Solucao impares Calculo 6ed SEC.7.1a7.4

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7 TECHNIQUES OF INTEGRATION

7.1 Integration by Parts

1. Letx = ln {, gy = {2g{ gx =_{1g{, y =1₃{3. Then by Equation 2, U {2_{ln { g{ = (ln {)}1 3{3 U 1 3{3 ₁ { g{ = 1 3{3 ln { 13 U {2_{g{ =} 1 3{3 ln { 13 ₁ 3{3 + F = 1 3{3 ln { 19{3+ F or 1₃{3ln { 1₃+ F

Note: A mnemonic device which is helpful for selectingxwhen using integration by parts is the LIATE principle of precedence forx: Logarithmic

Inverse trigonometric Algebraic

Trigonometric Exponential

If the integrand has several factors, then we try to choose among them axwhich appears as high as possible on the list. For example, inU{h2{g{ the integrand is{h2{, which is the product of an algebraic function ({) and an exponential function (h2{). SinceAlgebraic appears beforeExponential, we choosex = {. Sometimes the integration turns out to be similar regardless of the selection ofxandgy, but it is advisable to refer to LIATE when in doubt.

3. Letx = {, gy = cos 5{ g{ gx = g{, y = 1₅sin 5{. Then by Equation 2, U

{ cos 5{ g{ = 1

5{ sin 5{ U ₁

5sin 5{ g{ =15{ sin 5{ +251 cos 5{ + F.

5. Letx = u, gy = hu@2gu gx = gu, y = 2hu@2. ThenUuhu@2gu = 2uhu@2U2hu@2gu = 2uhu@2 4hu@2+ F. 7. Letx = {2,gy = sin { g{ gx = 2{ g{ and y = 1cos {. Then

L =U{2 _{sin { g{ =}1

{2cos { +2 U

{ cos { g{ (B). Next let X = {, gY = cos { g{ gX = g{,

Y = 1 sin {, so U { cos { g{ = 1 { sin { 1 U sin { g{ = 1 { sin { +12cos { + F1. Substituting forU{ cos { g{ in (B), we get

L = 1

{2cos { +2 ₁

{ sin { +12cos { + F1= 1{2cos { +22{ sin { +23cos { + F, where F = 2F1. 9. Letx = ln(2{ + 1), gy = g{ gx = 2 2{ + 1g{, y = {. Then ] ln(2{ + 1) g{ = { ln(2{ + 1) ] _{2{ + 1}2{ g{ = { ln(2{ + 1) ] (2{ + 1) 1_{2{ + 1} g{ = { ln(2{ + 1) ] 1 1 2{ + 1 g{ = { ln(2{ + 1) { +1 2ln(2{ + 1) + F =1 2(2{ + 1) ln(2{ + 1) { + F 11. Letx = arctan 4w, gy = gw gx = 4 1 + (4w)2 gw =_{1 + 16w}4 2gw, y = w. Then ] arctan 4w gw = w arctan 4w ] 4w 1 + 16w2gw = w arctan 4w 1₈ ] 32w 1 + 16w2gw = w arctan 4w 18ln(1 + 16w2) + F. 295

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296 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION

13. Letx = w, gy = sec22w gw gx = gw, y = 1₂ tan 2w. Then U w sec2_{2w gw =}1 2w tan 2w 12 U tan 2w gw = 1 2w tan 2w 14ln |sec 2w| + F.

15. First letx = (ln {)2,gy = g{ gx = 2 ln { ·1_{g{, y = {. Then by Equation 2, L =U(ln {)2_{g{ = {(ln {)}2₂U_{{ ln { ·}1 {g{ = {(ln {)2 2 U ln { g{. Next let X = ln {, gY = g{ gX = 1@{ g{, Y = { to getUln { g{ = { ln { U{ · (1@{) g{ = { ln { Ug{ = { ln { { + F1. Thus, L = {(ln {)2_{2({ ln { { + F} 1) = {(ln {)2 2{ ln { + 2{ + F, where F = 2F1. 17. First letx = sin 3, gy = h2g gx = 3 cos 3 g, y = 1₂h2. Then

L =Uh2_{sin 3 g =} 1

2h2sin 3 32 U

h2_{cos 3 g. Next let X = cos 3, gY = h}2_{g gX = 3 sin 3 g,} Y =1 2h2to get U h2_{cos 3 g =}1 2h2cos 3 +32 U

h2_{sin 3 g. Substituting in the previous formula gives} L = 1 2h2sin 3 34h2cos 3 94 U h2_{sin 3 g =} 1 2h2sin 3 34h2cos 3 94L 13

4L = 12h2sin 3 34h2cos 3 + F1. Hence,L= 131h2(2 sin 3 3 cos 3) + F, where F =134F1. 19. Letx = w, gy = sin 3w gw gx = gw, y = 1₃cos 3w. Then

U 0 w sin 3w gw = 1 3w cos 3w 0 +13 U 0 cos 3w gw = ₁ 3 0 +1 9 sin 3w₀ = 3= 21. Letx = w, gy = cosh w gw gx = gw> y = sinh w. Then

U1

0 w cosh w gw =

w sinh w1₀U₀1sinh w gw = (sinh 1 sinh 0) cosh w1₀= sinh 1 (cosh 1 cosh 0) = sinh 1 cosh 1 + 1.

We can use the denitions ofsinh and cosh to write the answer in terms of h: sinh 1 cosh 1 + 1 =1 2(h1 h1) 12(h1+ h1) + 1 = h1+ 1 = 1 1@h. 23. Letx = ln {, gy = {2g{ gx = 1 {g{, y = {1. By (6), ] 2 1 ln { {2 g{ = ln {_{ 2 1+ ] 2 1 { 2_{g{ =}1 2ln 2 + ln 1 + 1_{ 2 1= 1 2ln 2 + 0 12+ 1 = 1212ln 2. 25. Letx = |, gy = g| h2| = h2|g| gx = g|, y = 12h2|. Then ] 1 0 | h2| g| = k 1 2|h2| l1 0+ 1 2 ] 1 0 h 2|_{g| =}1 2h2+ 0 1 4 k h2|l1 0= 1 2h214h2+41 =1434h2. 27. Letx = cos1{, gy = g{ gx = g{ 1 {2,y = {. Then L =]1@2 0 cos 1_{{ g{ =}_{{ cos}1_{1@2 0 + ] 1@2 0 { g{ 1 {2 =12 ·3 + ] 3@4 1 w 1@21 2gw , wherew = 1 {2 gw = 2{ g{. Thus, L = 6 +12 U1 3@4w1@2gw = 6 + w1_3@4= 6 + 1 3 2 = 16 + 6 33.

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SECTION 7.1 INTEGRATION BY PARTS ¤ 297 29.Letx = ln (sin {), gy = cos { g{ gx = cos {

sin {g{, y = sin {. Then

L =Ucos { ln(sin {) g{ = sin { ln(sin {) Ucos { g{ = sin { ln(sin {) sin { + F.

Another method: Substitute w = sin {, so gw = cos { g{. Then L =Uln w gw = w ln w w + F (see Example 2) and so

L = sin { (ln sin { 1) + F. 31.Letx = (ln {)2,gy = {4g{ gx = 2 ln { { g{, y = { 5 5. By (6), ] 2 1 { 4_{(ln {)}2_{g{ =}{5 5 (ln {)2 2 1 2 ] 2 1 {4 5 ln { g{ =325(ln 2)2 0 2 ]2 1 {4 5 ln { g{. LetX = ln {, gY = { 4 5 g{ gX = 1{g{, Y = { 5 25. Then ] 2 1 {4 5 ln { g{ = {5 25ln { 2 1 ] 2 1 {4 25g{ =3225ln 2 0 {5 125 2 1= 32 25ln 2 ₃₂ 1251251 . SoU₁2{4(ln {)2g{ =32₅(ln 2)2 2₂₅32ln 2 ₁₂₅31= 32₅(ln 2)264₂₅ln 2 +₁₂₅62. 33.Let| ={, so that g| =1₂{1@2g{ = 1 2{g{ = 12|g{. Thus, U

cos{ g{ =Ucos | (2| g|) = 2U| cos | g|. Now use parts withx = |, gy = cos | g|, gx = g|, y = sin | to getU| cos | g| = | sin | Usin | g| = | sin | + cos | + F₁, soUcos{ g{ = 2| sin | + 2 cos | + F = 2{ sin{ + 2 cos{ + F.

35.Let{ = 2, so thatg{ = 2 g. Thus, ] /2 3_cos2_{g =}] /2 2_cos2_·1 2(2 g) =12 ]

/2{ cos { g{. Now use parts withx = {, gy = cos { g{, gx = g{, y = sin { to get

1 2 ] /2{ cos { g{ = 1 2 { sin {_/2] /2sin { g{ = 1 2 { sin { + cos {_/2 = 1 2( sin + cos ) 12 2sin2 + cos2 =1 2( · 0 1) 12 2 · 1 + 0 = 1 2 4 37.Let| = 1 + {> so that g| = g{. Thus,U{ ln(1 + {) g{ =U(| 1) ln | g|. Now use parts with x = ln |> gy = (| 1) g|,

gx = 1 |g|, y =12|2 | to get U (| 1) ln | g| =1 2|2 | ln | U 1 2| 1 g| = 1 2|(| 2) ln | 14|2+ | + F = 1 2(1 + {)({ 1) ln(1 + {) 14(1 + {)2+ 1 + { + F, which can be written as 1₂({2 1) ln(1 + {) 1₄{2+1₂{ +3₄ + F.

