Strategy for Integration

Home , Continuous function, Elementary function

As we have seen, integration is more challenging than differentiation. In ﬁnding the deriv- ative of a function it is obvious which differentiation formula we should apply. But it may not be obvious which technique we should use to integrate a given function. Until now individual techniques have been applied in each section. For instance, we usually used substitution in Exercises 5.5, integration by parts in Exercises 5.6, and partial fractions in Exercises 5.7 and Appendix 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 strategy that you may ﬁnd useful. A prerequisite for strategy 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 they are easily derived. Formula 19 can be avoided by using partial fractions, and trigonometric substitutions can be used in place of Formula 20.

Table of Integration Formulas Constants of integration have been omitted.

x n1 1 1.y x n dx n 1 2. y dx ln x n 1 x a x 3.y e x dx e x 4. y a x dx ln a

5.y sin x dx cos x 6. y cos x dx sin x

7.y sec2x dx tan x 8. y csc2x dx cot x

9.y sec x tan x dx sec x 10. y csc x cot x dx csc x

11.y sec x dx ln sec x tan x 12. y csc x dx ln csc x cot x

13.y tan x dx ln sec x 14. y cot x dx ln sin x

15.y sinh x dx cosh x 16. y cosh x dx sinh x

dx 1 x dx x 17.y tan 1 18. y sin1 x 2 a 2 a a sa 2 x 2 a

dx 1 x a dx *19.y ln *20. y ln x sx 2 a 2 x 2 a 2 2a x a sx 2 a 2

1 2 ■ STRATEGY FOR INTEGRATION

Once you are armed with these basic integration formulas, if you don’t immediately see how to attack a given integral, you might try the following four-step strategy.

1. Simplify the Integrand if Possible Sometimes the use of algebraic manipulation or trigonometric identities will simplify the integrand and make the method of integration obvious. Here are some examples:

y sx (1 sx) dx y (sx x) dx

tan sin y d y cos2 d sec2 cos 1 y sin cos d 2 y sin 2 d

y sin x cos x2 dx y sin2x 2 sin x cos x cos2x dx

y 1 2 sin x cos x dx

2. Look for an Obvious Substitution Try to ﬁnd some function u tx in the integrand whose differential du tx dx also occurs, apart from a constant factor. For instance, in the integral x y dx x 2 1

we notice that if u x 2 1 , then du 2x dx . Therefore, we use the substitution u x 2 1 instead of the method of partial fractions.

3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led to the solution, then we take a look at the form of the integrand f x . (a) Trigonometric functions. If f x is a product of powers of sin x and cos x , of tan x and sec x , or of cot x and csc x , 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 procedure involving partial fractions in Section 5.7 and Appendix G. (c) Integration by parts. If f x is a product of a power of x (or a polynomial) and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing u and dv according to the advice given in Section 5.6. If you look at the functions in Exer- cises 5.6, you will see that most of them are the type just described. (d) Radicals. Particular kinds of substitutions are recommended when certain radicals appear. (i) If sx 2 a 2 occurs, we use a trigonometric substitution according to the table in Additional Topics: Trigonometric Substitution. (ii) If sn ax b occurs, we use the rationalizing substitution u sn ax b. More generally, this sometimes works for sn tx .

4. Try Again If the ﬁrst three steps have not produced the answer, remember that there are basically only two methods of integration: substitution and parts. (a) Try substitution. Even if no substitution is obvious (Step 2), some inspiration or ingenuity (or even desperation) may suggest an appropriate substitution. (b) Try parts. Although integration by parts is used most of the time on products 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 tan1x ,sin1x , and ln x , and these are all inverse functions. (c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the denominator or using trigonometric identities) may be useful in transforming STRATEGY FOR INTEGRATION ■ 3

the integral into an easier form. These manipulations may be more substantial than in Step 1 and may involve some ingenuity. Here is an example:

dx 1 1 cos x 1 cos x y y dx y dx 1 cos x 1 cos x 1 cos x 1 cos2x

1 cos x cos x y dx y csc2x dx sin2x sin2x

(d) Relate the problem to previous problems. When you have built up some expe- rience in integration, you may be able to use a method on a given integral that is similar to a method you have already used on a previous integral. Or you may even be able to express the given integral in terms of a previous one. For instance, x tan2x sec x dx is a challenging integral, but if we make use of the identity tan2x sec2x 1, we can write

y tan2x sec x dx y sec3x dx y sec x dx

and if x sec3x dx has previously been evaluated (see Example 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 types, or it might combine integration by parts with one or more substitutions. In the following examples we indicate a method of attack but do not fully work out the integral.

