Làm chủ môn toán trong 30 ngày đại số giải tích 12 quyển 2 (tt)
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LE HONG DUC (Chu bien)
fl6 Hoang Ha - Dao Thi N goc Ha
LAMCHiJ
TRONG
NGAY
QUYEN 2
NBA XUAT BAN D,;J HQC QUOC GIA HA N()I
NANG LUC
•
--==========: DTB < 5,0
5,0 < DTB < 6,5
6,5 < DTB < 8,0
DTB > 8,0
On ~P lfi ki~n thuc cfi
Moi ban quay lai IQ trinh 1
L9 trinh
1
Test I
ClurO'IIC mot
Apd\Jllll91riu1
Kiin thfrc + bai t~p b6 sung IQ trlnh 2
Moi ban quay l~i IQ trinh 2
Test 2
Chll'OII& mol
KiSn thuc + bai tijp b6 sung I~ trinh 3
Moi ban quay l~ lq trinh 3
ldt qui< 8
Lq trinh 3
Test 3
K~t qui ~8
Ap ditag 19 criall 2
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(Phat hanh sach SPBook), theo hop d6ng chuyen nhuong giira Cong ty
TNHH Dich V\l Van h6a Sil pham va cac tac gia Le H6ng Due (Chu bien),
D6 Hoang Ha, Dao Thi Ng9c Ha.
Ban quyen sach thUQC v~ Cong ty NHH
'
,
Bat cu sao chep nao ' khong,duoc S\1' dong y cua Cong ty , ~
[ch VI}
Van b6a SU' pham deu la bat hop phap va vi pham Luat Xuat ban Vist Nam,
'
'
'
luat Ban quyen
Quoc
te, va Cong iroc Berne ve' ban quyen
va SO' hiru tri tue.
V oi sir menh tao nen nguon tai lieu tri thirc, hfru ich dS giup viec hoc ~P tro
nen d8 dang hon, SPBOOK luon mong muon duce cong tac cung cac tac
'
.,
'
,
gia, cac thay co giao tren ca mroc de co nhieu hon nira nhirng cuon sach hay
dUQ'C phat hanh va d€n voi d(k gia n6i chung va cac em hQC sinh n6i rieng
tren ca mroc.
'
'
,
,
Cac thay co giao, tac gia c6 nhu cau viet, xuat ban sach xin vui long lien h~
voi chung toi qua:
Difn thnai: 024. 385. 00012 - 0988. 852. 781
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LOIN6IBAU
CHU'ONG Ill:
NGUYEN HAM, Ti CH PHAN
vA UNG
Dt)NG
9
BAI 1: Nguyen ham
10
BAI 2: Mc)t s6 phuong phap tim nguyen ham. .
56
BAI 3: Tich phan
94
BAI 4: M9t s6 phuong phap tinh tich phan
BAI 5: (fog dung tich phan
de tinh di¢n tich hinh pb~ng
BAI 6: (fog dung tich phan dS tinh thS tich v~t thS
OAP so - HlfONG DAN - LOI GIAI
CHUONG IV:
SO PIIU'C
BAI 1:
se phirc.
210
230
245
313
314
.
BAI 2: Can b~c hai cua s6 phirc va phuong trlnh b~c bai
BAI 3: Dang hrong giac cua s6 plnrc va irng dung
OAP
122
so - HlfONG
DAN - LCH GIAI
340
364
388
Trang5.J
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wi trng dung
BAI4-------------------------------......
MOT s6 PHUONG PHAP TiNH TicH PHAN j
[
A. KIEN THCJC CAN NHO
1. Phuong phap d6i bi@n
s6
Co so cua phirong phap d6i bien s6 la cong thirc sau:
b
p
•
a
J f[u(x)Ju'(x)dx = J f(u)du,
v6i a= u(a) va J3 = u(b).
Tu do, chung ta th§y co hai plnrong phap d6i bi€n:
Phuong phdp I: £)~ tinh tich phan:
b
I=
J g(x)dx
a
Ta thuc hi~n cac buoc:
Buac I: Chon:
+ Phan tich g(x)dx
= f [u(x)]u'(x)dx = f [u(x)]d[u(x)].
+ f)~t u = u(x).
Buac 2: Thuc men phep d6i c~n:
+ Vm x = a thi u = u(a).
+ V6i x = b thl u = u(b).
Buac 3: Khi d6:
b
u(b)
a
u(a)
J g(x)dx = J
f(u)du.
Phuong phap 2: £)€ tinh tich phan:
b
I=
J f(x)dx,
v6i gia tbi~t ham s6 f(x) lien tuc tren [a, b]
a
ta
thuc hien theo cac buoc:
Buac 1: Chon x =
Butrc 2: Liy vi phan dx =
Buac 3: Ta lua chon m9t trong hai hmrng:
HuO'llg 1: N~u tinh duoc cac can a va J3 nrong img theo a va b (voi a=q>(a) va b= q>(J3))
thi ta duce:
I=
J: f( q>(t)).
Huong 2: N~u khong tinh duce d~ dang cac c~ nrong irng theo a va b thi ta lua
chon viec xac dinh nguyen ham, tir d6 suy ra gia tri cua rich phan xac dinh
(trong tnrong hop nay thanh ham s6 cua x).
L
Trang 122
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sa cua t
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,
I
,
;
;1
I
Gitii tlch I 2 - Quydn 2
C/111 y: f>S minh hoa vi~c lua chon mot trong hai huong tren, ta co vi du:
1/2
f f(x)dx, vi~c hra chon An phu x = sint,-
a. Voi I=
0
1t ~
t ~
2
1t
cho phep ta hra chon huong 1,
2
boi khi do:
+ Voi x = 0, suy rat= 0.