In Exercises 39–42, leti({)denote the integrand andI ({)its antiderivative (withF = 0). 39.Letx = 2{ + 3, gy = h{g{ gx = 2 g{, y = h{. Then

U

(2{ + 3)h{_{g{ = (2{ + 3)h}{₂U_h{_{g{ = (2{ + 3)h}{_2h{_{+ F} = (2{ + 1) h{_{+ F}

We see from the graph that this is reasonable, sinceI has a minimum where i changes from negative to positive.

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298 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 41. Letx =1₂{2,gy = 2{1 + {2g{ gx = { g{, y =2₃(1 + {2)3@2. Then U {3_{1 + {}2_{g{ =}1 2{2 k 2 3(1 + {2)3@2 l 2 3 U {(1 + {2₎3@2_g{ =1 3{2(1 + {2)3@223 ·25 ·12(1 + {2)5@2+ F =1 3{2(1 + {2)3@2152(1 + {2)5@2+ F

Another method: Use substitution with x = 1 + {2to get1₅(1 + {2)5@21₃(1 + {2)3@2+ F. 43. (a) Takeq = 2 in Example 6 to get

]

sin2_{{ g{ = 1}

2cos { sin { + 12 ]

1 g{ = {₂ sin 2{₄ + F. (b)Usin4{ g{ = 1₄cos { sin3{ +3₄Usin2{ g{ = 1₄cos { sin3{ +3₈{ ₁₆3 sin 2{ + F. 45. (a) From Example 6,

] sinq_{{ g{ = 1} qcos { sinq1{ + q 1q ] sinq2_{{ g{. Using (6),} ] @2 0 sin q_{{ g{ =}_{cos { sin}q1{ q @2 0 + q 1q ] @2 0 sin q2_{{ g{} = (0 0) + q 1_q ] @2 0 sin q2_{{ g{ = q 1} q ] @2 0 sin q2_{{ g{}

(b) Usingq = 3 in part (a), we haveU₀@2sin3{ g{ = 2₃U₀@2sin { g{ =2₃cos {@2₀ = 2₃. Usingq = 5 in part (a), we haveU₀@2sin5{ g{ = 4₅U₀@2sin3{ g{ =4₅ ·2₃ =₁₅8.

(c) The formula holds forq = 1 (that is, 2q + 1 = 3) by (b). Assume it holds for some n 1. Then ] @2 0 sin 2n+1_{{ g{ = 2 · 4 · 6 · · · (2n)} 3 · 5 · 7 · · · (2n + 1). By Example 6, ] @2 0 sin 2n+3_{{ g{ = 2n + 2} 2n + 3 ] @2 0 sin 2n+1_{{ g{ = 2n + 2} 2n + 3· 2 · 4 · 6 · · · (2n)3 · 5 · 7 · · · (2n + 1) = _{3 · 5 · 7 · · · (2n + 1)[2 (n + 1) + 1]}2 · 4 · 6 · · · (2n)[2 (n + 1)] ,

so the formula holds forq = n + 1. By induction, the formula holds for all q 1. 47. Letx = (ln {)q,gy = g{ gx = q(ln {)q1(g{@{), y = {. By Equation 2,

U

(ln {)q_{g{ = {(ln {)}qU_{q{(ln {)}q1_{(g{@{) = {(ln {)}q_qU_{(ln {)}q1_g{.

49. Utanq_{{ g{ =}U_tanq2_{{ tan}2_{{ g{ =}U_tanq2_{{ (sec}2_{{ 1) g{ =}U_tanq2_{{ sec}2_{{ g{}U_tanq2_{{ g{} = L Utanq2_{{ g{.}

Letx = tanq2{, gy = sec2{ g{ gx = (q 2) tanq3{ sec2{ g{, y = tan {. Then, by Equation 2, L = tanq1_{{ (q 2)}U_tanq2_{{ sec}2_{{ g{}

1L = tanq1_{{ (q 2)L} (q 1)L = tanq1_{

L = tan_{q 1}q1{ Returning to the original integral,Utanq{ g{ = tan

q1_{

q 1

U

tanq2_{{ g{.}

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SECTION 7.1 INTEGRATION BY PARTS ¤ 299 51.By repeated applications of the reduction formula in Exercise 47,

U (ln {)3_{g{ = { (ln {)}3₃U_{(ln {)}2_{g{ = {(ln {)}3₃_{{(ln {)}2₂U_{(ln {)}1_g{ = { (ln {)3_{3{(ln {)}2_{+ 6}_{{(ln {)}1₁U_{(ln {)}0_g{ = { (ln {)3_{3{(ln {)}2_{+ 6{ ln { 6}U_{1 g{ = { (ln {)}3_{3{(ln {)}2_{+ 6{ ln { 6{ + F} 53.Area=U₀5{h0=4{g{. Let x = {, gy = h0=4{g{ gx = g{, y = 2=5h0=4{_{. Then} area=2=5{h0=4{5₀+ 2=5U₀5h0=4{g{ = 12=5h2_{+ 0 + 2=5}_2=5h0=4{5 0 = 12=5h2_6=25(h2_{1) = 6=25 18=75h}2 _or 25 4 754h2 55.The curves| = { sin { and | = ({ 2)2intersect atd 1=04748 and

e 2=87307, so area=U_de[{ sin { ({ 2)2] g{ ={ cos { + sin { 1 3({ 2)3 e d [by Example 1] 2=81358 0=63075 = 2=18283

57.Y =U₀12{ cos({@2) g{. Let x = {, gy = cos({@2) g{ gx = g{, y = 2

sin({@2). Y = 2 2 { sin { 2 1 0 2 · 2 ] 1 0 sin { 2 g{ = 2 2 0 4 2 cos { 2 1 0= 4 + 8(0 1) = 4 8 . 59.Volume=U₁0 2(1 {)h{g{. Let x = 1 {, gy = h{g{ gx = g{, y = h{. Y = 2(1 {)(h{₎0 1 2 U0 1h{g{ = 2 ({ 1)(h{_{) + h}{0 1= 2 {h{0 1= 2(0 + h) = 2h 61.The average value ofi({) = {2ln { on the interval [1> 3] is iave= 1

3 1 ] 3 1 { 2_{ln { g{ =}1 2L. Letx = ln {, gy = {2g{ gx = (1@{) g{, y =1₃{3. SoL =1₃{3ln {3₁U₁3 1₃{2g{ = (9 ln 3 0) 1₉{33₁= 9 ln 3 3 1₉= 9 ln 3 26₉. Thus,iave= 1₂L =1₂ 9 ln 3 26 9 =9 2ln 3 139.

63.Sincey(w) A 0 for all w, the desired distance is v(w) =U₀wy(z) gz =U₀wz2hzgz.

First letx = z2,gy = hzgz gx = 2z gz, y = hz. Thenv(w) =z2hzw₀+ 2U₀wzhzgz. Next letX = z, gY = hzgz gX = gz, Y = hz. Then

v(w) = w2_hw_{+ 2}_zhzw 0+ Uw 0hzgz = w2_hw_{+ 2}_whw_{+ 0 +}_hzw 0 = w2_hw_{+ 2(wh}w_hw_{+ 1) = w}2_hw_2whw_2hw_{+ 2 = 2 h}w_(w2_{+ 2w + 2) meters} 65.ForL =U₁4{i00({) g{, let x = {, gy = i00({) g{ gx = g{, y = i0({). Then

L ={i0_({)4 1

U4

1 i0({) g{ = 4i0(4) 1 · i0(1) [i(4) i(1)] = 4 · 3 1 · 5 (7 2) = 12 5 5 = 2. We used the fact thati00is continuous to guarantee thatL exists.

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67. Using the formula for volumes of rotation and the gure, we see that

Volume=U₀ge2g| U₀fd2g| U_fg[j(|)]2 g| = e2g d2f U_fg[j(|)]2 g|. Let | = i({), which givesg| = i0({) g{ and j(|) = {, so that Y = e2g d2f U_de{2i0({) g{.

Now integrate by parts withx = {2, andgy = i0({) g{ gx = 2{ g{, y = i({), and Ue d{2i0({) g{ = {2_i({)e d Ue

d2{ i({) g{ = e2i(e) d2i(d) Ue

d2{ i({) g{, but i(d) = f and i(e) = g Y = e2_{g d}2_fk_e2_{g d}2_fUe

d2{ i({) g{ l

=U_de2{ i({) g{.

