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Unformatted text preview: 236 CHAPTER 4 * INTEaaarroN TECHNIQUES The integral is evaluated as follows: aÂ»: 1 1  cosr _
m : w*ï¬‚dx Multrplv by l.
1! coax 1+ rose: 1â€” cosx l â€” cos): . .
: winds Simpitfy.
1 .. cos x
/'1 " cos}; _ 3' . a
: Tip; 1â€” cos a: = state:
am a:
1 cost
2 f _ 2 d]: â€” f . 2 air Split up the fraction.
_ sin a: sin a ll sin a sin a: 1 case:
[cscï¬xdx _. foscrcotartir csca 2 ,cota: *2 â€œcot a: + csc .r. l C. Integrate using Table 4.1.
Reiared Exercises 37â€”40 *( The techniques illustrated in this section are designed to transform or simplify an integrand
before you apply a specific method. In fact, these ideas may help you recognize the best
method to use. Keep them in mind as you learn new integration methods and improve your integration skills. SECTION 4.1 EXERCISES Review Questions 15â€”22. Subtle substitutions Evaiaare the foiiawiag integrate
1. What change of variables would you use for the integral Ex 6 .322: d
".6 o â€” I
for â€” 7.3;) an 15 f3. .. genre 1 Â£2: .. 4.9â€”: r
2. Before integrating, how would you rewrite the integrand of 1 . 2 2
e __ a ? 9 ln s: . sinâ€”ix
fat 2) ï¬x 17. f ( )ar 18. f 5 ex
3. What trigonometric identity is useful in evaluating fainzacixâ€˜?  1 it: cos .x
s a
. . . . . a: â€” 2.3: + 4 4 . +
4. Describe a first step in integrating /â€”x:1m ex. 19. /Ci1: .1: ex 2â€œ M ex
511115 o Ve3+a2+4
. . . . . 10 air  till!
5. Describe a first step in 1ntegratu1g W air. 21. _ '  22â€˜ / . .
.r â€”4.r+5 .r'l+1 yâ€”1+},â€”s .rmâ€” 23cl1+liltr2 +1 1 23â€”23. Splitting fractions Evaiaare Ihefoiiowing iategrais.
3r 6. Describe a first step in integrating / 9. 5/2 # iii,
23. f â€œI: 2 ab: 24. / â€œ1â€”31,: ax
Basic Skills it + â€˜4 4 '3â€˜
7â€”14. Substitution Review Evaiaare rite faiiawiag iategrais. _ _2
. star + taut 4 + e I
six . 25. â€”â€”â€”â€”2 a: 26. Tit
7 fme 3. / (9a:  2)'"3 ex G05 i e
(3 ' 5r.)
2 â€” 3a: 3.1: + 1 27. â€”â€”â€”~â€”ax .
ï¬afii 2 2
. V â€” Vs â€”
a. / sin (2a m E) ex 10. fame: I x â€œ1'
CI 29â€”32. Division with rational functions Evaiaare the following ï¬‚ air iategrais.
1 2x
11./â€˜11 (ix 12. w 3+2 4.12242
.1? â€”5 4 " a: 29. six 30. ii":
a: + 4 2 it â€”~ 1
gr ext/EH
13/ ex 14 f3 riy iii.2 6â€”14
E. + 1 ' e 32 ex
V3; 31 f r + 1 [I2 + 4 . 4.1 Basic Approaches 237 3366. Completing the square Evaiaate the following integrate. _ 56. Different methods
air 2 I a. Evaluate feet a csc2 a: tit using the substitution at = cot x.
33' /m 34* A miâ€œ in. Evaluate j cotecsc2 rat using the substitution a = csc .r. c. Reconcile the resuits in parts (a) and (b). air 57. Different methods
2
at a. Evaluate / tixusing the substitution a = at + 1. d6
iii/WM 36. feye3?â€”
w27_69â€”92 x+2x+1 37â€”40.. Multiply by 1 Evaluate thefette wing integrals. .1: + 1
as 1  i: xi . _ .. . _ _ _
37. m 33. We]: _ h. Evaluate a: + 1 {it after f1rstperfortn1nglcng division
sin ._ a:
on the integrand.
39 '51" 40 {:53 e. Reconcile the results in parts (a) and (b).
1' secs 1. i 1 "cacti 53. Different substitutions ' eh:
Further Explorations a. Show that fmg = sinâ€”1&3: â€” l) + Cusing either
41. Enplain why or why not Determine whether the following state V X â€”â€˜ I '
ments are true and give an explanation or counterexample. ' H _ _ _ l
aw2.t letaâ€”xâ€”â€”.
3 3 a 2
a. a tit = egttt + â€”d.r. tix _
_ a + 4 itâ€œ 4 b. Show that â€”â€”2 = 2 sin_1 a: + C usinga = Vi.
11. Long division simplifies the evaluation of the integral â€œvâ€ x â€˜â€œ x
a .
we; c. Prove the identity 2 sinâ€˜1 .1:  sin[(235 â€” i) = 3.