tan3x EXAMPLE 1 y dx cos3x In Step 1 we rewrite the integral:

tan3x y dx y tan3x sec3x dx cos3x

The integral is now of the form x tanmx secnx dx with m odd, so we can use the advice in Additional Topics: Trigonometric Integrals. Alternatively, if in Step 1 we had written

tan3x sin3x 1 sin3x y dx y dx y dx cos3x cos3x cos3x cos6x then we could have continued as follows with the substitution u cos x :

sin3x 1 cos2x 1 u 2 y dx y sin x dx y du cos6x cos6x u 6

u 2 1 y du y u 4 u 6 du u 6

s EXAMPLE 2 y e x dx

According to Step 3(d)(ii) we substitute u sx . Then x u 2 , so dx 2u du and

y e sx dx 2 y ue u du 4 ■ STRATEGY FOR INTEGRATION

The integrand is now a product of u and the transcendental function e u so it can be inte- grated by parts.

x 5 1 EXAMPLE 3 y dx x 3 3x 2 10x No algebraic simpliﬁcation or substitution is obvious, so Steps 1 and 2 don’t apply here. The integrand is a rational function so we apply the procedure of Appendix G, remem- bering that the ﬁrst step is to divide.

dx EXAMPLE 4 y xsln x Here Step 2 is all that is needed. We substitute u ln x because its differential is du dxx,which occurs in the integral.

1 x EXAMPLE 5 y dx 1 x Although the rationalizing substitution

1 x u 1 x

works here [Step 3(d)(ii)], it leads to a very complicated rational function. An easier method is to do some algebraic manipulation [either as Step 1 or as Step 4(c)]. Multiply- ing numerator and denominator by s1 x , we have

1 x 1 x y dx y dx 1 x s1 x 2 1 x y dx y dx s1 x 2 s1 x 2 sin1x s1 x 2 C

Can We Integrate All Continuous Functions?

The question arises: Will our strategy for integration enable us to ﬁnd the integral of every 2 continuous function? For example, can we use it to evaluate x e x dx ? 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 elementary functions. These are the polynomials, rational functions, power functions x a , exponential functions a x , logarithmic functions, trigonometric and inverse trigonometric functions, hyperbolic and inverse hyperbolic functions, and all functions that can be obtained from these by the ﬁve operations of addition, subtraction, multiplication, division, and compo- sition. For instance, the function

x 2 1 f x lncosh x xe sin 2x x 3 2x 1

is an elementary function. If f is an elementary function, then f is an elementary function but x f x dx need not 2 be an elementary function. Consider f x e x . Since f is continuous, its integral exists, and if we deﬁne the function F by

x 2 Fx y e t dt 0 STRATEGY FOR INTEGRATION ■ 5

then we know from Part 1 of the Fundamental Theorem of Calculus that

2 Fx e x

2 Thus,f x e x has an antiderivative F , but it has been proved that F is not an elementary function. This means that no matter how hard we try, we will never succeed in evalu- 2 ating x e x dx in terms of the functions we know. (In Chapter 8, however, we will see how 2 to express x e x dx as an inﬁnite series.) The same can be said of the following integrals: e x y dx y sinx 2 dx y cose x dx x

1 sin x y sx 3 1 dx y dx y dx ln x x

In fact, the majority of elementary functions don’t have elementary antiderivatives. You may be assured, though, that the integrals in the following exercises are all elementary functions.

Exercises

27. y cot x lnsin x dx 28. y sin sat dt A Click here for answers. S Click here for solutions.