7.2 Trigonometric Integrals

The symbols=s and=c indicate the use of the substitutions{x = sin {> gx = cos { g{}and{x = cos {> gx = sin { g{}, respectively. 1. Usin3_{{ cos}2_{{ g{ =}U_sin2_{{ cos}2_{{ sin { g{ =}U_{(1 cos}2_{{) cos}2_{{ sin { g{} c

=U(1 x2_)x2_(gx) =U(x2_1)x2_{gx =}U_(x4_x2_{) gx =} 1

5x513x3+ F = 15cos5{ 13cos3{ + F 3. U_@23@4_sin5_{{ cos}3_{{ g{ =}U3@4

@2 sin5{ cos2{ cos { g{ = U3@4

@2 sin5{ (1 sin2{) cos { g{ s =U₁2@2x5_{(1 x}2_{) gx} =U₁2@2(x5_x7_{) gx =}1 6x618x8 2@2 1 = 1@8 6 1@168 1 6 18 = 11 384 5. Let| = {, so g| = g{ and U sin2_{({) cos}5_{({) g{ =} 1 U sin2_{| cos}5_{| g| =} 1 U

sin2_{| cos}4_{| cos | g|} = 1

U

sin2_{| (1 sin}2_|)2 _{cos | g|} s = 1 U x2_{(1 x}2₎2_{gx =} 1 U (x2_2x4_{+ x}6_{) gx} = 1 ₁ 3x325x5+17x7 + F = 1

3sin3| 52 sin5| +71 sin7| + F

= 1

3sin3({) 52 sin5({) +71 sin7({) + F 7. U₀@2_cos2_{g =}U@2

0 12(1 + cos 2) g [half-angle identity] =1 2 +1 2sin 2 @2 0 =12 2 + 0 (0 + 0)= 4 9. U₀sin4_{(3w) gw =}U 0 sin2_(3w)2 _{gw =}U 0 ₁ 2(1 cos 6w) 2 _{gw =}1 4 U 0 (1 2 cos 6w + cos26w) gw =1 4 U 0 1 2 cos 6w +1 2(1 + cos 12w) gw = 1 4 U 0 ₃ 2 2 cos 6w +12cos 12w gw =1 4 ₃ 2w 13sin 6w +241 sin 12w 0 =14 ₃ 2 0 + 0 (0 0 + 0)=3 8 11. U(1 + cos )2_{g =}U_{(1 + 2 cos + cos}2_{) g = + 2 sin +}1

2 U

(1 + cos 2) g = + 2 sin +1

2 +14sin 2 + F = 32 + 2 sin +14sin 2 + F 13. U₀@2sin2_{{ cos}2_{{ g{ =}U@2 0 14(4 sin2{ cos2{) g{ = U@2 0 14(2 sin { cos {)2g{ =14 U@2 0 sin22{ g{ =1 4 U@2 0 12(1 cos 4{) g{ = 18 U@2 0 (1 cos 4{) g{ = 18 { 1 4sin 4{ @2 0 = 18 2 = 16

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SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 301 15. ] cos5 sin g = ] cos4 sin cos g = ] 1 sin22 sin cos g s =] (1 x 2)2 x gx =] 1 2x_x_1@22+ x4gx =] (x1@2_2x3@2_{+ x}7@2_{) gx = 2x}1@24 5x5@2+29x9@2+ F = 2 45x1@2(45 18x2+ 5x4) + F = 452

sin (45 18 sin2_{+ 5 sin}4_{) + F} 17.

]

cos2_{{ tan}3_{{ g{ =}] sin3{ cos { g{ c =] (1 x2_x)(gx)=] 1_x + x gx = ln |x| +1 2x2+ F = 12cos2{ ln |cos {| + F 19. ] _{cos { + sin 2{} sin { g{ =

] _{cos { + 2 sin { cos {}

sin { g{ = ] _{cos {} sin {g{ + ] 2 cos { g{ s =] 1_xgx + 2 sin { = ln |x| + 2 sin { + F = ln |sin {| + 2 sin { + F

Or: Use the formulaUcot { g{ = ln |sin {| + F.

21.Letx = tan {, gx = sec2{ g{. ThenUsec2{ tan { g{ =Ux gx =1₂x2+ F = 1₂tan2{ + F.

Or: Let y = sec {, gy = sec { tan { g{. ThenUsec2{ tan { g{ =Uy gy = ₂1y2+ F = 1₂sec2{ + F.

23.Utan2_{{ g{ =}U_(sec2_{{ 1) g{ = tan { { + F}

25.Usec6_{w gw =}U_sec4_{w · sec}2_{w gw =}U_(tan2_{w + 1)}2_sec2_{w gw =}U_(x2_{+ 1)}2_gx _[_{x = tan w, gx = sec}2_{w gw]} =U(x4_{+ 2x}2_{+ 1) gx =}1

5x5+23x3+ x + F = 15tan5w +23tan3w + tan w + F 27.U₀@3tan5_{{ sec}4_{{ g{ =}U@3

0 tan5{ (tan2{ + 1) sec2{ g{ = U 3 0 x5(x2+ 1) gx [x = tan {, gx = sec2{ g{] =U₀3(x7_{+ x}5_{) gx =}1 8x8+16x6 3 0 = 818 +276 =818 +92 =818 +368 = 1178 Alternate solution: U@3 0 tan5{ sec4{ g{ = U@3

0 tan4{ sec3{ sec { tan { g{ = U@3

0 (sec2{ 1)2 sec3{ sec { tan { g{ =U₁2(x2₁₎2_x3_{gx [x = sec {, gx = sec { tan { g{] =}U2

1(x4 2x2+ 1)x3gx =U₁2(x7_2x5_{+ x}3_{) gx =}1 8x813x6+14x4 2 1= 32 64 3 + 4 1 813+14 = 117 8 29.Utan3_{{ sec { g{ =}U_tan2_{{ sec { tan { g{ =}U_(sec2_{{ 1) sec { tan { g{}

=U(x2_{1) gx [x = sec {, gx = sec { tan { g{] =}1

3x3 x + F =13sec3{ sec { + F 31.U_tan5_{{ g{ =}U_(sec2_{{ 1)}2_{tan { g{ =}U_sec4_{{ tan { g{ 2}U_sec2_{{ tan { g{ +}U_{tan { g{}

=Usec3_{{ sec { tan { g{ 2}U_{tan { sec}2_{{ g{ +}U_{tan { g{} =1

4sec4{ tan2{ + ln |sec {| + F [or 14sec4{ sec2{ + ln |sec {| + F ] 33.

] tan3 cos4g =

]

tan3_sec4_{g =}] _tan3_{· (tan}2_{+ 1) · sec}2_g =Ux3_(x2_{+ 1) gx} _[_{x = tan , gx = sec}2_g] =U(x5_{+ x}3_{) gx =} 1

6x6+14x4+ F = 16tan6 +14tan4 + F

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35. Letx = {> gy = sec { tan { g{ gx = g{> y = sec {. Then U

{ sec { tan { g{ = { sec { Usec { g{ = { sec { ln |sec { + tan {| + F. 37. U_@6@2cot2_{{ g{ =}U@2 @6(csc2{ 1) g{ = cot { {@2_@6=0 2 3 6 =3 3 39. U_cot3_csc3_{g =}U_cot2_csc2_{· csc cot g =}U_(csc2_{1) csc}2_{· csc cot g}

=U(x2_1)x2_{· (gx)} _[_{x = csc , gx = csc cot g]} =U(x2_x4_{) gx =} 1

3x315x5+ F = 13csc3 15csc5 + F

41. L =] csc { g{ =] csc { (csc { cot {)_{csc { cot {} g{ =] csc { cot { + csc_{csc { cot {} 2{g{. Let x = csc { cot { gx = ( csc { cot { + csc2_{{) g{. Then L =}U_{gx@x = ln |x| = ln |csc { cot {| + F.}

43. Usin 8{ cos 5{ g{2a =U 1 2[sin(8{ 5{) + sin(8{ + 5{)] g{ = 12 U sin 3{ g{ +1 2 U sin 13{ g{ = 1 6cos 3{ 261 cos 13{ + F 45. U_{sin 5 sin g}2b =U 1 2[cos(5 ) cos(5 + )] g = 12 U cos 4 g 1 2 U cos 6 g =1 8sin 4 121 sin 6 + F 47. ] 1 tan2_{ sec2_{ g{ = ]

cos2_{{ sin}2_{_{g{ =}] _{cos 2{ g{ = 1}

2sin 2{ + F

49. Letx = tan(w2) gx = 2w sec2(w2) gw. ThenUw sec2(w2) tan4(w2) gw =Ux41₂gx=₁₀1x5+ F = ₁₀1 tan5(w2) + F. In Exercises 51–54, leti({)denote the integrand andI ({)its antiderivative (withF = 0).

51. Letx = {2, so thatgx = 2{ g{. Then U { sin2_({2_{) g{ =}U_sin2_x1 2gx =1 2 U ₁ 2(1 cos 2x) gx = 1 4 x 1 2sin 2x + F = 1 4x 14 ₁ 2· 2 sin x cos x + F = 1 4{214sin({2) cos({2) + F

We see from the graph that this is reasonable, sinceI increases where i is positive and I decreases where i is negative. Note also thati is an odd function and I is an even function.

53. U_{sin 3{ sin 6{ g{ =}U 1 2[cos(3{ 6{) cos(3{ + 6{)] g{ =1 2 U (cos 3{ cos 9{) g{ =1 6sin 3{ 181 sin 9{ + F

Notice thati({) = 0 whenever I has a horizontal tangent.