'.
1
7t
+ V O'l x = - suy ra t = - .
2'
6
1/3
J f(x)dx,
b. V6i I=
vi~c lua chon An phu x = sint, -
1t ~
2
0
t ~~ta thuong lua chon huong 2, boi
2
khi d6:
+ Voi x = 0, suy rat= 0.
+ V &i x =
i,
ta khong chi ra dtroc
s6 do g6c t.
2. Phuong phap tich phan tirng phin
CO' so cua phucmg phap tich phan tung phfu1 lA cong tlnrc sau:
J u(x).v'(x).dx
b
b
=
u(x).v(x)I.
J v(x).u'(x).dx.
b
-
(
1)
b
B~ sit dung ( l) trong vi~c tinh tich phan I=
J f (x)dx
ta thuc hien cac bircc:
a
Buac 1: Bien d6i rich phan ban d~u v€ dang:
b
b
=
I= Jt(x)dx
a
Jf1(x).f2(x)dx.
•
Bu&c 2: E>~t:
u = f1 (x)
{ dv = fi{x)dx
=>
{du
v ·
Bu&c 3: Khi d6:
I = UV
I
b
b
a -
f vdu .
•
Chu y: Khi SU' dung phuong phap tich phan
rung phan dS tinh tich
phan chung ta c§n tuan thu
cac nguyen tic sau:
1. Lua chon phep d~t dv sao cho v diroc xac djnh m{>t each da dang.
b
f
2. Tich phan vdu duoc xac dinh m9t each d~ dang hem so voi I.
a
3. Chung ta c!n nhc cac dang
CCY
ban sau:
Dang 1: Tich phan I= Jxci.lnxdx, voi aeR\{-1} khi do d~t u = lnx.
Dang 2: Tich phan 1 = JP(x)ecixdx (hoac I= JP(x)eQJCdx) v&i P la m9t da thirc thuoc R[X) va
aeR" khi c16 d~t u = P(x).
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wi trng dung
Dang 3: Tich phan I= JP(x)sina.xdx (hojic JP(x)cosa.xdx) voi P la mc}t da thirc thuoc R[X]
va a.eR" khi d6 d~t u = P(x).
Dang 4: Tich phan I = feaxcosbxdx (hoac feaxsin(bx)dx vci a, b f. 0 khi d6 d~t u
=
cos(bx)
(ho~c u = sin(bx)).
B. PHAN LO,.;.I v A PHUONG PHAP GIA.I cAc D,.;.NG TOAN
Phuong phap d8i bil11 d{lng 1
Phuong phap
Sfr dung ki€n thirc trong phuong phap 1 cua phin phuong phap d6i bi~n s6.
Vi d1.1 l: Tinh tich phan:
I
rt/3
I=
11/6
d
cosx. x
.
sin 2 x - 5 sin x + 6
f>~t t = sinx, suy ra dt = cosx.dx
f>6i c~:
1t
l
6
2
7t
.fj
•x=-=>t=-.
•x=-=>t=-.
3
2
Taco:
cosxdx
dt
sin? x-5sinx + 6
t2 -5t+ 6
------=
dt
=---(t-2)(t-3)
= (~ ~)dt=
+
t-3
, d,o:
tu
{A+
B=0
-2A-3B=l
Suy ra:
~
t-2
[(A+B)t-2A-3B]dt
(t-2)(t-3)
{AB=-1
=l .
(-1- __ l_)
cos xdx
=
sin 2 x - 5 sin x + 6
t- 3
t-
dt.
2
Khid6:
I=
J"r(-1- 112
Lrrang
t-3
124
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_l_)dt =
t-2
lnl t-311.[J,2
t-2
1/2
= ln 3(6-.fi).
5(4-.fi)
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Sung
Cac em hQC sinh hay:
l(lO:
+ Tao ra m<)t bai toan nrong nr.
+ Tao ra cau hoi trllc nghiem tu n9i dung vi du tren.
C/111 j: Ti~p theo, chung ta xet vi du voi viec hra chon
•
Vi di} 2: Tinh tlch phan I =
D~t t = 1
+ sin2x,
J cos x.si_n
rt/2
•
3
0
1 +sm x
2
x.
dx
An phu t bing ca m§u s6.
.
suy ra dt = 2sinx.cosx.dx.
Df>i c~:
• x=O=>t=
1.
7t
• x= - => t=2.
2
Tac6:
cosx.sin3 xdx = sin2x.cosx.sinxdx
l+sin2x
I+ sin 2 x
= (t-l)dt
2t
=
_!_(t -!)dt.
2
t
Khi d6:
1
2
I= -
2
1
f Cl--)dt
t
1
1
=-(t-lnltl)I
2
1
2
2
= -(l-ln2).
1
Sang too: Cac em hQC sinh hay:
+ Tao ra mot bai toan nrong tu.
+ Tao
•vidi} 3:
ra cau hoi trllc nghiern tu n(>i dung vi du tren.
dx
J 2sinsmx2+x.cos'
.
x
rt/6
Tinh tich phan I=
0
•
2
D~t t = 2sin2x + cos-x, suy ra dt = sin2xdx
J:'\,t.
"
uo1
can:
• x=O=>t=l.
7t
5
.x=-=>t=-.
