y 1 x dx ln x y a x dx 3. y e x dx e x 15. y sinh x dx cosh x y cos x dx sin x y csc 2 x dx cot x 7. y sec 2 x dx tan x 9. y sec x tan x dx sec x

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1 Strateg for Integration As we have seen, integration is more challenging than differentiation. In finding the derivative of a function it is obvious which differentiation formula we should appl. But it ma not be obvious which technique we should use to integrate a given function. Until now indivial techniques have been applied in each section. For instance, we usuall used substitution in Eercises 5.5, integration b parts in Eercises 5.6, and partial fractions in Eercises 5.7 and Appendi G. But in this section we present a collection of miscellaneous integrals in random order and the main challenge is to recognize which technique or formula to use. No hard and fast rules can be given as to which method applies in a given situation, but we give some advice on strateg that ou ma find useful. A prerequisite for strateg selection is a knowledge of the basic integration formulas. In the following table we have collected the integrals from our previous list together with several additional formulas that we have learned in this chapter. Most of them should be memorized. It is useful to know them all, but the ones marked with an asterisk need not be memorized since the are easil derived. Formula 9 can be avoided b using partial fractions, and trigonometric substitutions can be used in place of Formula. Table of Integration Formulas Constants of integration have been omitted. n. n n. n 3. e e sin cos 6. ln a a ln a cos sin 7. sec tan 8. csc cot 9. sec tan sec. sec ln sec tan.. tan ln sec csc cot csc csc ln csc cot cot ln sin 5. sinh cosh 6. cosh sinh a a tan a *9. *. a a ln a a sin sa a s a ln s a

2 STRATEGY FOR INTEGRATION Once ou are armed with these basic integration formulas, if ou don t immediatel see how to attack a given integral, ou might tr the following four-step strateg.. Simplif the Integrand if Possible Sometimes the use of algebraic manipulation or trigonometric identities will simplif the integrand and make the method of integration obvious. Here are some eamples: tan sec s ( s) (s ) d sin cos d cos sin cos d sin d sin cos sin sin cos cos sin cos. Look for an Obvious Substitution Tr to find some function u t in the integrand whose differential t also occurs, apart from a constant factor. For instance, in the integral we notice that if u, then. Therefore, we use the substitution u instead of the method of partial fractions. 3. Classif the Integrand According to Its Form If Steps and have not led to the solution, then we take a look at the form of the integrand f. (a) Trigonometric functions. If f is a proct of powers of sin and cos, of tan and sec, or of cot and csc, then we use the substitutions recommended in Section 5.7 and Additional Topics: Trigonometric Integrals. (b) Rational functions. If f is a rational function, we use the procere involving partial fractions in Section 5.7 and Appendi G. (c) Integration b parts. If f is a proct of a power of (or a polnomial) and a transcendental function (such as a trigonometric, eponential, or logarithmic function), then we tr integration b parts, choosing u and dv according to the advice given in Section 5.6. If ou look at the functions in Eercises 5.6, ou will see that most of them are the tpe just described. (d) Radicals. Particular kinds of substitutions are recommended when certain radicals appear. (i) If s a occurs, we use a trigonometric substitution according to the table in Additional Topics: Trigonometric Substitution. (ii) If s n a b occurs, we use the rationalizing substitution u s n a b. More generall, this sometimes works for s n t. 4. Tr Again If the first three steps have not proced the answer, remember that there are basicall onl two methods of integration: substitution and parts. (a) Tr substitution. Even if no substitution is obvious (Step ), some inspiration or ingenuit (or even desperation) ma suggest an appropriate substitution. (b) Tr parts. Although integration b parts is used most of the time on procts of the form described in Step 3(c), it is sometimes effective on single functions. Looking at Section 5.6, we see that it works on tan, sin, and ln, and these are all inverse functions. (c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the denominator or using trigonometric identities) ma be useful in transforming

3 STRATEGY FOR INTEGRATION 3 the integral into an easier form. These manipulations ma be more substantial than in Step and ma involve some ingenuit. Here is an eample: cos cos cos cos cos cos cos sin csc cos sin (d) Relate the problem to previous problems. When ou have built up some eperience in integration, ou ma be able to use a method on a given integral that is similar to a method ou have alread used on a previous integral. Or ou ma even be able to epress the given integral in terms of a previous one. For instance, tan sec is a challenging integral, but if we make use of the identit tan sec, we can write and if sec 3 has previousl been evaluated (see Eample 8 in Additional Topics: Trigonometric Integrals), then that calculation can be used in the present problem. (e) Use several methods. Sometimes two or three methods are required to evaluate an integral. The evaluation could involve several successive substitutions of different tpes, or it might combine integration b parts with one or more substitutions. In the following eamples we indicate a method of attack but do not full work out the integral. EXAMPLE tan3 cos 3 In Step we rewrite the integral: tan sec sec 3 sec tan3 cos 3 tan 3 sec 3 The integral is now of the form tan m sec n with m odd, so we can use the advice in Additional Topics: Trigonometric Integrals. Alternativel, if in Step we had written tan3 cos 3 sin3 cos 3 cos 3 sin3 cos 6 then we could have continued as follows with the substitution u cos : sin3 cos 6 cos sin u cos 6 u 6 u u 6 u 4 u 6 EXAMPLE e s According to Step 3(d)(ii) we substitute u s. Then u, so u and e s ue u

