ϫ y x x |||| It’s customary to write x 1 dx as x dx. ෇ x ln x Ϫ y dx |||| Check the answer by differentiating it. ෇ x ln x Ϫ x ϩ C Integration by parts is effective in this example because the derivative of the function f ͑x͒ ෇ ln x is simpler than f . EXAMPLE 3 Find y t 2et dt. SOLUTION Notice that t 2 becomes simpler when differentiated (whereas et is unchanged when differentiated or integrated), so we choose u ෇ t 2 dv ෇ et dt Then du ෇ 2tdt v ෇ et Integration by parts gives 2 t 2 t t 3 y t e dt ෇ t e Ϫ 2 y te dt The integral that we obtained,x tet dt , is simpler than the original integral but is still not obvious. Therefore, we use integration by parts a second time, this time with u ෇ t and 478 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION dv ෇ et dt. Then du ෇ dt,v ෇ et , and y tet dt ෇ tet Ϫ y et dt ෇ tet Ϫ et ϩ C Putting this in Equation 3, we get 2 t 2 t t y t e dt ෇ t e Ϫ 2 y te dt 2 t t t ෇ t e Ϫ 2͑te Ϫ e ϩ C͒ 2 t t t ෇ t e Ϫ 2te ϩ 2e ϩ C1 where C1 ෇ Ϫ2C EXAMPLE 4 Evaluate y e x sin x dx. |||| An easier method, using complex numbers, SOLUTION Neither e x nor sin x becomes simpler when differentiated, but we try choosing is given in Exercise 48 in Appendix G. u ෇ e x and dv ෇ sin x dx anyway. Then du ෇ e x dx and v ෇ Ϫcos x, so integration by parts gives 4 y e x sin x dx ෇ Ϫe x cos x ϩ y e x cos x dx The integral that we have obtained,x e x cos x dx , is no simpler than the original one, but at least it’s no more difficult. Having had success in the preceding example integrating by parts twice, we persevere and integrate by parts again. This time we use u ෇ e x and dv ෇ cos x dx. Then du ෇ e x dx,v ෇ sin x , and 5 y e x cos x dx ෇ e x sin x Ϫ y e x sin x dx At first glance, it appears as if we have accomplished nothing because we have arrived at x |||| Figure 1 illustrates Example 4 by x e sin x dx, which is where we started. However, if we put Equation 5 into Equation 4 showing the graphs of f ͑x͒ ෇ e x sin x and we get ͑ ͒ ෇ 1 x͑ Ϫ ͒ F x 2 e sin x cos x . As a visual check on our work, notice that f ͑x͒ ෇ 0 when F has x x x x a maximum or minimum. y e sin x dx ෇ Ϫe cos x ϩ e sin x Ϫ y e sin x dx 12 This can be regarded as an equation to be solved for the unknown integral. Adding x ex sin x dx to both sides, we obtain F f 2 y e x sin x dx ෇ Ϫe x cos x ϩ e x sin x _3 6 Dividing by 2 and adding the constant of integration, we get _4 x ෇ 1 x͑ Ϫ ͒ ϩ y e sin x dx 2 e sin x cos x C FIGURE 1 If we combine the formula for integration by parts with Part 2 of the Fundamental Theorem of Calculus, we can evaluate definite integrals by parts. Evaluating both sides of Formula 1 between a and b, assuming f Ј and tЈ are continuous, and using the Fundamental SECTION 7.1 INTEGRATION BY PARTS ❙❙❙❙ 479 Theorem, we obtain b ͑ ͒tЈ͑ ͒ ෇ ͑ ͒t͑ ͒ b Ϫ b t͑ ͒ Ј͑ ͒ 6 y f x x dx f x x ]a y x f x dx a a 1 Ϫ EXAMPLE 5 Calculate y tan 1xdx. 0 SOLUTION Let u ෇ tanϪ1x dv ෇ dx dx Then du ෇ v ෇ x 1 ϩ x 2 So Formula 6 gives 1 1 1 x Ϫ1 ෇ Ϫ1 Ϫ y tan xdx x tan x]0 y 2 dx 0 0 1 ϩ x 1 x ෇ ؒ Ϫ1 Ϫ ؒ Ϫ1 Ϫ 1 tan 1 0 tan 0 y 2 dx 0 1 ϩ x |||| Ϫ1 ജ ജ ␲ Since tan x 0 for x 0, the integral in ෇ Ϫ 1 x Example 5 can be interpreted as the area of the y 2 dx 4 0 1 ϩ x region shown in Figure 2.
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