6
4
Khid6:
514
I=
dt
J-t
= In It 1
S/4
1
5
= In4.
l
Sang tqo: Cac em hQC sinh hay:
+ Tao ra m<)t bai toan nrong tu,
+ Tao ra cau hoi trllc nghiem tu n<)i dung vi du tren.
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MALODA.VN - KHO SÁCH QUÝ, THI HẾT BÍ
Chuang Ill - Nguyen ham, tlch phiin
Vi du 4: Tinh tich phan I =
'}2 ~
ex +2
O
r.--=
..2
...Jex + 2 => r
¥
f)~t t =
.
2tdt
ex + 2, suy ra: 2tdt = e'dx ~ dx = -2t -2
=
D6ic~:
• x=O=>t=JJ.
• x = ln2 => t = 2.
Khid6:
wi trng dung
•
~~r
= _1_ ln
= _1_ In (2-J2)( Jj + J2)
2J2 ~.,13
2J2 (2+J2)(fi-J2)'
I= J2~
.J3t2-2
Sang tpo: Cac em hQC sinh hay:
+ Tao ra mQt bai toan tucmg nr.
+ Tao ra cau h6i tr&c nghiem tu n9i dung vi du tren .
.J3
Vi di} 5: Tinh tich phan I=
J x .J1 + x dx.
5
2
0
D~t t = .J1 + x2 , suy ra: t2 = 1 + x2 => 2tdt = 2xdx
D6i c~n:
• x=O=>t=
• x = fi
Khi d6:
1.
=> t = 2.
2
2
I= J(t2-t)2t2dt=
t
Sang
(
J(t6-2t4+t2)dt=
f(J.O:
1
25+-t31 ) 2 =-.
848
-t7--t
7
t
5
3
1
105
Cac em hoc sinh hay:
+ Tao ra m9t bai toan tuong tu,
+ Tao ra cau hoi tr&c nghiem tu nQi dung vi du tren.
11/2
Vi du 6: Tinh tich phan I=
J .Jcosx.sin
3
x.dx.
0
D~t t = .Jcosx thl t2 = cosx suy ra 2t.dt = +sinx.dx.
D6ic~:
• x=O=>t=l.
• x =-
1t
2
=> t = 0.
Lrrang126
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Ta co:
.Jcos x sirr'x.dx = sin'x .Jcos x .sinx.dx
=
= (1 - cos'x) .Jcos x sinx.dx
(1 - t4).t.< - 2tdt) = 2(t6 - r)dt.
J
Kb 1. db : I = 2 0(t 6 - t2 )dt = 2( -1 t7 - -1 .3
r) Io = -s .
1
7
3
I
21
Sang tao: Cac em hQC sinh hay:
+ Tao ra met bai toan tuong tu.
+ Tao ra cfiu hoi tr&c nghiem tu n<)i dung vi d1,1 tren.
Ji
J ~x +I .
• Vi dy 7: Tinh tich phan I=
JJX
D~t t =
,2
r
=
2
..f x2 + 1 suy ra:
2
x + 1 => t.dt
=
x.dx ~ dx
t.dt
= -
x
=>
tdt
d:x
dt
= -2- = -2x t
t -1
x~
•
D6i cin:
= .J3 => t = 2.
x = ./8 => t = 3.
• x
•
Khi d6: I=
J .z., = .!. lnl t- l 1
-1
2 t2
2
t
+1
3
2
=
.!.1n I.
2
2
Sang tao: Cac em hQC sinh hay:
+ Tao ra m(:>t bai toan nrong nr,
+ Tao ra cau hoi tric nghiem tu n9i dung vi d1,1 tren .
Vi di} 8: Tinh tich phan I
=
f ~3d .
.fi
3
o
Datt=
·
1 + x2
+ 1 => t3 = x2 + 1 => 3t2dt = 2xdx <:::> dx =
3t2dt
2x
.
r.~· can:
"
:uo1
.x=O=:>t=l.
• x=
Ta c6:
Ji => t = 2.
x'dx
x3 .3t2dt
=
Khi d6: I=
2xt
J (t -t)dt
2
l
4
I
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3
3
2
2
= -t(t'- l)dt = -(t4
=
I
2
(f-~)
5
di 6. 0-
2
2
I
-
t)dt
141
20
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MALODA.VN - KHO SÁCH QUÝ, THI HẾT BÍ
Chuang Ill - Nguyen ham, tlch phiin
Sung
l(lO:
Cac em
hQC
wi trng dung
sinh hay:
+ Tao ra m<)t bai toan nrong nr.
+ Tao ra cau hoi trllc nghiem tu n9i dung vi du tren.
, ' y:' T.1..
' cac
' vii d\l v01
'. dang I =
C,nu
rep th eo, c h'ung ta xet
j
• Vi d\l 9: Tinh tich phan I = ,j
0
dx
f
dx
.j(x +a)ex + b)
.
.
(x + l)(x + 2)
a. V 6i a = 0 va b = 2.
b. Voi a=- 5 va b =- 3.
x+1 >0
a. Vi xe(O; 2) nen: {
.
x+2>0
fl~t t = J;:+I. +
dt =
(
.Jx + 2 suy ra:
1
+
1
) dx = ( ,J;+i + ~
)dx <=>
dx
2J;:+i
2.Jx + 2
2.j(x + l)ex + 2)
.Jex+ l)ex + 2)
2dt
t
fl6i c~n:
• x=l=>t=l+.J2.
• x=2=>t=2+./3.
