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SupIntegrationTech8-0

# SupIntegrationTech8-0 - incnnmnns on A inrnnnnrron OVERVIEW...

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Unformatted text preview: incnnmnns on ' A * inrnnnnrron OVERVIEW The Fundamental Theorem connects antiderivatives and the definite integral. Evaluating the indefinite integral / re) dx I is equivalent to finding a function P such that F’(x) = ﬂat), and then adding an arbitrary constant C: / fem : in) + c. In this chapter we study a number of important techniques for finding indefinite integrals of more complicated functions than those seen before. The goal of this chapter is to show how to change unfamiliar integrals into integrals we can recognize, find in a table, or evaluate with a computer. We also extend the idea of the definite integral to improper integrals for which the integrand may be unbounded over the interval of inte— gration, or the interval itself may no longer be finite. haste integration Feminine _ I To help us in the search for finding indefinite integrals, it is useﬁil to build up a table of integral formulas by inverting formulas for-derivatives, as we have done in previous chap- ters. Then we try to match any integral that confronts us against one of the standard types. This usually involves a certain amount of algebraic manipulation as well as use of the Sub— stitution Rule. Recall the Substitution Rule from Section 5.5: f figengemx: f may where u = g(x) is a differentiable ﬁanction whose range is an interval I and f is continuous on I. Success in integration often hinges on the ability to spot what part of the integrand should be called a in order that one will also have du, so that a known formula can be applied. This means that the first requirement for skill in integration is a thorough mastery of the formulas for differentiation. 553 554 Chapter 8: Techniques of Integration Table 8.1 shows the basic forms of integrals we have evaluated so far. 111 this section we present several algebraic or substitution methods to help us use this table. There is a more extensive table at the back of the book; we discuss its use in Sectidn 8.6. YABLE 8.1 Basic integration formulas du=u+C cotudu—ln|sinu|f+C 1,, —]n + kdu = ku + C (anynumber k) lcscul C (du+dv)—/du+/dv un+1 undu:m+c (niil) +C a u = adu km (a > (La \\\ sinhudu : coshu + C coshudu = sinhu + C cosudu = Sinu + C seczudu tanu + C est:2 3; du —cot u + .C secutanudu secu + C cscucotudu — —cscu + C \\\\\\\ tanua'u *ln |cosu| + C / f f f / f f f in Isecul + C We often have to rewrite an integral to match it to a standard formula. EXAMPLE 1 Making a Simplifying Substitution Evaluate 2.x—9 Vx2—9x+l 8.1 Basic Integration Formulas 555 Solution 2x — 9 dx K fciu r , x2 _ u : 7 9x + 1, Vx2_9x+1 V1: 'duﬁ(2x79)dx. # fat—“20’s; : tam/2)“ = C Table 8.1Formula4, (ml/2) .3. 1 withn= —1/2 : 2111/25 + C "':'2m+c EXAMPLE 2 Completing the Square Evaiuate f dx V 31: — x2 Solution We complete the square to simplify the denominator: 8x—x2=e(x2*8x)=e(x2*8x+16716) e(_x2*8x+16)+16=167(xi4)2. IE Then = a=4,§12={x'*4)= -\ 1‘12 _ “2 du 5 d): i . 