1–80 Evaluate the integral. 5 3w 1 2 29. y dw 30. y x 2 4x dx sin x sec x 0 w 2 2 1. y dx 2. y tan3 d tan x 1 x s2x 1 31. y dx 32. y dx 2 2t x 1 x 2x 3 3. y 2 dt 4. y dx 0 t 3 s3 x 4

arctan y 2 1 4 cot x 1 e 33. y s3 2x x 2 dx 34. y dx 5. y 2 dy 6. y x csc x cot x dx 4 4 cot x 1 1 y

1 3 4 4 x 1 8 7. y r ln r dr 8. y 2 dx 35. y x sin x dx 36. y sin 4x cos 3x dx 1 0 x 4x 5 1 x 1 x 9. y dx 10. y dx 4 2 2 4 5 3 2 4 2 37. y cos tan d 38. y tan sec d x 4x 5 x x 1 0 0

11. y sin3 cos5 d 12. y sin x coscos x dx x 1 39. y dx 40. y dy 1 x 2 s1 x 2 s4y 2 4y 3 dx s1 ln x 13. y 2 32 14. y dx 1 x x ln x 41. y tan2 d 42. y x 2 tan 1x dx 2 12 x s22 x 15. y dx 16. y dx 0 s1 x 2 0 s1 x 2 43. y e xs1 e x dx 44. y s1 e x dx e2t 17. y x sin2x dx 18. y dt 4t 1 e x 3 1 e 45. y x 5e x dx 46. y dx x 3 x 19. y e xe dx 20. y esx dx 1 e x a x 47. y dx 48. y dx 21. y t 3e2t dt 22. y x sin1x dx x 2 a 2 x 4 a 4

1 1 1 8 2 23. y (1 sx) dx 24. y lnx 1 dx 49. y dx 50. y 2 dx 0 xs4x 1 x s4x 1

3x 2 2 3x 2 2 1 dx 25. y dx 26. y dx 51. y dx 52. y x 2 2x 8 x 3 2x 8 xs4x 2 1 xx 4 1 6 ■ STRATEGY FOR INTEGRATION

2x 2 2 e lnx 1 53. y x sinh mx dx 54. y x sin x dx 69. y dx 70. y dx 1 e x x 2 1 x ln x s 55. y dx 56. y dx x t s s 2 71. y dx 72. y dt x 4 4 x 1 x 1 x 4 4x 2 3 1 s3 t

s3 2 1 dx 57. y x x c dx 58. y x ln 1 x dx 73. y dx 74. y x 2x 2 4 e x e x 1 1 59. y dx 60. y dx 2 e 3x e x x s3 x 75. y sin x sin 2x sin 3x dx 76. y x bx sin 2x dx x 4 x 3 s 61. y dx 62. y dx x sec x cos 2x 10 10 77. y dx 78. y dx x 16 x 1 1 x 3 sin x sec x sx 3 ln tan x 63. y sxe dx 64. y dx 2 sin x cos x 79. y x sin x cos x dx 80. y dx 4 sin x cos x sin4 x cos4 x 3 1 3 u 1 ■■■■■■■■■■■■■ 65. y dx 66. y 3 2 du sx 1 sx 2 u u 2 2 81. The functions y e x and y x 2e x don’t have elementary 2 x 2 3 arctanst 1 antiderivatives, but y 2x 1e does. Evaluate 2 67. y dt 68. y x x dx 2 x 1 st 1 2e e x 2x 1e dx. STRATEGY FOR INTEGRATION ■ 7