55. iave =₂1 U

sin2{ cos3{ g{ =21 U

sin2{ (1 sin2{) cos { g{

= 1 2 U0 0 x2(1 x2) gx [wherex = sin {] = 0

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SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 303 57. D =] @4 @4(cos 2_{{ sin}2_{{) g{ =}]@4 @4cos 2{ g{ = 2] @4 0 cos 2{ g{ = 2 1 2sin 2{ @4 0 = sin 2{@4₀ = 1 0 = 1

59. It seems from the graph thatU₀2cos3{ g{ = 0, since the area below the

{-axis and above the graph looks about equal to the area above the axis and below the graph. By Example 1, the integral issin { 1₃sin3{2₀ = 0. Note that due to symmetry, the integral of any odd power ofsin { or cos { between limits which differ by2q (q any integer) is 0.

61.Using disks,Y =U_@2 sin2{ g{ = U_@2 1₂(1 cos 2{) g{ = 1₂{ 1₄sin 2{ @2= 2 0 4 + 0 =2 4 63.Using washers, Y =U₀@4(1 sin {)2_{(1 cos {)}2_g{ = U@4 0

(1 2 sin { + sin2_{{) (1 2 cos { + cos}2_{)_g{ = U₀@4(2 cos { 2 sin { + sin2_{{ cos}2_{{) g{}

= U@4

0 (2 cos { 2 sin { cos 2{) g{ = 2 sin { + 2 cos { 1 2sin 2{ @4 0 = 2 +2 1 2 (0 + 2 0)= 22 5 2 65.v = i(w) =Uw

0sin $x cos2$x gx. Let | = cos $x g| = $ sin $x gx. Then v = 1 $ Ucos $w 1 |2g| = $1 ₁ 3|3 cos $w 1 =3$1 (1 cos3$w). 67.Just note that the integrand is odd [i({) = i({)].

Or: If p 6= q, calculate ] sin p{ cos q{ g{ = ] 1 2[sin(p q){ + sin(p + q){] g{ = 1₂ cos(p q){_{p q} cos(p + q){_{p + q} = 0 Ifp = q, then the rst term in each set of brackets is zero.

69.U cos p{ cos q{ g{ =U 1 2[cos(p q){ + cos(p + q){] g{. Ifp 6= q, this is equal to 1 2 sin(p q){ p q + sin(p + q){p + q = 0. Ifp = q, we getU 1₂[1 + cos(p + q){] g{ =1₂{+ sin(p + q){ 2(p + q) = + 0 = .

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7.3 Trigonometric Substitution

1. Let{ = 3 sec , where 0 ? ₂ or ? 3₂. Then g{ = 3 sec tan g and

{2_{9 =}_{9 sec}2_{9 =}s_9(sec2_{1) =}_{9 tan}2 = 3 |tan | = 3 tan for the relevant values of . ]

1

{2_{2₉g{ = ]

1

9 sec2_{· 3 tan}3 sec tan g = 19 ] cos g =1 9sin + F = 1₉ {2₉ { + F

Note that sec( + ) = sec , so the gure is sufcient for the case ?3₂ . 3. Let{ = 3 tan , where ₂ ? ?₂. Theng{ = 3 sec2 g and

{2_{+ 9 =}_{9 tan}2_{+ 9 =}s_9(tan2_{+ 1) =}_{9 sec}2 = 3 |sec | = 3 sec for the relevant values of . ] {3 {2_{+ 9}g{ = ] 33_tan3 3 sec 3 sec2 g = 33 ]

tan3_{sec g = 3}3] _tan2_{tan sec g}

= 33U_(sec2_{1) tan sec g = 3}3U_(x2_{1) gx} _[_{x = sec , gx = sec tan g]} = 331 3x3 x + F = 331 3sec3 sec + F = 33 % 1 3 {2_{+ 9}3@2 33 {2_{+ 9} 3 & + F =1 3({2+ 9)3@2 9 {2_{+ 9 + F or} 1 3({2 18) {2_{+ 9 + F} 5. Letw = sec , so gw = sec tan g, w =2 =₄, andw = 2 = ₃. Then

] 2 2 1 w3_w2₁gw = ] @3 @4 1

sec3_tansec tan g = ] @3 @4 1 sec2g = ] @3 @4 cos 2_g =U_@4@31 2(1 + cos 2) g = 12 +1 2sin 2 @3 @4 = 1 2 k 3 +12 3 2 4 +12· 1 l =1 2 12+ 3 4 12 = 24+ 3 8 14 7. Let{ = 5 sin , so g{ = 5 cos g. Then

] ₁

{2_{25 {}2 g{ =

U 1

52_sin2_{· 5 cos}5 cos g = 1₂₅ ] csc2_g = 1₂₅cot + F = 1₂₅ 25 {2 { + F

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SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 305 9.Let{ = 4 tan , where ₂ ? ? ₂. Theng{ = 4 sec2 g and

{2_{+ 16 =}_{16 tan}2_{+ 16 =}s_16(tan2_{+ 1)} =16 sec2_{= 4 |sec |}

= 4 sec for the relevant values of . ] g{ {2_{+ 16} = ] 4 sec2_g 4 sec = ]

sec g = ln |sec + tan | + F1= ln {2_{+ 16} 4 + {4 + F1 = ln{2+ 16 + { ln|4| + F₁_{= ln}_{2_{+ 16 + {}_{+ F, where F = F}₁_{ln 4.} (Since{2+ 16 + { A 0, we don’t need the absolute value.)

11.Let2{ = sin , where ₂ ₂. Then{ =1₂sin , g{ =1₂cos g, and1 4{2=s1 (2{)2= cos . U 1 4{2_{g{ =}U_cos1 2cos g =1 4 U (1 + cos 2) g =1 4 +1 2sin 2 + F = 1 4( + sin cos ) + F =1 4 sin1_{(2{) + 2{}_{1 4{}2_{+ F} 13.Let{ = 3 sec , where 0 ?₂ or ? 3₂ . Then

g{ = 3 sec tan g and{2_{9 = 3 tan , so} ]

{2₉

{3 g{ =

]

3 tan

27 sec33 sec tan g = 1₃ ] tan2 sec2g = 1 3 U sin2_{g =} 1 3 U ₁

2(1 cos 2) g = 16 121 sin 2 + F = 16 16sin cos + F = 1₆sec1{ 3 1₆ {2₉ { 3{+ F = 16sec1{3 {2₉ 2{2 + F

15.Let{ = d sin , g{ = d cos g, { = 0 = 0 and { = d =₂. Then Ud

0 {2

d2_{2_{g{ =}U@2

0 d2sin2 (d cos ) d cos g = d4 U@2 0 sin2 cos2 g = d4 U@2 0 ₁ 2(2 sin cos ) 2 _g = d₄4] @2 0 sin 2_{2 g = d}4 4 ] @2 0 1 2(1 cos 4) g = d 4 8 1₄sin 4 @2 0 = d₈4k₂ 0 0l= ₁₆d4 17.Letx = {2 7, so gx = 2{ g{. Then ] { {2₇g{ = 12 ] 1 xgx = 12· 2 x + F =s{2_{7 + F.}

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19. Let{ = tan , where ₂ ? ? ₂. Theng{ = sec2 g and1 + {2= sec , so ] 1 + {2 { g{ = ] sec tan sec2 g = ] sec tan (1 + tan2) g =U(csc + sec tan ) g

= ln |csc cot | + sec + F [by Exercise 7.2.41] = ln 1 + {2 { 1{ +1 + {₁ 2 + F = ln 1 + {2₁ { +1 + {2_{+ F} 21. Let{ = 3₅sin , so g{ =3₅cos g, { = 0 = 0, and { = 0=6 = ₂. Then

] 0=6 0 {2 9 25{2 g{ = ] @2 0 ₃ 5 2_sin2 3 cos 3 5cos g = 9₁₂₅ ] @2 0 sin 2_g = 9 125 U@2 0 12(1 cos 2) g =2509 1 2sin 2 @2 0 = 9 250 2 0 0= 9 500 23. _{5 + 4{ {}2_{= ({}2_{4{ + 4) + 9 = ({ 2)}2_{+ 9. Let} { 2 = 3 sin , 2 2, sog{ = 3 cos g. Then U 5 + 4{ {2_{g{ =}U s_{9 ({ 2)}2_{g{ =}U s_{9 9 sin}2_{3 cos g} =U 9 cos2_{3 cos g =}U_{9 cos}2_g

=9 2 U (1 + cos 2) g = 9 2 +1 2sin 2 + F =9

2 +94sin 2 + F = 92 +94(2 sin cos ) + F = 9₂sin1{ 2 3 + 9₂· { 2₃ · 5 + 4{ {2 3 + F = 9₂sin1{ 2 3 + 1₂({ 2)5 + 4{ {2_{+ F} 25. {2_{+ { + 1 =}_{2_{+ { +}1 4 +3 4 = { +1 2 2₊3 2 2 . Let { +1 2 = 3 2 tan , so g{ = 3 2 sec2 g and {2_{+ { + 1 =}3 2 sec . Then [continued]