7] u — sm ('5) + C Table 8.1,Formu1a 18 sin‘1 (x14) + C. EXAMPLE 3 Expanding a Power and Using a Trigonometric Identity Evaluate [(secx + tanx)2dx. Solution We expand the integrand and get (seex + tanx)2 = seczx + ZSecxtanx + tanzx. The first two terms on the right-hand side of this equation are familiar; we can integrate them at once. How about tan2 x? There is an identity that connects it with see2 x: 2 tanzx -l-1z see x, 2 tan2x= see 36- I. 556 Chapter 8: Techniques of Integration x r 3 3x + 2l3x2 — 7x 3x2 + 2x —9x 7 9x -6 +6 We replace tanzx by see2 x i 1 and get I /(seex + tanx)2dx : f (seezx + 2secxtanx + seczx *.1)cix 2 secéxailr+2 ,secxtanxdx—jhdx 2tanx + ZSecx e x JF'C. EXAMPLE 4 Eliminating a Square Root .,, Evaluate 7:14 I f V 1 + cos 4x‘dx. 0 , Solution We use the identity 1 + cos 26 00529: 2 , 1+00526=2c0326. With 6 = 2x, this identity becomes 1+ cos4x = 2cos22x. Hence, 1114 .7r/4 V1 + cos4xdx—f WVcoszzxdx 0 ‘ 0 17,14 7 i |c052xldx will” 0 On [0, C0521 2 0, so Eeos2x| : cost. 17/4 1 W] (:05 2x dx 0 - W/ 4 Table 8.1, Formula 7, with W Ismzx u:2xanddu=2dx 2 0 =\/5%—o xvi 7' EXAMPLE 5 Reducing an Improper Fraction Evaluate 3x2 7— 7x 3x+2 056' Solution The integrand is an improper fraction (degree of numerator greater than or- equal to degree of denominator). To integrate it, we divide ﬁrst, getting a quotient plus a- remainder that is a proper fraction: 3x2—7x_ __ _ 6 3x+2 —x 3+3x+2' Therefore, 3x2—7x _ A 6 _2Cf_ 3x+2 (ix) x 3+ dx#2 3x+2ln|3x+2l+C. 8.1 Basic Integration Formulas 557 Reducing an improper fraction by long division (Example 5) does not always lead to an expression we can integrate directly. We see what to do about that in Section 8.5. EXAMPLE 6 Separating a Fraction Evaluate 3x + 2 ‘ dx V 1 — 3:2 Solution 'We first separate the integrand to get 3x + 2 dx xax + 2f ax m V1 — x2 « V1 — x2 _ In the first of these new integrals, we substitute ‘ 33 u = l — 12 du — Alxdx, and' a 7 ’12d 3 xdx :3/W__2/u_y2du V1“);2 V1: 2 =—3Vl—x2+C1 The second of the new integrals is a standard form, dx - 71 2/w=2s1n x+C2. W Combining these results and renaming C1 + C2 as C gives 3x+2 he‘dx=i3V1—x2+251nlx+C. a VlmJCZ 1), The final example of this section calculates an important integral by the algebraic technique of multiplying the integrand by a form of 1 to change the integrand into one we can integrate. .1, HISTORICAL BIOGRAPHY EXAMPLE 7 Integral of y = sew—Multiplying by a Form of 1 fs'ecxdx. George David Birkhoff Evaluate (1884—1944) [secxdx—f(secx)(1)dx=fsecx-de secx + tanx : seczx + secxtanx ' secx + tanx _ ﬂ :1 = tan): + secx, u du = (seclx + secxtanx)dx :ln|u| +C:1n]secx+tanxj +C. 558 Chapter 8: Techniques of Integration With cosecants and cotangents in place of secants and tangents, the method of Exam- ple 7 leads to a companion formula for the integral of the cosecant (see Exercise 95). TABLE 8.2 The secant and cosecant integrals 1. secudu lnisecu + C 2. cscudu— #]n|cscu+cotul_,+ C .,, '— Procedures for Matching integrals to Basic Formulas PROCEDURE EXAMPLE . . . . 