Answers

3 45. 1 x 3 1ex C S Click here for solutions. 3 47. ln sx 2 a 2 tan1xa C 1. sin x ln csc x cot x C s4x 1 1 s4x 2 1 1 49. ln C 51. ln C 3. 4 ln 9 5. e4 e4 s4x 1 1 2x 243 242 1 2 1 2 2 3 7. 5 ln 3 25 9. 2 ln x 4x 5 tan x 2 C 53. 1 m x cosh mx 2 m x sinh mx 2 m cosh mx C 1 8 1 6 1 4 1 6 1 8 s s 11. 8 cos 6 cos C (or 4 sin 3 sin 8 sin C) 55. 3 ln( x 1 3) ln( x 1 1) C 3 73 3 43 s 2 1 s 57. x c cx c C 13. x 1 x C 15. 1 2 3 7 4 1 2 1 1 2 x 1 x x 17. 4 x 2 x sin x cos x 4 sin x C 59. e 2 ln e 1 e 1 C 1 1 1 or x 2 x sin 2x cos 2x C 1 1 1 5 s sx ( 4 4 8 ) 61. 20 tan (4 x ) C 63. 2(x 2 x 2)e C e x 1 t 19. e C 21. e 2 4t 3 6t 2 6t 3 C 2 32 32 8 65. 3 x 1 x C 4097 23 5 2 1 23.45 25. 3x 3 ln x 4 3 ln x 2 C s x x 67. 3 3 2 ln 2 69. e ln 1 e C 1 2 2 27. 2 ln sin x C 29. 15 7 ln 7 1 2 1 2 71. 4 ln x 3 4 ln x 1 C 31. sin1x s1 x 2 C 73. 1 ln x 2 1 lnx 2 4 1 tan1x2 C 33. 2 sin1x 12 x 12s3 2x x 2 C 8 16 8 1 1 1 2 1 32 1 s 2 75. cos 6x cos 4x cos 2x C 77. tan x C 35. 0 37. 8 4 39. ln(1 1 x ) C 24 16 8 3 1 2 2 x32 1 3 1 1 3 x2 41. tan 2 ln sec C 43. 3 1 e C 79. 3 x sin x 3 cos x 9 cos x C 81. xe C 8 ■ STRATEGY FOR INTEGRATION

Solutions: Strategy for Integration

sin x +secx sin x sec x 1. dx = + dx = (cos x +cscx) dx =sinx +ln csc x cot x + C tan x tan x tan x | − | Z Z µ ¶ Z 2 1 1 1 2t − 2(u +3) − 2 6 6 − 3. 2 dt = 2 du [u = t 3, du = dt] = + 2 du = 2ln u 0 (t 3) 3 u − 3 u u | | − u 3 Z − Z− Z− µ ¶ · ¸− =(2ln1+6) (2 ln 3 + 2) = 4 2ln3or 4 ln 9 − − − 1 arctan y π/4 dy e u u π/4 π/4 π/4 − 5. Let u =arctany.Thendu = 2 2 dy = e du =[e ] π/4 = e e . 1+y ⇒ 1 1+y π/4 − − Z− Z− 4 3 u =lnr, dv = r dr, 3 3 3 4 1 1 4 243 1 5 7. r ln rdr dr 1 = r5 ln r r dr = ln 3 0 r du = v = r5  5 − 5 5 − − 25 Z1 r 5 · ¸1 Z1 · ¸1  = 243 ln 3 243 1 = 243 ln 3 242 5 − 25 − 25 5 − 25 x 1 (x 2) + 1 u ¡ 1 ¢ 9. − dx = − dx = + du [u = x 2, du = dx] x2 4x +5 (x 2)2 +1 u2 +1 u2 +1 − Z − Z − Z µ ¶ 1 2 1 1 2 1 = ln u +1 +tan− u + C = ln x 4x +5 +tan− (x 2) + C 2 2 − − 11. sin3 θ cos5 θ dθ = cos5¡θ sin2 θ¢sin θ dθ = cos5 θ (1¡ cos2 θ)( ¢sin θ) dθ − − − u =cosθ, R = R u5(1 u2) du R − − du = sin θ dθ · − ¸ = (Ru7 u5) du = 1 u8 1 u6 + C = 1 cos8 θ 1 cos6 θ + C − 8 − 6 8 − 6 Another solution: R sin3 θ cos5 θ dθ = sin3 θ (cos2 θ)2 cos θ dθ = sin3 θ (1 sin2 θ)2 cos θ dθ − u =sinθ, R = R u3(1 u2)2 du R = u3(1 2u2 + u4) du − du =cosθ dθ − · ¸ = R (u3 2u5 + u7) du = 1 u4 1 u6 + 1 uR 8 + C = 1 sin4 θ 1 sin6 θ + 1 sin8 θ + C − 4 − 3 8 4 − 3 8 13. Let x =sinθ,whereR π θ π .Thendx =cosθ dθ and − 2 ≤ ≤ 2 (1 x2)1/2 =cosθ,so − dx cos θ dθ = = sec2 θ dθ (1 x2)3/2 (cos θ)3 Z Z Z − x =tanθ + C = + C √1 x2 −