Khi d6:
2
I=
!
dx
2+Jj
dt
2+Jj
.Jex+ l)(x + 2) = 2 ,.~-1 = 2lnlt"'+.Ji
b. Vi xe( -5; -3) nen:
2 + J3
= 2ln l + J2 .
x + 1 <0
.
{ x+2<0
D~t t = .J-ex + l) + .j-(x + 2) suy ra:
d -[
1
l
t-
2.j-(x + 1)
2.j-(x + 2)
<=>
.:..
£) 01
dx
2dt
=-.j(x + l)(x + 2)
t
Jd _
x- -
[.j-ex+l)+.j-(x+2)]dx
2.j(x + t)ex + 2)
·
"
can:
• x=-5=>t=2+./3.
• x=-3:=>t=l+.J2.
Khi do:
-3
dx
I = L-.J;=(x=+=l=)e=x=+=2==-)= - 2
L
Trang 128
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it t
1+./1
dt
I
2 + fj
= 2lnltl i+../2 = 2ln 1 + J2
2+Jj
.
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miin TOAN trong 30 ngay - Giai tlch I 2 - Quydn 2
b
C/u,
y: Nhir vay, d~ tinh
tich phan: I=
f '1(x + a)(x
dx
,
+ f3)
a
ta dn xet xem trong (a, b) dAu cua cac x + a. va x + f3 la fun hay dirong. DiSu nay chung ta
phan tlch ky trong ph!n ti ch phan ham s6
ti bei c6 th~ sir dung bi~n d6i Euler:
vo
'1(x + l)(x +2)
= t(x +
se
1).
Sang tao: Cac em hQC sinh hay:
+ Tao ra mQt bai toan nrong tu,
+ Tao ra cau hoi tric nghicm nr nQi dung vi d1,1 tren.
•
Vi di} 10: Tinh tich phan:
I=
2r .
tt/2
2dx
2smx-cosx+l
.
x
1
Datt= tan-2, ta duce: dt = -2.
1
2 x
1
2
2dt
dx =- .(1 + tan - )dx =- .(1 + t )dx <=> dx = --2 •
v
cos" ~
2
2
1 +t
2
2
D6i c~:
·x=~=>t=l.
2
27t
• x=
=> t = fj .
3
Khid6:
4dt
I= JjJJ
1 + t2
- JjJ3 2dt -2 JjJ3 d(t + 1) - In
I _i!__ l-t2
+l
I l2 +2t
I (t+l)2
-:
1 + t2 1 + t2
1_!_]"3 = In
1+2)1
3-J?,
fj +2.
Sang tao: Cac em hQC sinh hay:
+ Tao ra mQt bai toan tucmg tu,
+ Tao ra cau hoi tric nghiern tu noi dung vi d1,1 tren.
Chu y: Ti~p theo, chung ta se xet mot vii vi d1,1 voi viec hra chon
in phu dua tren cac
tinh ch~t
cua ham s6 duoi d!u tich phan.
tt/2
Vi di} 11: Tinh tich phan I=
J
cosx
dx.
o -./8-2sin2 x
D~t t = sinx thi dt = cosxdx.
"
can:
• x = 0 => t= 0.
.l.·
f)01
rt
•X=-=>t=l.
2
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MALODA.VN - KHO SÁCH QUÝ, THI HẾT BÍ
Chuang Ill - Nguyen ham, tlch phiin
Khi do: I = ft
·
dt
=
_1_J1
dt
Ji o '14-t2
o Js-2t2
wi trng dung
•
DJt t = 2sinu thl dt = 2cosudu.
D<3i can:
• t = 0 => u = 0.
1t
• t=l=:>u=-.
Khi d6: I=
6
-·ho-T
=
2cosudu
J4-4sin2u
_l_T
J2
du = -•-ur/6 = ~-
J2 °
0
6h
NJ,~,, xet: Nhir v~y, n~u chung ta da c6 duce chut kinh nghiem thi co th~ chi
phep d<3i bien, cu th~:
= 2sinu
D~t sinx
l•
f) 01
can:
thi cosxdx
d.n thuc hien IDQt
= 2cosudu.
A
0 => u = 0.
• X =
1t
• x=-
2
Khi do: I=
=:>u=-.
f
2
tt/2
1t
6
f
d
1
cosu u --
o '18-8sin2 u
n/6
d
l
cosu u =-
hoJI-sin2u
1
f du =-ul~'6
11./6
lio
Ji
= ~-
6Ji
Sang tao: Cac em hQC sinh hay:
+ Tao ra m¢t bai toan nrong nr.
+ Tao ra cau hoi tr&c nghiem tu n9i dung vi du tren.
Vi dy 12: Tinh tich phan I =
I
4
Ox
x~
.
+x +l
f)~t t = x2 thi dt = 2xdx.
l·
,..
f) 01
can:
• x = 0 => t= 0.
• x=l=:>t=l.
. d'. I--- 1 JI dt
Kh IO
.
2 0 t2 +t+l
_
l
Datt+ ·
2
_ 1 Jl
dt
2 0 (t+.!.)2
2
--
J3
,
= -tanu
thi:
2
J3
+l.
du
4
.fj
2.
dt= -.-= -(1 + tan-u)du.
2 cos2 u
2
D6ic~:
• t=O=:>u=
• t=l::>u=-.
L
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6
1t
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Khi d6:
.fj
2
+ tan u)du
1 "'3 -(1
I=2
J
2
I tan
1116
u+
2
4
I
"'3
1
J du =-ul
.fj
1
=-
.fj
nt6
,.,3
1t
=-.