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SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 307 ] { {2_{+ { + 1}g{ = ] 3 2tan 12 3 2 sec 3 2 sec2 g =U 3 2 tan 12 sec g =U 3 2 tan sec g U ₁ 2sec g =3

2 sec 12ln |sec + tan | + F1 ={2_{+ { + 1}1 2ln23 {2_{+ { + 1 +}2 3 { +1 2 + F1 ={2_{+ { + 1}1 2ln23 {2_{+ { + 1 +}_{{ +}1 2 + F1 ={2_{+ { + 1}1 2ln2312ln {2_{+ { + 1 + { +}1 2 + F1 ={2_{+ { + 1}1 2ln {2_{+ { + 1 + { +}1 2 + F, where F = F11₂ln2₃ 27.{2_{+ 2{ = ({}2_{+ 2{ + 1) 1 = ({ + 1)}2_{1. Let { + 1 = 1 sec ,}

sog{ = sec tan g and{2+ 2{ = tan . Then U

{2_{+ 2{ g{ =}U_{tan (sec tan g) =}U_tan2_{sec g} =U(sec2_{1) sec g =}U_sec3_gU_{sec g} =1

2sec tan +12ln |sec + tan | ln |sec + tan | + F =1

2sec tan 12ln |sec + tan | + F =12({ + 1) {2_{+ 2{}1 2ln{ +1 + {2+ 2{ +F 29.Letx = {2,gx = 2{ g{. Then U {1 {4_{g{ =}U _{1 x}21 2gx = 1 2 U cos · cos g

wherex = sin , gx = cos g, ands1 x2= cos

= 1

2 U ₁

2(1 + cos 2) g =14 +18sin 2 + F = 14 +14sin cos + F = 1 4sin1x +14x 1 x2_{+ F =} 1 4sin1({2) +14{2 1 {4_{+ F} 31. (a) Let{ = d tan , where ₂ ? ? ₂. Then{2+ d2= d sec and

U g{

{2_{+ d}2 =

U d sec2_g d sec =

U

sec g = ln|sec + tan | + F1= ln {2_{+ d}2 d + {d + F1 = ln{ +{2_{+ d}2_{+ F where F = F}₁_{ln |d|}

(b) Let{ = d sinh w, so that g{ = d cosh w gw and{2+ d2 = d cosh w. Then

] _g{

{2_{+ d}2 =

] _{d cosh w gw}

d cosh w = w + F = sinh1{d+ F. 33.The average value ofi({) ={2 1@{ on the interval [1> 7] is

1 7 1 ] 7 1 {2₁ { g{ = 16 ] 0 tan

sec · sec tan g

where{ = sec , g{ = sec tan g, s

{2_{1 = tan , and = sec}1₇

= 1 6 U 0 tan2 g = 16 U 0(sec2 1) g = 16 tan ₀ = 1 6(tan ) = 16 48 sec1₇

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35. Area of4S RT = 1₂(u cos )(u sin ) =1₂u2sin cos . Area of region S TU =U_{u cos}u u2 {2g{. Let{ = u cos x g{ = u sin x gx for x ₂. Then we obtain

U

u2_{2_{g{ =}U_{u sin x (u sin x) gx = u}2U_sin2_{x gx =}1

2u2(x sin x cos x) + F = 1

2u2cos1({@u) +12{

u2_{2_{+ F} so area of regionS TU = 1₂u2cos1({@u) + {u2 {2u

u cos = 1

2

0 (u2_{+ u cos u sin )}₌1

2u2 12u2sin cos and thus,(area of sector S RU) = (area of 4S RT) + (area of region S TU) =1₂u2.

37. From the graph, it appears that the curve| = {24 {2and the line| = 2 { intersect at about { = d 0=81 and { = 2, with

{2_{4 {}2_{A 2 { on (d> 2). So the area bounded by the curve and the line is} D U2 d {2_{4 {}2_{(2 {)}_{g{ =}U2 d{2 4 {2_g{_2{1 2{2 2 d. To evaluate the integral, we put{ = 2 sin , where ₂ ₂. Then g{ = 2 cos g, { = 2 = sin1_{1 =} 2, and{ = d = = sin1(d@2) 0=416. So U2 d{2 4 {2_g{U@2

4 sin2 (2 cos )(2 cos g) = 4 U@2 sin22 g = 4 U@2 12(1 cos 4) g = 2 1 4sin 4 @2 = 2 2 0 1 4(0=996) 2=81 Thus,D 2=81 2 · 2 1₂ · 222d 1₂d2 2=10.

39. (a) Letw = d sin , gw = d cos g, w = 0 = 0 and w = { = sin1({@d). Then ] { 0 s d2_w2_{gw =} ] sin1({@d) 0 d cos (d cos g) = d2] sin 1_({@d) 0 cos 2_{g = d}2 2 ] sin1_({@d) 0 (1 + cos 2) g = d2 2 k +1 2sin 2 lsin1_({@d) 0 = d 2 2 k

+ sin cos lsin1({@d) 0 = d₂2 sin1{ d + {_d· d2_{2 d 0 = 1 2d2sin1({@d) +12{ d2_{2

(b) The integralU₀{d2 w2gw represents the area under the curve | =d2 w2between the vertical linesw = 0 and w = {. The gure shows that this area consists of a triangular region and a sector of the circlew2+ |2= d2. The triangular region has base{ and heightd2 {2, so its area is1₂{d2 {2. The sector has area1₂d2 =1₂d2sin1({@d).

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SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 309 41.Let the equation of the large circle be{2+ |2= U2. Then the equation of

the small circle is{2+ (| e)2= u2, wheree =U2 u2is the distance between the centers of the circles. The desired area is

D =Uu u e +u2_{2_U2_{2_g{ = 2U₀ue +u2_{2_U2_{2_g{ = 2U₀ue g{ + 2U₀uu2_{2_{g{ 2}Uu 0 U2_{2_g{

The rst integral is just2eu = 2uU2 u2. The second integral represents the area of a quarter-circle of radiusu, so its value is 1₄u2. To evaluate the other integral, note that

U

d2_{2_{g{ =}U_d2_cos2_{g [{ = d sin , g{ = d cos g] =}1 2d2 U (1 + cos 2) g = 1 2d2 +1 2sin 2 + F = 1 2d2( + sin cos ) + F = d2 2 arcsin{d + d2 2 {dd 2_{2 d + F = d 2 2 arcsin{d + { 2 d2_{2_{+ F} Thus, the desired area is

D = 2uU2_u2_{+ 2}1 4u2 U2_{arcsin({@U) + {}_U2_{2u 0 = 2uU2_u2₊1 2u2 U2_{arcsin(u@U) + u}_U2_u2_{= u}_U2_u2₊ 2u2 U2arcsin(u@U) 43.We use cylindrical shells and assume thatU A u. {2= u2 (| U)2 { = ±su2 (| U)2,

soj(|) = 2su2 (| U)2and Y =UU+u Uu 2| · 2 s u2_{(| U)}2_{g| =}Uu u4(x + U) u2_x2_gx _[_where_{x = | U}_] = 4U_uu xu2_x2_{gx + 4U}Uu u u2_x2_gx

wherex = u sin , gx = u cos g in the second integral

= 4k1 3(u2 x2)3@2 lu u+ 4U U@2 @2u2cos2 g = 43(0 0) + 4Uu2 U@2 @2cos2 g = 2Uu2U@2 @2(1 + cos 2) g = 2Uu2 +1 2sin 2 @2 @2= 22Uu2

Another method: Use washers instead of shells, so Y = 8UU₀usu2 |2g| as in Exercise 6.2.63(a), but evaluate the

integral using| = u sin .

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7.4 Integration of Rational Functions by Partial Fractions

1. (a) 2{ ({ + 3)(3{ + 1)= D{ + 3+3{ + 1E (b) 1 {3_{+ 2{}2_{+ {}= 1 {({2_{+ 2{ + 1)}= 1 {({ + 1)2 = D_{ + E_{{ + 1}+ F ({ + 1)2 3. (a) { 4_{+ 1} {5_{+ 4{}3 = {4_{+ 1} {3_({2_{+ 4)}= D_{ + E_{2 + F_{3 + G{ + H_{2_{+ 4} (b) 1 ({2₉₎2 = _{[({ + 3)({ 3)]}1 2 =_{({ + 3)}21_{({ 3)}2 = D_{{ + 3}+_{({ + 3)}E 2 + F_{{ 3}+_{({ 3)}G 2 5. (a) { 4 {4₁ = ({ 4_{1) + 1} {4₁ = 1 + 1

{4₁ [or use long division] = 1 +

1 ({2_1)({2_{+ 1)} = 1 +_{({ 1)({ + 1)({}1 ₂_{+ 1)} = 1 + D_{{ 1}+ E_{{ + 1}+ F{ + G_{₂_{+ 1} (b) w 4_{+ w}2_{+ 1} (w2_{+ 1)(w}2_{+ 4)}2 = Dw + E_w2_{+ 1} + Fw + G_w2_{+ 4} + Hw + I_(w2_{+ 4)}2 7. ] _{ { 6g{ = ] _{({ 6) + 6} { 6 g{ = ] 1 + 6_{{ 6} g{ = { + 6 ln |{ 6| + F 9. { 9