15. Let u =1 x2 du = 2xdx.Then − ⇒ − 1/2 3/4 1 1 x 1 1 1 1/2 1 1/2 1 √3 dx = du = u− du = 2 2u = √u =1 2 √1 x2 − 2 √u 2 3/4 3/4 − Z0 Z1 Z3/4 − h i £ ¤ u = x, 2 2 dv =sin xdx, 17. x sin xdx 2 1 1 1 "du = dx v = sin xdx= 2 (1 cos 2x) dx = 2 x 2 sin x cos x# R − − = 1 x2 1 x sin x cos xR 1 x 1 Rsin x cos x dx 2 − 2 − 2 − 2 1 2 1 1 2 1 2 1 2 1 1 2 = x x sin x cos x R¡x + sin x + C =¢ x x sin x cos x + sin x + C 2 − 2 − 4 4 4 − 2 4 1 2 Note: sin x cos xdx= sds= 2 s + C [where s =sinx, ds =cosxdx]. A slightlyR different methodR is to write x sin2 xdx= x 1 (1 cos 2x) dx = 1 xdx 1 x cos 2xdx.Ifwe · 2 − 2 − 2 evaluate the second integral by parts, weR arrive at the equivalentR answer 1 x2 1 x sinR 2x 1 cosR 2x + C. 4 − 4 − 8 STRATEGY FOR INTEGRATION ■ 9

x x x 19. Let u = ex.Then ex+e dx = ee ex dx = eu du = eu + C = ee + C.

3 2t 21. Integrate by parts threeR times, ﬁrstR with u = t , dvR = e− dt:

3 2t 1 3 2t 1 2 2t 1 3 2t 3 2 2t 1 2t t e− dt = t e− + 3t e− dt = t e− t e− + 3te− dt − 2 2 − 2 − 4 2 R R R 2t 1 3 3 2 3 2t 3 2t 2t 1 3 3 2 3 3 = e− t + t te− + e− dt = e− t + t + t + + C − 2 4 − 4 4 − 2 4 4 8 Z £ ¤ £ ¤ 1 2t 3 2 = e− 4t +6t +6t +3 + C − 8 23. Let u =1+√x.Thenx =(¡u 1)2, dx =2(u ¢ 1) du − − ⇒ 2 1 (1 + √x )8 dx = 2 u8 2(u 1) du =2 2 u9 u8 du = 1 u10 2 1 u9 0 1 · − 1 − 5 − · 9 1 R = R1024 1024 1 + 2 = R4097¡ ¢ £ ¤ 5 − 9 − 5 9 45 3x2 2 6x +22 A B 25. − =3+ =3+ + 6x +22=A(x +2)+B(x 4). Setting x2 2x 8 (x 4)(x +2) x 4 x +2 ⇒ − − − − − x =4gives 46 = 6A,soA = 23 . Setting x = 2 gives 10 = 6B,soB = 5 .Now 3 − − − 3 3x2 2 23/3 5/3 − dx = 3+ dx =3x + 23 ln x 4 5 ln x +2 + C. x2 2x 8 x 4 − x +2 3 | − | − 3 | | Z − − Z µ − ¶ 27. Let u =ln(sinx).Thendu =cotxdx cot x ln(sin x) dx = udu= 1 u2 + C = 1 [ln(sin x)]2 + C. ⇒ 2 2 5 3w 1 5 7 R 5 R 29. − dw = 3 dw = 3w 7ln w +2 0 w +2 0 − w +2 − | | 0 Z Z µ ¶ h i =15 7ln7+7ln2=15+7(ln2 ln 7) = 15 + 7 ln 2 − − 7 31. As in Example 5,

