hitung integral
Transcript of hitung integral
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ffiffis.
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k%d,&HaffiHM&ru
NTE HIT
UNG GRAL
ffiffi, ffi tr-$*'i.tS-ffi*ffi ffi ffi ffi ffiFiHliif,ii:, rii,i.I#:i*ii9rTl,?rWW riFx +,:i::iiisi'r.Bi.{;ii!"{i11,.if.i
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: .
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'*i 'N #.+,,.:: Am.I .rfi-E*#brl . 't.':&r= :t 'rl it 4:"
€- .
utculu*ooi f,1\}
I' 1., " l.: '"t: ] lttuluu gt utqrruutl'td :j,;
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HITUNG Drs.Wardiman
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INT'EGRA! Penerbit: Jln. Tamansiswa PT. HANINDITA Yogyakarta
101 B - GRAHA 6
WIDYA .0274l- 8N22,88193 Ttta lgtak : T. lran Disein kulit muka : Hanindita
@ Hak Pengarang dan Penerbit dilindungi oleh Undang-undang. Cetakan Ketiga, Juli 1987 axclt t olcli :
PT. HANINDITA OFFSET -
Yogyakarta lsi diluar tenggung iawab percetakan.
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UVINV9NSd V.I.VT t]
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'mq?ilet aPor?w-?Potzut uo*uap lonsas tuot uop 'tolns tuof todutos ouoqtapas tuo{
uop ,loos pos-lrrls lnt uollos qnq uoqnqtad untuoqwaw 1n7un lrlltrl-lulnt
! dt
nto, orror)'l!#T::#r# loosloos totoltd-toluad
unlmsala(uant uolop lmlotuaw
un1 ,otutttltly !u! nlng 'umlas
lq uqo1 ,n lp nlnq totu8uay tuldwos ryluntad uop
lnun molas WolDlp uotlo ,orfpwe\tmu wtralnsatl-uotolnsfil otuomat wolop tw[. tunpl otlow'lottalur lutt
to;oselelucd qoruot-tlotuoc'otn[ uoluaqrp apota,u-e?opu noro ,yoay1to4 tuo{ 'onqwad opodaq Wsorl owydt t1o{.uoq-ryKwq uotldom
-eol qeto 'Ws-loos uoryosapfuaw
uqdutouatl opod uolyoq ruorl pteq rfiry uotuep uqueqp tuoi{, uotlopsal-uoqolos?rl uolntaqwed
rytwn uqwntyp otnl wdop 'olqoqa oilsrsntry-Dttsrsoqow tuo( 'pttatul tuu1t1
{uxal qoples nlopd-totwed
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'nqaslet uotulnserl-uotolnsarl '!u!
Zg6l ,ouolof,Soy slpuad uDnuDI uowrP,toA['ao tuoleuau pdop ottlas
unry olrsrsotry-Dilsrsaqow qep -mlopd uqqn,qwa'tuot'rlotuauey
umts qofmsrrs wP o,,.qsoqoy{-otttsTsoqotu
uoVu?qwau r?ns ouofit?t so1tlt1t
-o*rlsoqilt -ur, ?potinu -qa[ucut -1oos
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ISI HVIJVO N
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u?ur?lBtl uvrNvDNad vJ.v) uYrcvo tst u T gvg
I Snvtlnu*snnnu UVSV(I
'IVUCATNT 'l nlueuel4qle-6a1u1 'Z serEelulsnutu-snunu Fos-leos "" ":" "':""
":" ' I St t ""lBos-lBos ""' 8z {v8 ll rsvucarNMooJ,3n 0€
Inlueq 'B p.6.a1ul pup x xP : 'q
le6alul utuls x usoc pup {ntueq
: ,,,sr i ures x xp uup
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u,Elc 0g x ucesoc s€ :
x xp 't 1u.6e1ul I uup
{nlueg ,, 9t, xru s03 xu xp f u* xp xu uls *.'u "o, f xp xu soc xur
6t, '?, ;u-Ea1u1
uutuep uu>;uun8Buatu Isnlqsgns ' 'B Enlllsqns l$unJ
JDqBfie ' '" .g ....p1embuoEpguuEuapFnlnsqns O, gV 'c prEeluJ lrep ;sun3 r;eccd l?uors8.r 'p prEalul
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lrep r6un;-r$un; [BuorsuJn 6t Ig .'e . lsunc Jsl.toIsBJ uBp uls x uBp soc x '
" ., ....[8os_Fos ..... aL €L u
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BAB III
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INTEGRASI FUNGSI HIPERBOLIK l. Rumus-rumus dasar fungsi hiperbolik 2. Rumus-rumus dasar integral dari fungsi hiperbolik 3.