({ + 5)({ 2)= D{ + 5+ E{ 2. Multiply both sides by({ + 5)({ 2) to get { 9 = D({ 2) + E({ + 5)(), or equivalently,{ 9 = (D + E){ 2D + 5E. Equating coefcients of { on each side of the equation gives us 1 = D + E (1) and equating constants gives us9 = 2D + 5E (2). Adding two times (1) to (2) gives us 7 = 7E E = 1 and hence,D = 2. [Alternatively, to nd the coefcients D and E, we may use substitution as follows: substitute 2 for { in () to get7 = 7E E = 1, then substitute 5 for { in () to get 14 = 7D D = 2.] Thus,

] { 9 ({ + 5)({ 2)g{ = ] 2 { + 5+ 1{ 2 g{ = 2 ln |{ + 5| ln |{ 2| + F. 11. 1

{2₁= _{({ + 1)({ 1)}1 = D_{{ + 1}+ E_{{ 1}. Multiply both sides by({ + 1)({ 1) to get 1 = D({ 1) + E({ + 1). Substituting1 for { gives 1 = 2E E =1₂. Substituting1 for { gives 1 = 2D D = 1₂. Thus,

] 3 2 1 {2₁g{ = ] 3 2 1@2 { + 1+ 1@2{ 1 g{ =1 2ln |{ + 1| +12ln |{ 1| 3 2 =1 2ln 4 +12ln 2 1 2ln 3 +12ln 1 =1 2(ln 2 + ln 3 ln 4) or 1₂ln3₂ 13. ] d{ {2_e{g{ = ] d{ {({ e)g{ = ] d { eg{ = d ln |{ e| + F

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SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 311 15. {3 2{2 4 {3_2{2 = 1 + 4 {2_{({ 2)}. Write 4

{2_{({ 2)} = D_{ + E_{2 + F_{{ 2}. Multiplying both sides by{2({ 2) gives 4 = D{({ 2) + E({ 2) + F{2_{. Substituting}_{0 for { gives 4 = 2E E = 2. Substituting 2 for { gives} 4 = 4F F = 1. Equating coefcients of {2_{, we get}_{0 = D + F, so D = 1. Thus,}

] 4 3 {3_2{2₄ {3_2{2 g{ = ] 4 3 1 + 1_{+ 2_{₂ 1_{{ 2} g{ = { + ln |{| 2_{ ln |{ 2| 4 3 =4 + ln 4 1 2 ln 2 3 + ln 3 2 3 0 =7 6 + ln23 17. 4|2 7| 12 |(| + 2)(| 3) = D| + E| + 2+ F| 3 4|2 7| 12 = D(| + 2)(| 3) + E|(| 3) + F|(| + 2). Setting | = 0 gives 12 = 6D, so D = 2. Setting | = 2 gives 18 = 10E, so E = 9

5. Setting| = 3 gives 3 = 15F, so F = 15. Now ] 2 1 4|2_{7| 12} |(| + 2)(| 3)g| = ] 2 1 2 |+ 9@5| + 2+ 1@5| 3 g| =2 ln ||| +9 5ln || + 2| +15ln || 3| 2 1 = 2 ln 2 +9 5ln 4 +15ln 1 2 ln 1 95ln 3 15ln 2 = 2 ln 2 +18 5 ln 2 15ln 2 95ln 3 = 275 ln 2 95ln 3 = 95(3 ln 2 ln 3) = 95ln83 19. 1 ({ + 5)2_{({ 1)} = D_{{ + 5}+_{({ + 5)}E 2 + F_{{ 1} 1 = D({ + 5)({ 1) + E({ 1) + F({ + 5)2. Setting{ = 5 gives 1 = 6E, so E = 1₆. Setting{ = 1 gives 1 = 36F, so F =₃₆1. Setting{ = 2 gives 1 = D(3)(3) + E(3) + F32_{= 9D 3E + 9F = 9D +}1 2 +14 = 9D + 34, so9D = 14 andD = 361. Now ] ₁ ({ + 5)2_{({ 1)}g{ = ] _1@36 { + 5 ({ + 5)1@6 2 + 1@36_{{ 1} g{ = 1₃₆ln |{ + 5| +_{6({ + 5)}1 + 1₃₆ln |{ 1| + F. 21. _{ {2_{+ 4 {}₃ + 0{2 _{+ 0{ + 4} {3 _{+ 4{} 4{ + 4 By long division,{ 3_{+ 4} {2_{+ 4} = { + 4{ + 4_{2_{+ 4} . Thus, ] {3_{+ 4} {2_{+ 4}g{ = ] { + 4{ + 4_{₂_{+ 4} g{ =] { 4{_{₂_{+ 4}+_{₂_{+ 2}4 ₂ g{ = 1₂{2_{4 · 1} 2ln{2+ 4 + 4 · 12tan1{2 + F = 1₂{2_{2 ln({}2_{+ 4) + 2 tan}1{ 2 + F 23. 5{2+ 3{ 2 {3_{+ 2{}2 = 5{ 2_{+ 3{ 2} {2_{({ + 2)} = D_{ + E_{2 + F_{{ + 2}. Multiply by{2({ + 2) to get5{2+ 3{ 2 = D{({ + 2) + E({ + 2) + F{2. Set{ = 2 to get F = 3, and take { = 0 to get E = 1. Equating the coefcients of {2_gives_{5 = D + F D = 2. So} ] _5{2_{+ 3{ 2} {3_{+ 2{}2 g{ = ] ₂ { 1{2 + 3_{{ + 2} g{ = 2 ln |{| + 1_{+ 3 ln |{ + 2| + F.

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N 7.47.4 ININ

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25. 10

({ 1)({2_{+ 9)} = D_{{ 1}+ E{ + F_{2_{+ 9} . Multiply both sides by({ 1)

{2_{+ 9}_{to get}

10 = D{2_{+ 9}_{+ (E{ + F)({ 1) (B). Substituting 1 for { gives 10 = 10D D = 1. Substituting 0 for { gives} 10 = 9D F F = 9(1) 10 = 1. The coefcients of the {2_{-terms in (}_{B) must be equal, so 0 = D + E} E = 1= Thus, ] 10 ({ 1)({2_{+ 9)}g{ = ] 1 { 1+ { 1{2_{+ 9} g{ = ] 1 { 1{2{_{+ 9}_{21_{+ 9} g{ = ln|{ 1| 1 2ln({2+ 9) 13tan1 _{ 3 + F In the second term we used the substitutionx = {2+ 9 and in the last term we used Formula 10. 27. {3+ {2+ 2{ + 1

({2_{+ 1)({}2_{+ 2)} = D{ + E_{2_{+ 1} + F{ + G_{2_{+ 2} . Multiply both sides by

{2_{+ 1}_{2_{+ 2}_{to get} {3_{+ {}2_{+ 2{ + 1 = (D{ + E)}_{2_{+ 2}_{+ (F{ + G)}_{2_{+ 1}

{3_{+ {}2_{+ 2{ + 1 =}_D{3_{+ E{}2_{+ 2D{ + 2E}₊_F{3_{+ G{}2_{+ F{ + G}

{3_{+ {}2_{+ 2{ + 1 = (D + F){}3_{+ (E + G){}2_{+ (2D + F){ + (2E + G). Comparing coefcients gives us the following} system of equations:

D + F = 1 (1) E + G = 1 (2)

2D + F = 2 (3) 2E + G = 1 (4)

Subtracting equation(1) from equation (3) gives us D = 1, so F = 0. Subtracting equation (2) from equation (4) gives us E = 0, so G = 1. Thus, L =] {_({3₂+ {_{+ 1)({}2+ 2{ + 1₂_{+ 2)} g{ =] _{₂{_{+ 1}+_{₂1_{+ 2} g{. For] _{₂{_{+ 1}g{, let x = {2_{+ 1} sogx = 2{ g{ and then ] _{ {2_{+ 1}g{ = 1₂ ] ₁ xgx = 12ln |x| + F = 12ln {2_{+ 1}_{+ F. For}] 1 {2_{+ 2}g{, use Formula 10 withd =2. So ] 1 {2_{+ 2}g{ = ] 1 {2₊₂2g{ = 1₂tan 1{ 2+ F. Thus,L = 1 2ln {2_{+ 1}_{+ 1} 2tan 1{ 2+ F. 29. ] { + 4 {2_{+ 2{ + 5}g{ = ] { + 1 {2_{+ 2{ + 5}g{ + ] 3 {2_{+ 2{ + 5}g{ = 1₂ ] (2{ + 2) g{ {2_{+ 2{ + 5}+ ] 3 g{ ({ + 1)2_{+ 4} = 1₂ln{2+ 2{ + 5 + 3] 2 gx 4(x2_{+ 1)} where{ + 1 = 2x, andg{ = 2 gx = 1₂ln({2_{+ 2{ + 5) + 3} 2tan1x + F = 12ln({2+ 2{ + 5) + 32tan1 { + 1 2 + F 31. 1 {3₁=_{({ 1)({}12_{+ { + 1)} = D_{{ 1}+ E{ + F_{2_{+ { + 1} 1 = D {2_{+ { + 1}_{+ (E{ + F)({ 1).}