1+x √1+x √1+x 1+x dx xdx dx = dx = dx = + 1 x √1 x · √1+x √1 x2 √1 x2 √1 x2 Z r − Z − Z − Z − Z − 1 =sin− x 1 x2 + C − − Another method: Substitute u = (1p + x)/(1 x). − p 33. 3 2x x2 = (x2 +2x +1)+4=4 (x +1)2.Let − − − − x +1=2sinθ,where π θ π .Thendx =2cosθ dθ and − 2 ≤ ≤ 2 √3 2x x2 dx = 4 (x +1)2 dx = 4 4sin2 θ 2cosθ dθ − − − − 2 R =4R pcos θ dθ =2 (1 +R cosp 2θ) dθ =2θR+sin2θ + C =2R θ +2sinθ cos θ + C 2 1 x +1 x +1 √3 2x x =2sin− +2 − − + C 2 · 2 · 2 µ ¶ 1 x +1 x +1 2 =2sin− + √3 2x x + C 2 2 − − µ ¶ 35. Because f(x)=x8 sin x is the product of an even function and an odd function, it is odd. Therefore,

1 8 1 x sin xdx=0 [by (5.5.7)(b)]. − R 10 ■ STRATEGY FOR INTEGRATION

37. π/4 cos2 θ tan2 θ dθ = π/4 sin2 θ dθ = π/4 1 (1 cos 2θ) dθ = 1 θ 1 sin 2θ π/4 0 0 0 2 − 2 − 4 0 R = R π 1 (0 0)R = π 1 £ ¤ 8 − 4 − − 8 − 4 39. Let u =1 x2.Thendu¡ = 2xdx¢ − − ⇒ xdx 1 du vdv = = [v = √u, u = v2, du =2vdv] 1 x2 + √1 x2 − 2 u + √u − v2 + v Z − − Z Z dv = = ln v +1 + C = ln 1 x2 +1 + C − v +1 − | | − − Z ³p ´ 41. Let u = θ, dv =tan2 θ dθ = sec2 θ 1 dθ du = dθ and v =tanθ θ.So − ⇒ − θ tan2 θ dθ = θ(tan¡ θ θ) ¢ (tan θ θ) dθ = θ tan θ θ2 ln sec θ + 1 θ2 + C − − − − − | | 2 R = θ tan θ 1 θ2 Rln sec θ + C − 2 − | | 43. Let u =1+ex,sothatdu = ex dx.Then

x x 1/2 2 3/2 2 x 3/2 e √1+e dx = u du = 3 u + C = 3 (1 + e ) + C. ROr: Let u = √1+eRx,sothatu2 =1+ex and 2udu = ex dx.Then ex√1+ex dx = u 2udu = 2u2du = 2 u3 + C = 2 (1 + ex)3/2 + C. · 3 3 R 3 R 2 R 5 x3 1 t 45. Let t = x .Thendt =3x dx I = x e− dx = te− dt. Now integrate by parts with u = t, ⇒ 3 t 1 t 1 t 1 t 1 t 1 x3 3 dv = e− dt: I = te− + e− dt =R te− e− +R C = e− x +1 + C. − 3 3 − 3 − 3 − 3

x + a 1 2xdx R dx 1 2 2 1 ¡ 1 x ¢ 47. dx = + a = ln x + a + a tan− + C x2 + a2 2 x2 + a2 x2 + a2 2 · a a Z Z Z 2 2 1 ¡ ¢ ³ ´ =ln√x + a +tan− (x/a)+C