3x4 + x 2
I: air = 1n lsina: + 1  + C I (Source: The Coiiege Mathemties Jettraat 32, 5, Nov 2001)
â€˜ sins: + l. â€˜ _ *
iii an i Applications
d' at h in E + Câ€œ 59. Area of a region between curves Find the area of the region
. a
.t 1
42â€”54. Miscellaneous integrals Use the appreaches discussed in this PDUï¬‚dï¬‚d by the curves )2 _ x3 _ 3.x: and 3'" __ .113 _ 3.x Dâ€ the
seetien to evaluate thefeiiewing integrals. mmï¬ï¬‚l [2r 4}
a dx {:1 x {it}. Area of a region between curves Find the area of the entire
42. m 43. / Tw tit . a3 8x
4 1 # Vii _1x + 2x + 2 region bounded by the curves y = 2 and y = a
x + i .1: + l
44 ft m at}; 45 fsiu x sin 22: [ix 61. Volumes of solids Consider the region R bounded by the graph of
' .3. â€˜ f{..t) = we? + 1 and the itâ€”Eixli} on the interval [a a]. a. Find the volume of the solid formed when R is revolved about we . .
46. / V1 + casaxax 47. L theeem
i} xii? + 1329 1'). Find the volume of the solid formed when R is revolved about
the yards.
1
43_ / d_;; 49â€˜ / 4:13"... ix 62. Volumes of solids Consider the region R bounded by the graph of
D 4 â€” v3 3:2 + 6.131? + 13 1
ï¬‚at) = + 2 and the xâ€”aais on the interval [0, 3].
.1:
11'er .1:
50. / 3V1 + sin 2): (1x 51. faxâ€”EM (ix a. Find the volume of the solid fanned when R is revolved about
{3' 3 + 253: + 1 the .taJtis. 11. Find the volume of the solid formed when R is revolved about 1:13 3 2 h .
52. V l 11 cos dear 53. / eâ€”ZHâ€”Hâ€”gx t e yearns.
II] 1 .1? + 2x + 1 63. Are length Find the length of the curve )2 = am ï¬‚ï¬‚ the interval [07. 1]. (Hint: Write the arc length integral and let a
2
54. fmtï¬‚ â€œ2 "T" 1 + (3)3 ï¬‚.)
gs+3s+33+l '
titl. Surface area Find the area of the surface generated when the re 55. Different substitutions gion bounded by the graph of y = at + if" on the interval
a. Evaluate ftan a: sec,3 :1: air using the substitution a w tan .t. [0; III 2] is revolved about the #398115.
b. Evaluate f tan a secirait using the substitution at w sec 3:. c. Reconciie the results in parts (a) and (b). 238 Cnanraa 4  Inraeaarren Tecnmeens as. Surface area Let ï¬x) = Va + 1. Find the area ef the surface QUâ€˜C" CHECK ANSWERS generated when the regien beunded by the graph eff en the 1. Let it == 6 + 5x. 2. Write the integrand its 133â€™; 2 + 2.7â€”1.
interval [0, l] is revelved abeut the râ€”aaisi * _ . _ . 2
3. Use leng divisien te write the integrand as l + ,
66. Skydiving A skydiver in free fall subject in gravitatienal I *" I
acceleratien and air resistance has a velecity given by 4. (r + 3)2 + 7 all aâ€œ  1 , . .
v(r) : v7( m. + 1 ) where vT is the terminal velecity and e 3e 0
E is a physical censtant. Find the distance that the skydiver falls
1â€˜ aftertsecends,which is d(r) = / v(y)dy.
e 42~'&â€t,egraâ€œÂ°â€_byliaï¬rt5W The Substitutien Rule (Sectien 3.5) arises when we reverse the Chain Rule fer derivatives. In this sectien, we empley a similar strategy and reverse the Prcduct Rule fer derivatives. The result is an integratien technique called integrerien by parts. Te illustrate the imper
tance ef integratien by parts, censider the indeï¬nite integrals [exdx=ex+C and [nexdx=? The first integral is an elementary integral that we have alreadyr enceuntered, The secend
integral is enly slightly differentâ€”and yet: the appearance ef the preduct areâ€œ: in the inte grand makes this integral (at the mement) impessible te evaluate. Integratien by parts is
ideally suited fer evaluating integrals ef precincts ef functiens. Integration by Parts fer Indefinite Integrals
Given twe differentiable functiens a and v, the Preduct Rule states that granite) â€”â€” mean) + were By integrating beth sides, we can write this rule in terms ef an indeï¬nite integral: a(x)v(x) = f(ttâ€™(r)v(;r] + a(r)vâ€™(:t))rix. Rearranging this eapressien in the ferm /n('r)wÂ£r = n(x)v(x) â€” /v(r)nâ€™(r) ab: (iv cits leads te the basic relatienship fer integrnrien by parts. It is expressed cempactly by eating
that dis = a â€™ (r) ab: and rip = v â€™ (x) doc. Suppressing the independent variable x, we have fadv=ave~ [vein The integral f a dv is viewed as the given integral, and we use integratien by parts te ex press it in terms ef a new integral fr rain. The technique is successful if the new integral
can be evaluated. Snppese that a and v are differentiable functiens. Then [adv = av  fvdn. Integratien by Parts ' 


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 Fall '15
 S. Tanu Halim
 Math, Assignment Chapter 4.1