Take{ = 1 to get D =1₃. Equating coefcients of{2and then comparing the constant terms, we get0 = 1₃+ E, 1 =1₃ F,

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SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 313 soE = 1₃,F = 2₃ ] ₁ {3₁g{ = ] 1 3 { 1g{ + ] 1 3{ 23 {2_{+ { + 1}g{ =13ln |{ 1| 1₃ ] _{{ + 2} {2_{+ { + 1}g{ = 1 3ln |{ 1| 1₃ ] { + 1@2 {2_{+ { + 1}g{ 1₃ ] (3@2) g{ ({ + 1@2)2_{+ 3@4} = 1 3ln |{ 1| 16ln {2_{+ { + 1}1 2 2 3 tan1 # { +1 2 32 $ + N = 1 3ln |{ 1| 16ln({2+ { + 1) 13tan1 1 3(2{ + 1) + N

33.Letx = {4+ 4{2+ 3> so that gx = (4{3+ 8{) g{ = 4({3+ 2{) g{, { = 0 x = 3, and { = 1 x = 8. Then ] 1 0 {3_{+ 2{} {4_{+ 4{}2_{+ 3}g{ = ] 8 3 1 x 1 4gx = 1 4 ln |x|8₃= 1 4(ln 8 ln 3) = 14ln 83. 35. 1 {({2_{+ 4)}2 = D_{ + E{ + F_{2_{+ 4} + G{ + H_({2_{+ 4)}2 1 = D({2+ 4)2+ (E{ + F){({2+ 4) + (G{ + H){. Setting { = 0 gives1 = 16D, so D =₁₆1. Now compare coefcients.

1 = 1 16({4+ 8{2+ 16) + (E{2+ F{)({2+ 4) + G{2+ H{ 1 = 1 16{4+12{2+ 1 + E{4+ F{3+ 4E{2+ 4F{ + G{2+ H{ 1 =1 16+ E {4_{+ F{}3₊1 2+ 4E + G {2_{+ (4F + H){ + 1} SoE +₁₆1 = 0 E = ₁₆1,F = 0, 1₂+ 4E + G = 0 G = 1₄, and4F + H = 0 H = 0. Thus, ] _g{ {({2_{+ 4)}2 = ] 1 16 { + 1 16{ {2_{+ 4}+ 1 4{ ({2_{+ 4)}2 g{ = 1₁₆ln |{| 1₁₆· 1₂ln{2+ 4 1 4 1₂ 1 {2_{+ 4}+ F = 1₁₆ln |{| 1₃₂ln({2_{+ 4) +} 1 8({2_{+ 4)}+ F 37. {2 3{ + 7 ({2_{4{ + 6)}2 = D{ + E_{2_{4{ + 6}+_({2F{ + G_{4{ + 6)}2 {2 3{ + 7 = (D{ + E)({2 4{ + 6) + F{ + G {2_{3{ + 7 = D{}3_{+ (4D + E){}2_{+ (6D 4E + F){ + (6E + G). So D = 0, 4D + E = 1 E = 1,} 6D 4E + F = 3 F = 1, 6E + G = 7 G = 1. Thus, L =] _({{₂2_{4{ + 6)} 3{ + 7₂g{ =] _{₂_{4{ + 6}1 +_({₂_{4{ + 6)}{ + 1 ₂ g{ =] _{({ 2)}1₂_{+ 2}g{ +] _({₂_{4{ + 6)}{ 2 ₂g{ +] _({₂_{4{ + 6)}3 ₂ g{ = L1+ L2+ L3. [continued]

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314 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION L1= ] 1 ({ 2)2₊₂2g{ = 12tan 1{ 2 2 + F1 L2= 1₂ ] _{2{ 4} ({2_{4{ + 6)}2g{ = 1₂ ] ₁ x2 gx = 1₂ 1_x + F2= _2({₂_{4{ + 6)}1 + F2 L3= 3 ] ₁ k ({ 2)2₊₂2l2g{ = 3 ] ₁ [2(tan2_{+ 1)]}2 2 sec2_g { 2 =s2 tan , g{ =s2 sec2_g = 3 2 4 ] _sec2 sec4g = 3 2 4 ] cos2_{g = 3}2 4 ] ₁ 2(1 + cos 2) g = 3 2 8 +1 2sin 2 + F3= 3 2 8 tan1 { 2 2 + 3 2 8 ₁ 2 · 2 sin cos + F3 = 3 2 8 tan1 { 2 2 + 3 2 8 ·{2{ 2_{4{ + 6}· 2 {2_{4{ + 6}+ F3 = 3 2 8 tan1 { 2 2 +_4({3({ 2)₂_{4{ + 6)}+ F3 So L = L1+ L2+ L3 [F = F1+ F2+ F3] = 1 2tan1 { 2 2 +_2({₂_{4{ + 6)}1 + 3 2 8 tan1 { 2 2 +_4({3({ 2)₂_{4{ + 6)}+ F = 42 8 + 3 2 8 tan1{ 2 2 + 3({ 2) 2_4({₂_{4{ + 6)}+ F = 7 2 8 tan1 { 2 2 +_4({₂3{ 8_{4{ + 6)}+ F 39. Letx ={ + 1. Then { = x2 1, g{ = 2x gx ] _g{ {{ + 1 = ] _{2x gx} (x2_{1) x} = 2 ] _gx x2₁= ln x 1_{x + 1} + F = ln{ + 1 1 { + 1 + 1 + F. 41. Letx ={, so x2= { and g{ = 2x gx. Thus,

] 16 9 _{ { 4g{ = ] 4 3 x x2₄2x gx = 2 ] 4 3 x2 x2₄gx = 2 ] 4 3 1 +_x₂4₄

gx [by long division] = 2 + 8] 4

3

gx

(x + 2)(x 2) (B)

Multiply 1

(x + 2)(x 2) = Dx + 2+ Ex 2by(x + 2)(x 2) to get 1 = D(x 2) + E(x + 2). Equating coefcients we getD + E = 0 and 2D + 2E = 1. Solving gives us E =1₄ andD = 1₄, so 1

(x + 2)(x 2) = 1@4x + 2+ 1@4x 2and (B) is 2 + 8]4 3 1@4 x + 2+ 1@4x 2 gx = 2 + 8k1 4ln |x + 2| +14ln |x 2| l4 3= 2 + k 2 ln |x 2| 2 ln |x + 2|l4 3 = 2 + 2 lnx 2_{x + 2} 4 3= 2 + 2 ln2 6 ln15 = 2 + 2 ln2@6 1@5 = 2 + 2 ln5 3 or 2 + ln ₅ 3 2_{= 2 + ln}25 9

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SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 315 43.Letx =3{2+ 1. Then {2= x3 1, 2{ g{ = 3x2gx ] _{3_g{ 3 {2_{+ 1}= ] (x3₁₎3 2x2gx x = 32 ] (x4_{x) gx} = 3 10x534x2+ F = 103({2+ 1)5@334({2+ 1)2@3+ F

45.If we were to substitutex ={, then the square root would disappear but a cube root would remain. On the other hand, the substitutionx =3{ would eliminate the cube root but leave a square root. We can eliminate both roots by means of the substitutionx =6{. (Note that 6 is the least common multiple of 2 and 3.)

Letx =6{. Then { = x6, sog{ = 6x5gx and{ = x3,3{ = x2. Thus, ] g{ { 3_{= ] 6x5_gx x3_x2 = 6 ] x5 x2_{(x 1)}gx = 6 ] x3 x 1gx = 6] x2_{+ x + 1 + 1} x 1

gx [by long division] = 61 3x3+12x2+ x + ln |x 1| + F = 2{ + 33 { + 66 { + 6 ln6{ 1 + F 47.Letx = h{. Then{ = ln x, g{ = gx x ] _h2{_g{ h2{_{+ 3h}{_{+ 2}= ] _x2_(gx@x) x2_{+ 3x + 2}= ] _{x gx} (x + 1)(x + 2)= ] ₁ x + 1+ 2x + 2 gx = 2 ln |x + 2| ln |x + 1| + F = ln (h_h{_{+ 2)_{+ 1}2 + F

49.Letx = tan w, so that gx = sec2w gw. Then

] _sec2_w tan2_{w + 3 tan w + 2}gw = ] ₁ x2_{+ 3x + 2}gx = ] ₁ (x + 1)(x + 2)gx. Now 1 (x + 1)(x + 2) = Dx + 1+ Ex + 2 1 = D(x + 2) + E(x + 1). Settingx = 2 gives 1 = E, so E = 1. Setting x = 1 gives 1 = D. Thus, ] 1 (x + 1)(x + 2)gx = ] 1 x + 1 1x + 2 gx = ln |x + 1|ln |x + 2|+F = ln |tan w + 1|ln |tan w + 2|+F. 51.Letx = ln({2 { + 2), gy = g{. Then gx = 2{ 1