4 4 STRATEGY FOR INTEGRATION The integrand is now a proct of u and the transcendental function e u so it can be integrated b parts. EXAMPLE No algebraic simplification or substitution is obvious, so Steps and don t appl here. The integrand is a rational function so we appl the procere of Appendi G, remembering that the first step is to divide. EXAMPLE 4 sln Here Step is all that is needed. We substitute u ln because its differential is,which occurs in the integral. EXAMPLE 5 Although the rationalizing substitution u works here [Step 3(d)(ii)], it leads to a ver complicated rational function. An easier method is to do some algebraic manipulation [either as Step or as Step 4(c)]. Multipling numerator and denominator b s, we have s s s sin s C Can We Integrate All Continuous Functions? The question arises: Will our strateg for integration enable us to find the integral of ever continuous function? For eample, can we use it to evaluate e? The answer is no, at least not in terms of the functions that we are familiar with. The functions that we have been dealing with in this book are called elementar functions. These are the polnomials, rational functions, power functions a, eponential functions a, logarithmic functions, trigonometric and inverse trigonometric functions, hperbolic and inverse hperbolic functions, and all functions that can be obtained from these b the five operations of addition, subtraction, multiplication, division, and composition. For instance, the function f 3 is an elementar function. If f is an elementar function, then f is an elementar function but f need not be an elementar function. Consider f e. Since f is continuous, its integral eists, and if we define the function F b F e t dt ln cosh e sin

5 STRATEGY FOR INTEGRATION 5 then we know from Part of the Fundamental Theorem of Calculus that F e Thus, f e has an antiderivative F, but it has been proved that F is not an elementar function. This means that no matter how hard we tr, we will never succeed in evaluating e in terms of the functions we know. (In Chapter 8, however, we will see how to epress e as an infinite series.) The same can be said of the following integrals: e sin cos e s 3 ln sin In fact, the majorit of elementar functions don t have elementar antiderivatives. You ma be assured, though, that the integrals in the following eercises are all elementar functions. Eercises A Click here for answers. S Click here for solutions. 7. cot ln sin 8. sin sat dt 8 Evaluate the integral. sin sec. tan. 3. t t 3 dt 4. e arctan d tan 3 d s3 4 csc cot w 3. w dw s s 3 4 cot 4 4 cot 3 7. r 4 ln r dr sin 36. sin 4 cos cos tan d tan 5 sec 3 d. sin 3 cos 5 d s sin e e.. t 3 e t dt. sin cos cos s s ln ln e t dt 4t e e 3 s s sin 39. s tan d e s e e a a tan s e e e s4 4 3 d 4 a 4 3. ( s) 8 4. ln s4 s s4 4

6 6 STRATEGY FOR INTEGRATION 53. sinh m 54. sin e 69. e 7. ln s s s 3 c 58. ln e 3 e se s s s arctanst dt 68. st s 3 3 ln tan sin cos u 3 u 3 u e e sin sin sin b sin s sin cos The functions e and e don t have elementar antiderivatives, but e does. Evaluate e. ln st s 3 t dt e e sec cos sin sec sin cos sin 4 cos 4

7 STRATEGY FOR INTEGRATION 7 Answers S sin ln csc cot C ln 9 5. e e ln ln 4 5 tan C. 8 cos 8 6 cos 6 C (or 4 sin 4 3 sin 6 8 sin 8 C) 3. s C 5. s sin cos 4 sin C (or 4 4 sin 8 cos C) 9. e e C. 8 e t 4t 3 6t 6t 3 C 497 Click here for solutions ln ln C ln sin C ln sin s C sin s3 C ln( s ) C 4. C e 3 C tan ln sec e 3 C 47. ln s ln s4 ln a tan C s4 a C C m cosh m m sinh m m 3 cosh m C 3 ln(s 3) ln(s ) C c c c 4 3 C 59. e ln e e C 6. tan ( 4 5 ) C 63. ( s )e s C C s3 ln 69. e ln e C 7. 4 ln 3 4 ln C ln 6 ln 4 8 tan C tan 3 4 cos 6 6 cos 4 8 cos C C sin 3 3 cos 9 cos 3 C 8. e C