6J3
1('6
4
Sang t(JO: Cac em hoc sinh hay:
+ Tao ra m(:>t bai toan nrong
tu,
+ Tao ra cau h6i tric nghiem tu nQi dung vi d1,1 trcn.
=
Vi d\l 13: Tinh tich phan I
J .J9 + 3 x dx .
2
1
2
X
D~t x .fj = 3tant thl x = .fj tant, t e(D6i c~:
• x=l~t=-.
1t ,
2
2:.) suy ra dx =
2
~
cos
dt.
t
1t
6
•x=.fj~t=2:..
4
Khi d6:
+ 9tan x
I ~9tan
x
-
.fj
2
n/4
I-
2
- "'14
. -2-
t
COS
"'6
I
dt -
ttl6
"I'4
.fidt
COS
.
t.sm
2
l = ,,,,6
fj cos t.dt
(1-sin2 t).sin2 t ·
D~t v = cost, suy ra dv = - sintdt
A·
,..
D OlC~:
1t
1
6
2
1t
Ji
4
2
.t=-~v=-.
•t=-~v=-.
Khi d6:
f
t:
lt/4
I=
1116
v3cost.dt
. 2 t ) .sm. 2 t
(1 - sm
J (1v
= .fj ./212
112
1)
-2--2-
v -1
=
t:
Ji 12
f
112
(1
v3dv
-
dv = .fj
v 2) . v 2
=
.Ji /2
f .fie-) __ 1 _)dv
1
"2
v2
( ----1n
J t 1v-ll)Ji
v 2 v+l
12
v2 -
= .fj[2-
112
1 .fi.+2
J2 + -ln-----=-J.
2
3(2-.fi.)
Sang lflO: Cac em hQC sinh hay:
+ Tao ra m(:>t bai toan tuong
+ Tao ra cau h6i tric nghiem
Trai nghiim Ht sinh thai Hoc
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MALODA.VN - KHO SÁCH QUÝ, THI HẾT BÍ
Chuang Ill - Nguyen ham, tlch phiin
wi trng dung
Phuong phap i/Ji biJn dtJng 2
Phuong phap
Su dung lci~n tlnrc trong phtrong phap 2 cua phful phuong phap d6i bi~n
s6.
112
Vi du l: Tinh tich phan I=
J .J1-x dx.
2
0
Ta c6 th~ trlnh bay theo cac each sau:
Cach 1: f)~t x = sint, -
2:.
5t5
2
2:. suy ra dx = cost.dt
2
nA·
"
.uo1 can:
+V6ix=Othit=O
,.
l hl1t=-.1t
+ Vmx=-t
2
6
Khid6:
~6
~
J .J1-sin t.cost.dt
I=
=
2
0
J cos t.dt
2
0
1 +
= -(t
2
1 .
I "''6
-sin2t)
2
1~
=(l+cos2t).dt
2 0
J
1 1t + -).
J3
= -(2 6
0
4
Cacb 2: f>~t x = cost, te [O; 1t] suy ra dx = -sint.dt.
A·
f)01
"
can:
+ V oi x = 0 thi t =
n .
2
'.
l h'1t=-. 7t
+ Vmx=-t
2
3
Khi d6:
rt/3
J J1-cos t.sint.dt
I=-
2
11/3
J sini t.dt
=-
11/2
rt/2
}
1
2
2
.
=--(t- -sm2t)I
x/3
n/2
1
7t
l
x/3
2
r,/2
=--
J (1-cos2t)dt
J3
= -(- + -).
2 6
4
Sa11g l(JO: Cac em hQC sinh hay:
+ Tao ra m<)t bai toan tuong tu.
+ Tao ra cau hoi tric nghiem nr n{>i dung vi du tren.
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132
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Lam chu miin TOAN trong 30 ngay - Giai tlch I 2 - Quydn 2
2
Vi du 2: Tinh tich phan I=
I x ..f 4- x dx.
2
2
I
D~t x
= 2sint,
-
5: t 5:
1t
2
1t
2
suy ra dx
= 2costdt
D6i c~n:
+ V&i x = 1 thi t=
1t
6
+ V6i x = 2 thl l = 2:.
2
rr.12
rr.12
J 4sin t...f4-4sin t.2cost.dt=4
Khid6:I=
2
J sin' Zt.dt
2
1'.16
,ell
=2
rr.16
l .
)
=2 ( t--sm4t
«n
4
../3) .
=2 ( -1t + 3
1116
J (l-cos4t).dt
11(6
8
Sa11g t90: Cac em hQC sinh hay:
+ Tao ra mot bai toan nrong nr.
+ Tao ra cau b6i tric nghiem tu noi dung vi du tren.
Vi dy 3: Tinh tich phan I=
'Yo ~1-x .
Ta c6 thB trlnh bay theo cac each sau:
Cach 1: D~t x = sint, _,?: < t < 2: suy ra dx = cost.dt.
2
2
D6i c~n:
• x = 0 => t= 0.
1
1t
• x=-=:>t=-.
2
6
Khi do: I=
f .J1cos- sint.dt t = f coscostt.dt = f dt = tl:'6
rr./6
2
0
tt./6
rr./6
O
O
= 1t6.
Cdch 2: D~t x = cost, te(O; 1t) suy ra dx = -sint.dt.
f)f>i can:
+ vui x = o thi
t=
2
.l..'
+ Vvl
X = -1 thi1 t =
2
Khi do: I =-T
«n
Sang
1t .