{2_{{ + 2}g{, y = {, and (by integration by parts) ] ln({2_{{ + 2) g{ = { ln({}2_{{ + 2)}] 2{2 { {2_{{ + 2}g{ = { ln({2 { + 2) ] 2 + { 4 {2_{{ + 2} g{ = { ln({2_{{ + 2) 2{}] 12(2{ 1) {2_{{ + 2}g{ + 7₂ ] g{ ({ 1 2)2+74 = { ln({2_{{ + 2) 2{ 1} 2ln({2 { + 2) + 72 ] 7 2 gx 7 4(x2+ 1) 5 7 where{ 1 2= s 7 2 x, g{ =s7 2 gx, ({ 1 2)2+74=74(x2+ 1) 6 8 = ({ 1 2) ln({2 { + 2) 2{ + 7 tan1_{x + F} = ({ 1 2) ln({2 { + 2) 2{ + 7 tan12{ 1 7 + F

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53. From the graph, we see that the integral will be negative, and we guess

that the area is about the same as that of a rectangle with width2 and height0=3, so we estimate the integral to be (2 · 0=3) = 0=6. Now

1

{2_{2{ 3} =_{({ 3)({ + 1)}1 = D_{{ 3}+ E_{{ + 1}

1 = (D + E){ + D 3E, so D = E and D 3E = 1 D =1 4 andE = 1₄, so the integral becomes

] 2 0 g{ {2_{2{ 3}= 1₄ ] 2 0 g{ { 3 14 ] 2 0 g{ { + 1 = 14 k ln |{ 3| ln |{ + 1|l2 0= 14 ln{ 3_{{ + 1} 2 0 = 1 4 ln1 3 ln 3 = 1 2ln 3 0=55 55. ] _g{ {2_2{ = ] _g{ ({ 1)2₁= ] _gx x2₁ [putx = { 1] = 1₂lnx 1_{x + 1} + F [by Equation 6] =1₂ln{ 2_{ + F 57. (a) Ifw = tan{ 2 , then{

2 = tan1w. The gure gives cos{₂= 1 1 + w2 andsin { 2 = w 1 + w2. (b)cos { = cos 2 · { 2 = 2 cos2{ 2 1 = 2 1 1 + w2 2 1 = _{1 + w}2 ₂ 1 = 1 w_{1 + w}2₂ (c) { 2 = arctan w { = 2 arctan w g{ =1 + w2 2gw 59. Letw = tan({@2). Then, using the expressions in Exercise 57, we have

] 1 3 sin { 4 cos {g{ = ] 1 3 2w 1 + w2 4 1 w2 1 + w2 2 gw 1 + w2 = 2 ] gw 3(2w) 4(1 w2₎ = ] gw 2w2_{+ 3w 2} = ] gw (2w 1)(w + 2)= ] 2 52w 11 15w + 21

gw [using partial fractions] = 1

5 k

ln |2w 1| ln |w + 2|l+ F = 1₅ln2w 1_{w + 2} + F = 1₅ln2 tan ({@2) 1_{tan ({@2) + 2} + F 61. Letw = tan ({@2). Then, by Exercise 57,

] @2 0 sin 2{ 2 + cos { g{ = ] @2 0 2 sin { cos { 2 + cos { g{ = ] 1 0 2 · 2w_{1 + w}₂ · 1 w_{1 + w}2₂ 2 + 1 w_{1 + w}2₂ 2 1 + w2gw = ] 1 0 8w(1 w2₎ (1 + w2₎2 2(1 + w2_{) + (1 w}2₎gw =] 1 0 8w · 1 w2 (w2_{+ 3)(w}2_{+ 1)}2gw = L

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SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 317 If we now letx = w2, then 1 w

2 (w2_{+ 3)(w}2_{+ 1)}2 = 1 x (x + 3)(x + 1)2 = D_{x + 3}+ E_{x + 1}+ F (x + 1)2

1 x = D(x + 1)2_{+ E(x + 3)(x + 1) + F(x + 3). Set x = 1 to get 2 = 2F, so F = 1. Set x = 3 to get 4 = 4D, so} D = 1. Set x = 0 to get 1 = 1 + 3E + 3, so E = 1. So L =]1 0 8w w2_{+ 3} 8w_w2_{+ 1}+_(w28w_{+ 1)}2 gw = 4 ln(w2_{+ 3) 4 ln(w}2_{+ 1)} 4 w2_{+ 1} 1 0 = (4 ln 4 4 ln 2 2) (4 ln 3 0 4) = 8 ln 2 4 ln 2 4 ln 3 + 2 = 4 ln2 3 + 2 63.By long division, { 2_{+ 1} 3{ {2 = 1 + 3{ + 1_{3{ {}2. Now 3{ + 1

3{ {2 = 3{ + 1_{{(3 {)} = D_{ + E_{3 {} 3{ + 1 = D(3 {) + E{. Set { = 3 to get 10 = 3E, so E = 103. Set{ = 0 to get1 = 3D, so D =1₃. Thus, the area is

] 2 1 {2_{+ 1} 3{ {2g{ = ] 2 1 1 + 13 {+ 10 3 3 { g{ ={ +1 3ln |{| 103 ln |3 {| 2 1 =2 +1 3ln 2 0 1 + 0 10 3 ln 2 = 1 +11 3 ln 2 65. S + V

S [(u 1)S V] = DS +(u 1)S VE S + V = D [(u 1)S V] + ES = [(u 1)D + E] S DV (u 1)D + E = 1, D = 1 D = 1, E = u. Now w =] S + V S [(u 1)S V]gS = ] 1 S +(u 1)S Vu gS = ] gS S + uu 1 ] u 1 (u 1)S VgS sow = ln S + u

u 1ln|(u 1)S V| + F. Here u = 0=10 and V = 900, so w = ln S + 0=1

0=9ln|0=9S 900| + F = ln S 19ln(|1| |0=9S + 900|) = ln S 19ln(0=9S + 900) + F. Whenw = 0, S = 10,000, so 0 = ln 10,000 1₉ln(9900) + F. Thus, F = ln 10,000 +1₉ln 9900 [ 10=2326], so our equation becomes w = ln 10,000 ln S +1 9ln 9900 19ln(0=9S + 900) = ln 10 ,000 S + 19ln0=9S + 9009900 = ln 10,_S000+ 1₉ln_{0=1S + 100}1100 = ln 10,_S000+ 1₉ln 11_{S + 1000},000

67. (a) In Maple, we denei({), and then use convert(f,parfrac,x); to obtain

i({) = 24,110@4879_{5{ + 2} 668@323_{2{ + 1} 9438@80_{3{ 7},155+ (22,098{ + 48,935)@260,015_{₂_{+ { + 5} In Mathematica, we use the command Apart, and in Derive, we use Expand.

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318 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION (b) ] i({) g{ = 24,110 4879 ·15ln|5{ + 2| 668323·12ln|2{ + 1| 80,1559438 ·13ln |3{ 7| +_260,0151 ] 22,098{ +12 + 37,886 { +1 2 2 +19 4 g{ + F =24,110 4879 ·15ln|5{ + 2| 668323·12ln|2{ + 1| 80,1559438 ·13ln|3{ 7| + 1 260,015 22,098 ·1 2ln {2_{+ { + 5}_{+ 37,886 ·}t4 19tan1 1 19@4 { +1 2 + F =4822 4879ln|5{ + 2| 334323ln|2{ + 1| 80,1553146 ln|3{ 7| +260,01511,049 ln {2_{+ { + 5} + 75,772 260,01519tan1 k 1 19(2{ + 1) l + F Using a CAS, we get

4822 ln(5{ + 2) 4879 334 ln(2{ + 1)323 3146 ln(3{ 7)80,155 + 11,049 ln({_260,0152+ { + 5)+ 3988 19 260,015 tan1 19 19 (2{ + 1)

The main difference in this answer is that the absolute value signs and the constant of integration have been omitted. Also,

the fractions have been reduced and the denominators rationalized.

69. There are only nitely many values of{ where T({) = 0 (assuming that T is not the zero polynomial). At all other values of {, I ({)@T({) = J({)@T({), so I ({) = J({). In other words, the values of I and J agree at all except perhaps nitely many values of{. By continuity of I and J, the polynomials I and J must agree at those values of { too.

More explicitly: ifd is a value of { such that T(d) = 0, then T({) 6= 0 for all { sufciently close to d. Thus, I (d) = lim_{dI ({) [by continuity ofI ]

= lim_{dJ({) [wheneverT({) 6= 0]

= J(d) [by continuity ofJ]

7.5 Strategy for Integration

1. Letx = sin {, so that gx = cos { g{. ThenUcos {(1 + sin2{) g{ =U(1 + x2) gx = x +1₃x3+ F = sin { +1₃sin3{ + F. 3.

] _{sin { + sec {}

tan { g{ =

] _{sin {}

tan {+ sec {tan {

g{ =](cos { + csc {) g{ = sin { + ln |csc { cot {| + F 5. ] 2 0 2w (w 3)2 gw = ] 1 3 2(x + 3) x2 gx x = w 3, gx = gw =] 1 3 2 x+ 6x2 gx = 2 ln |x| 6_x 1 3 = (2 ln 1 + 6) (2 ln 3 + 2) = 4 2 ln 3 or 4 ln 9

7. Letx = arctan |. Then gx = g|

1 + |2 ] 1 1 harctan | 1 + |2 g| = ] @4 @4h x_{gx =}_hx@4 @4= h@4 h@4.