49. Let u = √4x +1 u2 =4x +1 2udu=4dx dx = 1 udu.So ⇒ ⇒ ⇒ 2 1 1 udu du u 1 dx = 2 =2 =2 1 ln − + C [by Formula 19] x √4x +1 1 (u2 1) u u2 1 2 u +1 Z Z 4 Z − ¯ ¯ − ¡ ¢ ¯ ¯ ¯ ¯ √4x +1 1 ¯ ¯ =ln − + C √4x +1+1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 51. Let 2x =tanθ x = 1 tan θ, dx = 1 sec2 θ dθ, √4x2 +1=secθ,so ⇒ 2 2 dx 1 sec2 θ dθ sec θ = 2 = dθ = csc θ dθ x √4x2 +1 1 tan θ sec θ tan θ Z Z 2 Z Z = ln csc θ +cotθ + C [or ln csc θ cot θ + C] − | | | − | √4x2 +1 1 √4x2 +1 1 = ln + + C or ln + C − 2x 2x 2x − 2x ¯ ¯ · ¯ ¯ ¸ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 ¯ 2 ¯ ¯ u = x2, dv = sinh(mx) dx, 53. x2 sinh(mx)dx = x2 cosh(mx) x cosh(mx) dx m − m du =2xdx v = 1 cosh(mx) Z Z " m # 1 2 1 U = x, dV =cosh(mx) dx, = x2 cosh(mx) 1 x sinh(mx) sinh(mx) dx m − m m − m dU = dx V = 1 sinh(mx) µ ¶ " m # 1 2 2 R = x2 cosh(mx) x sinh(mx)+ cosh(mx)+C m − m2 m3 STRATEGY FOR INTEGRATION ■ 11

55. Let u = √x +1.Thenx = u2 1 − ⇒ dx 2udu 1 3 = = − + du x +4+4√x +1 u2 +3+4u u +1 u +3 Z Z Z · ¸ =3ln u +3 ln u +1 + C =3ln √x +1+3 ln √x +1+1 + C | | − | | − ¡ ¢ ¡ ¢ 57. Let u = √3 x + c.Thenx = u3 c − ⇒ x √3 x + cdx= u3 c u 3u2 du =3 u6 cu3 du = 3 u7 3 cu4 + C − · − 7 − 4 R = R¡3 (x + c)7¢/3 3 c(x + c)4R¡/3 + C ¢ 7 − 4 59. Let u = ex.Thenx =lnu, dx = du/u ⇒ dx du/u du 1/2 1 1/2 = = = du e3x ex u3 u (u 1)u2(u +1) u 1 − u2 − u +1 Z − Z − Z − Z · − ¸ x 1 1 u 1 x 1 e 1 = + ln − + C = e− + ln − + C u 2 u +1 2 ex +1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 61. Let u = x5.Thendu =5x4 dx ¯ ¯ ¯ ¯ ⇒ 4 1 x dx 5 du 1 1 1 1 1 1 1 5 = = tan− u + C = tan− x + C. x10 +16 u2 +16 5 · 4 4 20 4 Z Z ¡ ¢ ¡ ¢ 1 63. Let y = √x so that dy = dx dx =2√xdy =2ydy.Then 2 √x ⇒ u =2y2, dv = ey dy, √xe√x dx = yey(2ydy)= 2y2ey dy du =4ydy v = ey Z Z Z · ¸ U =4y, dV = ey dy, =2y2ey 4yey dy − dU =4dy V = ey Z · ¸ =2y2ey 4yey 4ey dy =2y2ey 4yey +4ey + C − − − =2(y2 2y¡ +2)ey R+ C =2(¢ x 2 √x +2)e√x + C − −

dx 1 √x +1 √x 65. = − dx = √x +1 √x dx √x +1+√x √x +1+√x · √x +1 √x − Z Z µ ¶ Z − ¡ ¢ = 2 (x +1)3/2 x3/2 + C 3 − h i 67. Let u = √t.Thendu = dt/ 2√t ⇒ 3 √3 arctan √t ¡ 1 ¢ 1 1 2 √3 dt = tan− u (2 du)=2 u tan− u ln(1 + u ) [Example5inSection7.1] √t − 2 1 Z1 Z1 1 1 £ 1 1 ¤ =2 √3tan− √3 ln 4 tan− 1 ln 2 − 2 − − 2 =2£¡√3 π ln 2 π ¢ 1 ln¡ 2 = 2 √3 π ¢¤1 π ln 2 · 3 − − 4 − 2 3 − 2 − 69. Let u = ex.Thenx =ln£¡ u, dx = du/u¢ ¡ ¢¤ ⇒ e2x u2 du u 1 dx = = du = 1 du 1+ex 1+u u 1+u − 1+u Z Z Z Z µ ¶ = u ln 1+u + C = ex ln(1 + ex)+C − | | − 12 ■ STRATEGY FOR INTEGRATION