8 8 STRATEGY FOR INTEGRATION Solutions: Strateg for Integration. 3. sin +sec tan t (t 3) dt = = 3 µ sin tan + sec = (cos +csc) =sin +ln csc cot + C tan (u +3) u [u = t 3, = dt] = =(ln+6) ( ln 3 + ) = 4 ln3or 4 ln 9 3 π/4 µ u + 6 = ln u 6 u u 3 5. Let u =arctan. Then = d e arctan + + d = e u =[e u ] π/4 π/4 = eπ/4 e π/4. π/ r 4 ln rdr u =lnr, dv = r 4 dr, 3 3 = dr v = = r 5 r5 5 r5 ln r 5 r4 dr = ln 3 5 r5 = 43 5 ln = 43 4 ln µ ( ) + u = ( ) + = u + + [u =, = ] u + = ln u + +tan u + C = ln tan ( ) + C. R sin 3 θ cos 5 θ dθ = R cos 5 θ sin θ sin θ dθ = R cos 5 θ ( cos θ)( sin θ) dθ = R u =cosθ, u 5 ( u ) = sin θ dθ = R (u 7 u 5 ) = 8 u8 6 u6 + C = 8 cos8 θ 6 cos6 θ + C Another solution: R sin 3 θ cos 5 θ dθ = R sin 3 θ (cos θ) cos θ dθ = R sin 3 θ ( sin θ) cos θ dθ = R u =sinθ, u 3 ( u ) = R u 3 ( u + u 4 ) =cosθ dθ = R (u 3 u 5 + u 7 ) = 4 u4 3 u6 + 8 u8 + C = 4 sin4 θ 3 sin6 θ + 8 sin8 θ + C 3. Let =sinθ, where π θ π.then =cosθ dθ and ( ) / =cosθ,so cos θ dθ ( ) = 3/ (cos θ) = sec θ dθ 3 =tanθ + C = + C 5. Let u = =.Then / 7. R sin = " u =, = 3/4 u = 3/4 u / = hu /i 3/4 = u = 3 3/4 # dv =sin, v = R sin = R ( cos ) = sin cos = sin cos R sin cos = sin cos sin + C = 4 sin cos + 4 sin + C Note: R sin cos = R sds= s + C [where s =sin, ds =cos]. A slightl different method is to write R sin = R ( cos ) = R R cos.ifwe evaluate the second integral b parts, we arrive at the equivalent answer 4 4 sin cos + C. 8

9 STRATEGY FOR INTEGRATION 9 9. Let u = e. Then R e +e = R e e e = R e u = e u + C = e e + C.. Integrate b parts three times, first with u = t 3, dv = e t dt: R t 3 e t dt = t3 e t R R + 3t e t dt = t3 e t 3 4 t e t + 3te t dt = e t t3 + 3 t te t = 8 e t 4t 3 +6t +6t +3 + C 3. Let u =+.Then =(u ), =(u ) 5. e t dt = e t t t t C R ( + ) 8 = R u8 (u ) = R u 9 u 8 = 5 u u9 9 = = = ( 4)( +) =3+ A 4 + B + 6 +=A( +)+B( 4). Setting =4gives 46 = 6A, soa = 3. Setting = gives = 6B, sob = 5.Now µ 8 = 3+ 3/3 4 5/3 = ln ln + + C. 7. Let u =ln(sin). Then =cot R cot ln(sin ) = R u= u + C = [ln(sin )] + C w w + dw = 5 µ 3 7 h i 5 dw = 3w 7ln w + w + =5 7ln7+7ln=5+7(ln ln 7) = ln 7 3. As in Eample 5, r = + = = + + =sin p + C Another method: Substitute u = p ( + )/( ) = ( + +)+4=4 ( +).Let +=sinθ,where π θ π.then =cosθ dθ and R 3 = R p 4 ( +) = R p 4 4sin θ cosθ dθ =4 R cos θ dθ = R ( + cos θ) dθ =θ +sinθ + C =θ +sinθ cos θ + C µ + =sin C µ + =sin C 35. Because f() = 8 sin is the proct of an even function and an odd function, it is odd. Therefore, R 8 sin = [b (5.5.7)(b)].