2.x — 9 du Makmg a snnphfylng dx = f sz — 9x + l W substitution Completing the square V 8x * x2 = V_ 16 i (x — 4)2 Using a trigonometric (secx + tan x)2 = sec2 x + 2 secx tanx' + tan2 3: “Entity 3 sec2 x + 2 sec )5 tan x ‘+ (seczx m 1) : 23602)!“ + 2secxtanx fl Eliminating a square root \/l 4- cos 4x : \/2 cos2 2x : \/2 loos 2x1 . . 3x2 # 7x i i 6 Reducmgannnproper 3x + 2 — x 3 + #3): + 2 fraction ,,l . . 3x + 2 3x 2 Separating a fractmn # + \/1—x2 \/1—x2 \/1—;~:2 . .7 _ secx + tanx Multlplylng byaform ofl secx — seex secx + tanx seczx + sertanx secx + tanx Basic Substitutions Evaluate each integral in Exercises 1736 by using a substitution to re- d/«é~ 3 V sinv cos vdv duce it to standard form. /' 4. cot3 y csc2 y dy 16xdx 3005:de 1 léxdx 1T[\$56022 1. ' 2. 5./_ 6. dz O0 8x2+1 V1 +3sinx 8x2+2 .7/4 153112 8.1 Basic Integration Formulas 559 @ f dx 8 / dx trigeaometrﬁe identities , VHW + 1) x — V9; Evaluate each integral in Exercises 4346 by- using trigonometric ’ identities and substitutions to reduce it to standard form. 9. feet(3 * 7x)a.’x 10. fosc (77x k Unix ‘ _, .. 43. /(secx + cot’gtfdx .. 44. /(cscx * tanx)2dx c t 3 + 1 I ’ ‘ " . '" 11. [2005c (e6 + Daft? 12. f O ( x [Uddx , f 45. fcscx sin 3xdx secédt IQ fxsec (Jr2 — 5) dx " “in s ' _ 46. f (sin 33: cos 2.x 1 cos 3x sin 21:) (ix-~- " " " I 1 v_ g 15. — d 16. ‘ *dﬂ . / ems 7T) S f 92 m 0 Improper Freestone __ \fhﬁ ' W Evaluate each integral in Exercises 47—52 by reducing the improper L9 / 2x 9x2 dx I (sin y)e “my dy fraction and using a substitution (if necessary) to reduce it to standard 0 "/2 form. Wdt . 9. j‘em‘kec2 vdv 20. f e x x2 5 v; 47.. x+1dx 4s. x2+ldx . 2h” 3 3 3 2 Lf3x+1dx 22. x dx 2x [ 4x —7 49. ab: . dx @ \/2 x2 — l G}, . —1 2x + 3 2de f 26 4 3 _ 2 3 _ 2 23_ 24_ 10 d3 1‘ I + 162‘ 29 79 + 70 O 2%; 51. / W? + 4 d: 52. “23 _ 5 d6 ,5 A 9du 26. / 4d): . v . d f I + M i I + (2x + I}; Separating Fractions f l 6 1 Evaluate each integral in Exercises 53—56 by separating the fraction f / £137 28 d? and using a substitution (if necessary) to reduce it to standard form. Kx20 Vi—ax2 owes . . “‘ 2mg 2a: @[———“x dx 5 jxmﬂ+2mﬂaxw 2. ‘— wa ' Vlﬁxz ZxVx—l U Vl—s4 le—4lnzx W4 1 _ 1/2 1 '4- srnx 2 — 8x 6 dx dr 55. —T—dx 56. 2 dx 31_ 31 ____ 0 cos x 0 l + 4x xV25x2‘l rVr2—9 dx dy Multiplying by a Farm at 1 . 33' 31 + 91 34' f m Evaluate each integral in Exercises 57432 by multiplying by a form of m3 1 and using a substitution (if necessary) to reduce it to standard form. e dx lnxdx 35' xcos (lnx) 36' + 4 ln2 _ _1 ﬂ ‘1 1 x x x . _ 57"/14— sinxdx 58’ _/1 + cosxdx ﬁempletiug the Square 1 1 59. . d6 60. we"? Evaluate each integral in Exercises 37—42 by completing the square sec 6 + m 3 05° 9 + cm 6 and using a substitution to reduce it to standard form. 1 1 61' f1 — 58(3de 61/1 — csc dx @ [ZTMHZ 38' f42‘25xm x ' 1 x _ 2" + 2 x x Eliminating Square Reots "'“ d1 619 Evaluate each integral in Exercises 63—70 by eliminating the square Cy 40. V—:2 + 4: — 3 V26 e 62 ' root- 211' _ 11' 41.] dx ' 42. f “5‘ 61/ .fl—gﬂ'idx 64. / V1 .— costdx (x + 1)\/x2 + 2x (x — 2)\/;;2 e 4x + 3 o o 560 65. 67. 