Substitusi fungsi hiperbolik Soal-Soal BAB IV
RUMUS_RUMUS REDUKSI l. Rumus reduksi untuk Binomial Differensial Soal-Soal 2. Rurn us red 87 E7 88 9l 94 97 97 100 102 105 r06 160 lv tuk trigonbme tri k di fferensial Soal-Soal BAB V DAFTAR uksi un PENYELESAIAN SOAL_SOAL UJIAN PUSTAKA
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IgVSN
SNWNU.SNruNU UVSVO ]VHSSINI '1
]vu9:IrNt )tvt nrNlrHt ) trNHrcNt wucsrNt ( BIE6 J
(x) etlg tnqasTp .r (x) ntens alyu (1erta1u1 TrBp J nfuatr lrrp '(x) 1stun3 tuepas Eued ; @ (x) ledunduau nere lngesTp 1erta1u1 alTfe^1rap 'ueffii rtet '5 rd (x) - I '(x) nruCiiJr ) -
TrapuT 1arta1u1 {Bl -raf r,r.rerl ,r.ft + *P ,L tr*) nfens E* = - 5 lstunJ T
qafppe €x) + Eue( lertatuT t/ InqBtaITp IBpTt 1et8un1 t
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letaqas >1e1 ntuafral TrpP J =_(x) € qoluoc .'r*
: ' -tleTo = I$ )(D rx1 - (5 = e Z* ertegl'
ferta1ul IB? -ra?
nluof Tr?p J (x) _ E Zx ledep slTnlTp B;rcas umum ex
+ D tupfsuoc) Jo (uoltertaluT Inf,uO ueqele(uau lertaluy :1e1 nfualral TrBp J tue{ . Buerslp O -Tp 'tuerequae (x),
eTTni}p uetuap {nluaq : ,t (x) xP 'D+rx=*PZ, : Tpaf
rsvucSrNt snnnu-snwnu .z .t;'l";T'tr .T
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v( eJ C+(x1J=xP(x)t ,
'xp { (, ,rf (n+ xp = *on] (
)* (_ xe deltas
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4.
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5. 6. 7. 8. 9. 10. 11. tr: 13. t4. 15. L6. t7. 18. 19.
* ,l;' =
#T + c, m
I -r' F ="'lul*' fuou=#*" +c,a;'o, 1
) [.'ou=.u*.. f"t"udu=-cosu=c. ["."udu=slnu+c. frr"du=rnl"""ul*..
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f"., , du = ln lsln u[ * .. (""" I u du = ln.tse. 1"o".. u + tg Uf r - ctg 1"""' U dU = 1r, | "o""" U u du = cs u + c. 1"."""'udu=-crsu+c. t".. , tg U dU = sec U + C. ("o"." (du1u ) U ctg
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);z;;r lffi ('du U dU = - cosec U* -;
arc ts- {
c' = arc"in! + I l. C ul* C. C. C.
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3+ 'oz
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fr( T B 3Is cas 0 B 'TZ :J z'-Tfr NP = "1 'o.[=+1".
e+zn z B+ xA NP = +nl"r
B+z{l 5,, B+2fr NP = l"r
n * .,2 ,"ll - .rn
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* Z"\
rr"
"rB fi + '3 .SZ +
_e4
l(. l"t 0 + *l ', '92
- zr\ nf"r +
*f ', qoguoc-
rlotuoC uepese;aluad leos-leos ^*(', e)
xp = L* I + b + 'c
/t' e + '3 \, 3_Jt
-xP'- t=
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" EIZ- =c*!l\=xP t/1 e e/r* '3+
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*t, ,.\<r*' -
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5x * 3) dx =\rr*'ro*-.1, =T
x + *2*3*.'*c i o* u.! ,, - *)\l?o- =! *LtZ d* -\*t'' u* = tr*''' ,. !,r* + t6) dx = 3*3 + dx
-Z 5 L2*2 +5x (*3 * 2)\ * 2
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dx *512 * c.
+ a)2 o- = t(e*2+ -24x *16x+C.
' ,C.+4, u. =l(x + 5 - 4*-21 d* = kx +-' lL x +c. e. Illtunglah : ," .t k3 + z)2 3*2' d*; iul
\ 2.- dx; ( 'c)');{;r}
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(a*2 a* ; ta>. x Penyelesalan : Untuk mempermudah penyelesaian, dlmlsalkar, , *3 + 2 =
U, oaka dffferen- slal
darl U adalah : dU = 3 *2d*. (").( (*3 * z)2
t*2 d* =(u' )- ) ( )
4 *3 *2 ur= +u3 J *. = i,*' J +
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213 + c- "r.l(*1 ,",.t# + z)\ *20* = ; =tfut u, =3u''2 *c=i
I (*3 * z>-3.3 *2 d* = i
r*'*z:ttz ru-3 au *c. = i , -r"u-') * c = 3 3 *z>2
+c. (x ,., .,
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\/T *2 d* =+\ (*3 + z) -r/4 . 3 *2 d* = +\" -Ll4 ilJ - + ,u3to ) *C=
t_ 9 (*3 +
zt3l.4 +c.