10 STRATEGY FOR INTEGRATION 37. R π/4 cos θ tan θ dθ = R π/4 sin θ dθ = R π/4 = π 8 4 ( ) = π Let u =. Then = + = u + u = = ( cos θ) dθ = θ 4 sin θ π/4 vdv v + v [v = u, u = v, =vdv] dv ³p v + = ln v + + C = ln + + C 4. Let u = θ, dv =tan θ dθ = sec θ dθ = dθ and v =tanθ θ. So R θ tan θ dθ = θ(tan θ θ) R (tan θ θ) dθ = θ tan θ θ ln sec θ + θ + C = θ tan θ θ ln sec θ + C 43. Let u =+e,sothat = e. Then R e +e = R u / = 3 u3/ + C = 3 ( + e ) 3/ + C. Or: Let u = +e,sothatu =+e and u = e. Then R e +e = R u u = R u = 3 u3 + C = 3 ( + e ) 3/ + C. 45. Let t = 3. Thendt =3 I = R R 5 e 3 = 3 te t dt. Now integrate b parts with u = t, R dv = e t dt: I = 3 te t + 3 e t dt = 3 te t 3 e t + C = 3 e C. + a a = + a + a + a = ln + a + a a tan ³ + C a =ln + a +tan (/a)+c 49. Let u = 4 + u =4 + u=4 = u.so 4 + = u 4 (u ) u = =ln C 4 ++ u = u ln u + + C [b Formula 9] 5. Let =tanθ = tan θ, = sec θ dθ, 4 +=secθ,so 4 + = sec θ dθ = tan θ sec θ sec θ tan θ dθ = csc θ dθ = ln csc θ +cotθ + C [or ln csc θ cot θ + C] = ln or + C ln 4 + C sinh(m) = m cosh(m) " # u =, dv = sinh(m), cosh(m) m = v = cosh(m) m = m cosh(m) µ m m sinh(m) " # R U =, dv =cosh(m), sinh(m) m du = V = m sinh(m) = m cosh(m) m sinh(m)+ m 3 cosh(m)+c

11 STRATEGY FOR INTEGRATION 55. Let u = +.Then = u = 57. Let u = 3 + c. Then = u 3 c u u +3+4u = u u +3 =3ln u +3 ln u + + C =3ln ++3 ln ++ + C R 3 + c= R u 3 c u 3u =3 R u 6 cu 3 = 3 7 u7 3 4 cu4 + C 59. Let u = e.then =lnu, = /u e 3 e = = 3 7 ( + c)7/3 3 4 c( + c)4/3 + C /u u 3 u = / u u / u + (u )u (u +) = = u + ln u u + + C = e + ln e e + + C 6. Let u = 5.Then = = 5 u +6 = tan u + C = tan 5 + C Let = so that d = = d =d.then e = e (d)= e d = e 4e d U =4, du =4d u =, =4d dv = e d, V = e dv = e d, v = e = e 4e R 4e d = e 4e +4e + C =( +)e + C =( +)e + C = = 3 µ ++ h( +) 3/ 3/ i + C + = Let u = t.then = dt/ t 3 arctan t t dt = 3 tan u ( ) = u tan u ln( + u ) 3 = 3tan 3 ln 4 tan ln = 3 π 3 ln π 4 ln = Let u = e.then =lnu, = /u e +e = u +u u = 3 π π ln µ u +u = +u = u ln +u + C = e ln( + e )+C [Eample5inSection7.]

12 STRATEGY FOR INTEGRATION = ( +3)( +) = A + B +3 + C + D + =(A + B) + +(C + D) +3 = A 3 + B + A + B + C 3 + D +3C +3D =(A + C) 3 +(B + D) +(A +3C) +(B +3D) 73. A + C =, B + D =, A +3C =, B +3D = A =, C =, B =, D =.Thus, µ = +3 + ( )( +4) = A + B + C +4 + = 4 ln ln + + C or µ 4 ln + + C +3 =A +4 +(B + C)( ) = (A + B) +(C B) +(4A C). So=A + B = C B, =4A C. Setting =gives A = 8 B = 8 and C = 4.So µ ( )( +4) = = +4 8 = 8 6 ln 6 ln +4 8 tan (/) + C R sin sin sin 3 = R sin [cos( 3) cos( +3)] = R (sin cos sin cos 5) R R = 4 sin [sin( +5)+sin( 5)] = 8 cos R 4 (sin 6 sin 4) = 8 cos + 4 cos 6 6 cos 4 + C 77. Let u = 3/ so that u = 3 and = 3 / =. Then 3 + = 3 3 +u = 3 tan u + C = 3 tan ( 3/ )+C. 79. Let u =, dv =sin cos =, v = 3 sin3.then R sin cos = 3 sin3 R 3 sin3 = 3 R sin3 3 ( cos )sin = 3 sin3 + =cos, ( ) d 3 d = sin = 3 sin C = 3 sin3 + 3 cos 9 cos3 + C 8. The function =e does have an elementar antiderivative, so we ll use this fact to help evaluate the integral. R ( +)e = R e + R e = R (e ) + R e = e R e + R " # u =, dv =e, e = e + C = v = e