-1t .
3
sin t.dt = -Tsin t.dt _
«n sin t
.Ji - cos' t
-T
rr.11
dt =-
tl•'3
"'2
=- (
1t _ 1t)
3
2
=
1t.
6
Cac em hQC sinh hay:
+ Tao ra m{>t bai toan nrong tu,
+ Tao ra cau h6i tric nghiern tu n9i dung vi du tren.
t(JO:
Trai nghiim Ht sinh thai Hoc
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MALODA.VN - KHO SÁCH QUÝ, THI HẾT BÍ
Chuang Ill - Nguyen ham, tlch phiin
•
J• ~(l-x2)3
dx
b
Vi dy 4: Tinh tich phan I =
J3 .
a. V 6i a = 0 va b =
nu~t
a.
wi trng dung
b. Voi a= Ova b
2
= .!. .
3
·
x = smt,
- -1t < t < -1t suy ra dx = cost. dt.
2
2
... "
f) OIC~:
+ V6i x
= 0 thl t = 0.
+ VO'l,. x =
J3
-
hi t = -1t .
t
2
3
Khid6:
Iin
J
1 -_
dx
---;::::===
I
o \I (1-
J
1113
=
2 3
I
d
cost. t
o V (1-
X )
·
Slll
2
d
J cost. d = J--t= tantl"'J
o
11/3
=
t) 3
n/3
t
0
COS
3t
O COS
2t
=
J3.
b. Ta lira chon huong xac dinh nguyen ham:
[ J \I'(t-xdx
]
=f cos t.dt =f ~
= tant+c= sin t
cos3t
cosi t
cost
2)3
+ c.
x
J1-x2
X•Slnl
113
Khi d6: I=
+c-
1
x
J1-x2
0
-
2-fi. ·
Ch11 j:
1. Trong loi giai cua cau b)
sa d1 ta c6 ~(l-x )3 = cos3t va tant =
2
la boi:
x
.J1- x
2
--
1t
{.Jcos
1t
2
< t < - ~ cost> 0 ~
2
t
= cos t
.
cost=Jl-sin2t=Jl-x2
2
Tuy nhien, chung ta hoan toan c6 th~ thirc hien theo huong l, thijt vay:
-!
IIJ
1-
[
2. cAn d~
11
dx
]
~(l-x2)3 xcainl
-
...!.
0<:1 <:11 , vo, sin 11
y r~ng, dung
!
dt - tant II
cos2 t 1°
3
huong 2 th1 khi d6i bi€n
s8 ck don anh t H
x
= q>(t) dS d6i k€t qua tu t
sang x, con voi huong 1 thi khong cftn.
Sang tao: Cao em
bQC
sinh hay:
+ Tao ra mQt bai toan nrong nr.
+ Tao ra cau
L
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f
4/./i~dx
.. Vi dy 5: Tinh rich phan I=
x
x
2
2
sin2t'
Dat x = -·
0~t~
E>6i c~n:
• x=2=>t=~.
1t
suy ra dx =
-
4
•
3
•
4cos2t.dt
.
sin22t
-
4
4
7t
Jj
6
•x=-=>t=Khid6:
I
=_
4cos2t.dt
J
"'6
•
2
sin 2t
11/4
--
_
8
sin" 2t
1
,r/6 2cot
2t.
,r/4
--
J
J cosi Zt.dt =-- J (l+cos4t).dt
11/6
=-
2
rr./4
4cos2t.dt
2
sin 2t
8
sirr' 2t
1
11/6
1
.
2
rr./4
4
1t
,r/6
=--(t+ -sm4t)I
=x/4
24
Jj
- -.
16
Sang tqo: Cac em hQC sinh hay:
+ Tao ra m{>t bai toan tuong tu,
+ Tao ra cau hoi tric nghiem tu n{>i dung vi du tren.
Vi d~ 6: Tinh tich phan I =
21./i
dx
J ~ .
2 XX-)
_
1
E>~t x =
-.-
sin t
n
te(O; - ) suy ra dx =
,
2
-
cost
-.-dt.
sini t
E>f>i can:
1t
• x=l=>t=-.
2
2
7t
Jj
3
•X=-=>t=-.
Khid6:
1
tdt
1112 --.-2-cos
I=
J
sm t
rr./3_1_~
sin t
=-
1 -1
sin2 t
Trai nghiim Ht sinh thai Hoc
Website: maloda.vn
,r/2
J dt =-tn/3l
rr./2
7t
=-6.
11/3
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MALODA.VN - KHO SÁCH QUÝ, THI HẾT BÍ
Chuang Ill - Nguyen ham, tlch phiin
wi trng dung
Chu j: Cung co th~ sfr d1,U1g phep d6i bien t = _.!._ , b~g each vik
x
dx
21./5
I
I=
2
./512
1
x2~l-
dt
_
I
=
.
112 ~
x2
)., tiep
,I.
R01
t\JC
.
su• dung ph'ep d;..01 b'I.
ien t = sinu,
u e (0 ; 1t) , ta ducc:
2
1113
I=-
I du = - u
1113
1 1116
1t
=-
6
11/6
.
Sang tao: Cac em hoc sinh hay:
+ Tao ra mQt bai toan nrong nr.
+ Tao ra cau hoi trllc nghiem tu nQi dung vi du tren.
Vi d\17: Tinh tich phan I=
a.V6ia=J2
b
2d
a
2
J ~.
x -1
vab=.Js.
b.V6ia=-2vab=-Ji.