x x Ax + B Cx + D 71. = = + x4 +4x2 +3 (x2 +3)(x2 +1) x2 +3 x2 +1 ⇒

x =(Ax + B) x2 +1 +(Cx + D) x2 +3 = Ax3 + Bx2 + Ax + B + Cx3 + Dx2 +3Cx +3D

=(A + C)x3¡ +(B +¢ D)x2 +(A +3¡ C)x¢+(B¡ +3D) ¢ ¡ ¢ ⇒ A + C =0, B + D =0, A +3C =1, B +3D =0 A = 1 , C = 1 , B =0, D =0.Thus, ⇒ − 2 2 x 1 x 1 x dx = − 2 + 2 dx x4 +4x2 +3 x2 +3 x2 +1 Z Z µ ¶ 1 x2 +1 = 1 ln x2 +3 + 1 ln x2 +1 + C or ln + C − 4 4 4 x2 +3 µ ¶ ¡ ¢ ¡ ¢ 1 A Bx + C 73. = + (x 2)(x2 +4) x 2 x2 +4 ⇒ − − 1=A x2 +4 +(Bx + C)(x 2) = (A + B)x2 +(C 2B)x +(4A 2C).So0=A + B = C 2B, − − − − 1=4A¡ 2C.Setting¢ x =2gives A = 1 B = 1 and C = 1 .So − 8 ⇒ − 8 − 4 1 1 1 x 1 1 dx 1 2xdx 1 dx dx = 8 + − 8 − 4 dx = (x 2)(x2 +4) x 2 x2 +4 8 x 2 − 16 x2 +4− 4 x2 +4 Z − Z µ − ¶ Z − Z Z 1 1 2 1 1 = ln x 2 ln x +4 tan− (x/2) + C 8 | − | − 16 − 8 75. sin x sin 2x sin 3xdx = sin x 1 [cos(2x 3¡x) cos(2¢ x +3x)] dx = 1 (sin x cos x sin x cos 5x) dx · 2 − − 2 − 1 1 1 R = R sin 2xdx [sin(x +5x)+sin(x 5x)]Rdx 4 − 2 2 − 1 1 1 1 1 = R cos 2x (sinR 6x sin 4x) dx = cos 2x + cos 6x cos 4x + C − 8 − 4 − − 8 24 − 16 3/2 2 3 R1/2 77. Let u = x so that u = x and du = 3 x dx √xdx= 2 du.Then 2 ⇒ 3 2 √x 3 2 1 2 1 3/2 dx = du = tan− u + C = tan− (x )+C. 1+x3 1+u2 3 3 Z Z

79. Let u = x, dv =sin2 x cos xdx du = dx, v = 1 sin3 x.Then ⇒ 3 x sin2 x cos xdx = 1 x sin3 x 1 sin3 xdx= 1 x sin3 x 1 (1 cos2 x)sinxdx 3 − 3 3 − 3 − 1 1 y =cosx, R = x sin3 x +R (1 y2) dy R 3 3 − dy = sin xdx Z · − ¸ = 1 x sin3 x + 1 y 1 y3 + C = 1 x sin3 x + 1 cos x 1 cos3 x + C 3 3 − 9 3 3 − 9 2 81. The function y =2xex does have an elementary antiderivative, so we’ll use this fact to help evaluate the integral.

2 2 2 2 2 (2x2 +1)ex dx = 2x2ex dx + ex dx = x(2xex ) dx + ex dx

x2 R R x2 x2 R x2 R u = x, dvR=2xe dx, x2 = xe e dx + e dx 2 = xe + C − "du = dx v = ex # R R