69. ‘ Chapter 8: Techniques of Integration / V1 + cos 2: dt 07/2 0 / V1 7 weeds 17/4 / may W/‘l 0 66. / V1 + costdt 68. f V1 r sin26d9 17/2 0 70. / Vseczy — ldy “17/4 Assorted Integrations Evaluate each integral in Exercises 71—82 by using any technique you. think is appropriate. ' 71. 73 75 76 77. 79. 81. Bar/4 W 77/4 / (csc x — cotx)2 abc 72. / (sec x + 4 cos x)2 the 11/4 0 fcochsc(sin0)d6 74. f (1 + cot (x + lnx)dx [(cscx * secx)(sinx + cosx)dx f3sinh<§ + 1.15%); 6dy 78 / dx WU +y) ' xV4x2 a 1 f 76176 80 dx (x w I)\/x2 — 2x — 48 (2x +1)\/4x2 + 41 / see2 I tan (tan 1) dt 82/ ab“ x\/3 +352 trigonometric Powers 83. 84. a. Evaluate f cos3ede. (Hint: c0526 = 1 e sin26.) b. Evaluate f cossﬂdﬁ. (2. Without actually evaluating the integral, explain how you would evaluate cos9 6 d9. 3. Evaluate / sin39dt9. (Hint: sin26 = l — 00526.) b. Evaluate f sin56d6. 1:. Evaluate f sin79d6. 11. Without actually evaluating the integral, explain how you would evaluate f sin13 (9 d6. 3. Express f tan36d6 in terms of f tan6d0. Then evaluate f M3 Ede. (Hint'tan26 = seczt9 — l.) h. Express f tan5 6d6 in terms of f tan3 6 d6. c. Express f tan70d6 in terms of f tan5 61:16. d. Express f tanyﬁl 19 d6, where k is a positive integer, in terms offenZHBda. ‘ 3. Express f cot3 6 dB in terms of I out 6 d9. Then evaluate f 0013 6.19. (Hint: cot26 = csczﬂ — 1.) theory and tempos ‘ 87. 88. 89. 90. 91. 92. -~ 93. 94. 95. 96. b. Express f cot5 6 d6 in terms of f eot3 ads. (3. Express f Got] 0 dB in terms of / cots/6 .516. d. Express f cotZ'H'1 19 d6, where k is apositive integer, in terms off cot2ch1 8 d6. Area Find the area of region bounded above by y = 2 cosx and below byy = Secx, Mir/4 S x S 1r/4. Area Find the area of the ‘hiangularf’ region that is bounded from__above_,.an'd’ below by the curves y = escx and y = sin x, 77/6 5 x 5 17/2, and on the left by the linex : 17/6. Volume Find the volume of the solid generated by revolving the region in Exercise 87 about the x—axis. Volume Find the volume of the solid generated by revolving the region in Exercise 88 about the x—axis. Arc length Find the length of the curve y = In (cosx), 0 S x S 17/3. Are length Find the length of the curve y = ln(secx_), 0 s x S 7r/4. Centroid Find the centroid of the region bounded by the x-axis, the curvey = secx, and the linesx = —7r/4,x = 17/4. Centroid Find the centroid of the region that is bounded by the x-axis, the curve )2 : cscx, and the lines 11: = 7r/6, x = 517/6. The integral of csex Repeat the derivation in Example 7, using eofunctions, to show that fcscxdti“: kln [cscx + cotx| :t C. ‘ Using different substitutions Show that the integral 1 / ((xl — no + IDLE/in ,1 can be evaluated with any of the following substitutions. a. u = 1/0: + l) b. u = ((x — l)/(x + l))"fork z 1,1/2,1/3,—1/3,—2/3, and —l c. u = tan‘lx d. u : tanulVJt f. u = cos—1x e. u : tan—1((x _ 1)/2) g. u = cosh‘l x What is the value of the integral? (Source: “Problems and Solu- ' tions,” College Mathematics Journal, Vol. 21, No. 5 (Nov. 1990), pp. 425—426.) ...
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SupIntegrationTech8-0 - incnnmnns on A inrnnnnrron OVERVIEW...

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