Ji; Js]
a. D~t x =-.-1-, vi xe[
sin 2t
nen
Fi ~-.sin-2t1-~ Js <=>
~ s sin2t s ~
-n
v5
do d6, d~ thoa man tinh ch~t don anh ta chon O < t < ~ .
4
cos 2 tdt
, I
h h .
r
dinh
h,
Ta co, dx = - 2 sin!
, va ua c on ucmg xac .
nguyen am:
A
2t
f
x2dx
=-f
2dt
sin32t
.Jx2-1
=-f2(cos2t+sin2t)2dt
8sin3tcos3t
lf
l
1
2
( cott. -- 2 + tant. -+
)dt
4
sin t
cos2 t
sin tcos t
= - -
lf
2
1
(cott.-- I +tant.-- 1 + ---)
4
sin" t
cos' t
tant cos! t
=-
-
= - .!.. [ - fcott.d(cott) + ftant.d(tant) + 2J d(tgt)]
4
=-
tant
..!.. ( - ..!.. cot2t +
4
2
.!.. tan2t + 2lnjtantl)
= .!..x.Jx2 -1 - .!..1njx- .Jx2 -1
2
+ C=
2
2
.!. (cor't - tan2t) 8
..!..1njtantj + C
2
I+ C.
Khi d6:
1 rt=:
1
rrr: 1,/s =-(2v5-v2)+v5ln
l
t:
r:
r:
I=(-xvx--l--lnlx-vx--11)
2
2
Ji
2
Ji-1 .
Js-2
b. DJ nghi ban doc t¥ lam - Theo hirong d6i $1.
L
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Lam chu miin TOAN trong 30 ngay - Giai tlch I 2 - Quydn 2
C/111 y: Trong Joi giai cau a) so di ta c6:
cot2t- tan2t = 4x Jx2 -1 va tant = x - Jx2 -l
la boi:
cot2t-tan2t=
cos4t-sin4t
=
cosi r.sini t
4cos2t = 4J1-sin22t
sin22t
sin22t
tant= _si_n_t
= sin2t
= J-cos2t = 1-.J1-sin22t
sin2t
sin2t
cost
sin t cost
=-4-
~
sin2t
=x-
=4xJx2-I.
~~-i
Jx2-l.
Sang tao: Cac em hQC sinh hay:
+ Tao ra mot bai toan nrong nr.
+ Tao ra cau h6i tric nghiem tu noi dung vi du tren.
I
Vi d\l 8: Tinh tich phan I=
f xJl + x'dx .
0
1t
1t
2
2
I>~t x = tant, - - < t < - suy ra dx =
]:'\.. .
uOI
--dt .
cos'
t
~
can:
• x
= 0 :=) t = 0.
• x=l=:>t=-.
1t
4
Khid6:
I=
J tan t.Jl + tan
"14
2
dt
t.--2
cos t
O
= - "J14 d( cos4 t) = ~"14 = 2-.J2 -1
3 cos' t O
3
O cos t
Slz11g tao: Cac em hQC sinh hay:
+ Tao ra m9t bai toan urong nr.
+ Tao ra cau h6i tric nghiern nr n9i dung vi du tren.
.. Vi dy 9: Tinh tich phan I =
j J(l+x
dx
)3
a
a. V 6i a = 0 va b =
v
f)~tx=tant,--
.
2
J3 .
b.Vma=Ovab=2.
dt
1t
'It
2
cos'
t
a. I>6i c~n:
• x = 0 =:> t = 0.
•x=../3:=)t=..::..
3
Trai nghiim Ht sinh thai Hoc
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MALODA.VN - KHO SÁCH QUÝ, THI HẾT BÍ
Chuang Ill - Nguyen ham, tlch phiin
wi trng dung
dt
rr./3
J
=
.
~3
I
2 .
cos t.dt = smt O«n = ".,
0
b. Ta hra chon huong xac dinh nguyen ham:
I
dx
~(l+x2)3
=
2
I cos? tdt
2
cost
.
2
d.x
J
I
x
-'1===
+ C.
= cost.dt = smt + C =
'11
2
x
Khi d6: I= o~~;=(l=+=x=2)=3
=
+ x2 o -
Clu,j:
1. Trong Joi giai cau b) s6' di ta c6
I+x"
.Js.
Q
= cost va sint
l+x2
=
.Q
la boi:
l+x2
Jcos t = cost
2
1t
1t
2
2
-- < t < - => cost> 0 =>
{ sin t = tan t.cos t =
2. Phu011g phap tren duce ap dung
f
I=
dB giai
x
.
c=»
vl + x2
bai toan t6ng quat:
d.x
, voi keZ.
~(a2 + x2)2k+1
f(IO: Cac em hQC sinh hay:
+ Tao ra m9t bai toan tuong nr.
+ Tao ra cau hoi tr~c nghiem tu n9i dung vi d1,1 tren.
Su11g
Vi du. 10: Tinh tich phan I=
j ~a+a-xx d.x, voi a> 0.
_0
D~t x = a.cos2t, te(O;
; ] suy ra dx = -2a.sin2t.dt.
D6i can:
7t
• x=-a=:>t=-.
2
1t
.x=O=>t=-.
4
Khi d6:
I=
J
n,
"14
2
=--4a
L
· (- 2a.sin2t.dt)
a - a. cos 2t
a+ a cos2t
f cos
11./4
n/2
Trang 138
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2
t.dt=-2a
= - 2a
J I cou I .sin2t.dt
"14
11,?
f (l+cos2t)dt=-2a
11./2
11/4
(
t+21sin2t )"''2 =a ( 1- ~ ) .
,r./4
Trai nghi/m Hi sinh thai Hoc di 6.0-SPBook
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MALODA.VN - KHO SÁCH QUÝ, THI HẾT BÍ
Sung
Cac em hQC sinh hay:
+ Tao ra m<)t bai toan tuong tu,
+ Tao ra cau h6i tric nghiem tu n<)i dung vi d1,1 tren.
l(lO:
a+b
2
Vi dy 11: Tinh tich phan I=
J ~(x - a)(b-x)dx voi O
b.
Ja+b
4
f)~t x = a+ (b - a)sin2t, te [O; ; ] suy ra dx = (b - a).sin2t.dt.
.l.· ,..
f) OIC~:
3a + b
•X=--=>t=-.
4
rt
6
a+ b
1t
• x=--=:>t=-.
2
4
1114 (b
)2
(b
)2 11/4
Khi d6: I=
a sin22t.dt =
-a
(1-cos4t)dt
J ~
8
11/6
=
J
(b-a)2 (r-.!.sin4t)"'4
8
4
it/6
=
11/6
Jj)·
(b-a)2 (~8
12
8
Sang tao: Cac em hQC sinh hay:
+ Tao ra m9t
bai toan tucmg tu,
+ Tao ra cau h6i tric nghiern tu n<)i dung vi du tren.
Chu j: Cu6i cung, ngay ca voi tlch phan hfru ti chung ta ciing dn sir dung toi phuong phap d6i
bi8n. f)8 minh hoa ta xem xet m<)t vi du don gian nhimg rAt co ban sau .
•
Vi d\) 12: Tinb tich phan I=
-
f)~t x = tant, --
fOx ':+ 1 .
1t
< t < -1t suy ra dx
2
2
dt =(I+
cos2t
= --
2
tan t)dt.
B6i c~n:
• x = 0 thi t = 0.
• x
= 1 thi t =
1t .
4
+ tan z t )dt
11J14d
I "'4
Khl. do'·. I -- 11J'4<1
o
tan z t + l = o t = t o =
rt
4.
Sang t90: Cac em hQC sinh hay:
+ Tao ra m9t bai toan tucmg nr,
+ Tao ra cau h6i tric nghiern tu n9i dung vi du tren.
Trai nghiim Ht sinh thai Hoc
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di 6.0-SPBook
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MALODA.VN - KHO SÁCH QUÝ, THI HẾT BÍ
Chuang Ill - Nguyen ham, tlch phiin
wi trng dung
Nang Cao: Cac em hoc sinh c6 th~ tham khao them bai vi!t "Net il~p trong giai totin tich phiin
khi aua ky thu(it tach tich phan vao" cua Pham Trong Thu tren THTT thang 6/2016. Gioi thieu
tom t~t va cling voi muc dich d! cac em thiy day chinh la cac dang cau hoi tric nghiem nang cao:
I. cAcH THUC TACH TiCH PHAN
Tach tich phan thuong duoc th~y ap dung cho cac bai toan tich phan c6 phan thirc t6i gian. Sau
day
la met s6 cong tlnrc tach mang tinh ph6 bi€n.
1 • g ( x) = (
g(x)
f(x)
)(
x" -a x" -b
=-)-[
( ax + b
g(X ) =
E
N• ta di1 phAan tic
' h:
~(x) _ ~(x) ]·
x -a x -b
a-b
2. g(x) =
) , voi'· a, b E R ; m
f(x)
2
ex + dx +
r(
A1
(ax+ b)"
+
+
er
, vcn'· a, b, c, d
A2
E
R·, m, n
E
N• ta di1 p hAan tic
' h:.
A
+ ... +--m-+
(ax s- b)?"
ax+b
B1x+C1
B2x+C2
B0x+Cn
+
+ ... +--=-----=-(ex 2 + dx + e)" (cx2 + d.x + et-1
cx2 + dx + e
trong do, d~ tirn AP ... , Am; BP ... , B"; Cl' ... , C" duoc tim bfulg phuong phap d6ng nh~t thirc.
oi ab
, h:
3 . g ( x ) = a.sinx+b.cosx
.
, vv1
a, , c, d e R ta di1 p hAan tic
c.sm x + d.cos x
g(x) =A+ B. c.c~s x -d.sin x
c.sin x + d.cosx
trong d6, d! tim A, B duce tim b~ng phuong phap d6ng nh§t thirc.
= a.sinx+b.cosx+e
.
4• g ()x
c.sm x + d.cos x+h
g( x ) = A + B .
, vcn,. a, b , c, d , e, h e
R ta di phAan he
, h:
c.cosx-d.sinx
c
+-----c.sin x + d.cos x+h c.sin x +d.cos x+h
trong d6, d! tirn A, B, C duce fun bing phirong phap d6ng nhAt thirc.
II. M<)T
SO THI
01) 1\11NH HQA
s6 hfru
l. Tfnh tich phan ham
ti bing ky thoit tach tich phan
Thi du 1: Tinh tich phan:
dx
1-J-----,,.-(x + 2)(4x + 16x + 15)°
I
-
0
2
Binh hutmg: Ta c6 phan tich:
l
(x + 2)(4x2 +16x+15)
--------
L
Trang 140
Website: maloda.vn
=
1
(x+2)(2x+3)(2x+5)
A
B
C
=--+--+-x+2
2x+3
2x+5
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