View
50
Download
0
Category
Preview:
Citation preview
NGUYNHNG IP
N THI I HC
TCH PHN VNGDNG
z = 0.8 A
B
C
a
uv
F
G Cng Ty, nm 2014
to my family, my pippy andmy friends ( . )
2ndLATEX201401.1TCH PHN VNGDNG
Copyright 2014 by Nguyn Hng ip
LIMUXin bt u bng mt chuyn vui ton hc
Nhn ngy Nh gio Vit Nam, cc hc sinh c quy qun bn thy giody Ton. Gp li hc tr c, thy h hi: Thy rt mng l cc em u thnh t trong cuc sng. Trong cc ththy dy, c ci g sau ny cc em dng c khng ?Tt c hc sinh u im lng. Mt lc sau, c mt hc sinh rt r ni: Tha thy, cmt ln em i b b h th gi thi baym em xung nc.Em loay hoaymi khng bit lm th no vtm ln. Bng nhin em thyon dy thp v nh li cc bi ging ca thy. Em ly dy thp un thnhdu tch phn ri dng n ko m ln. Thy: ?!?!?!
Chuyn vui nhng cng c vn suy nhm. Tch phn c ng dng g? Ch cnchu kh ln google l c kha kh kt qu ( . ), nhng hc tch phn ch ly 1 imtrong k thi tuyn sinh th l ch hng ti ca i a s hc sinh. Trong cc nmgn y th im s phn tch phn khng cn l vn qu kh khn. Hy vng ti liunh ny gip ch c cho ai .
Th trn Vnh Bnh, ngy 06 thng 08 nm 20141
Nguyn Hng ip.
1Cn vi ngy na l i l Vu Lan nm Gip Ng.
Mc lc
LIMU iii
MC LC iv
I TCH PHN 11 Cc cng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1 Bng cc nguyn hm thng dng . . . . . . . . . . . . . . . . . . . . 21.2 Tch phn xc nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Phng php phn tch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Tch phn cha tr tuyt i, min, max . . . . . . . . . . . . . . . . . . . . . . 54 Phng php i bin s n gin . . . . . . . . . . . . . . . . . . . . . . . . 8
4.1 Dng cn thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84.2 Biu thc c cha cn bc khc nhau . . . . . . . . . . . . . . . . . . 114.3 Dng phn thc 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.4 Dng biu thc ly tha . . . . . . . . . . . . . . . . . . . . . . . . . . 134.5 Biu thc c logarit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5 i bin sang lng gic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145.1 Dng 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.2 Dng 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.3 Dng 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.4 Dng 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.5 Dng 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
6 Tch phn hm hu t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.1 Tch phn cha nh thc . . . . . . . . . . . . . . . . . . . . . . . . . . 236.2 Tch phn cha tam thc . . . . . . . . . . . . . . . . . . . . . . . . . . 236.3 Dng tng qut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
7 Tch phn hm lng gic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277.1 Cc cng thc lng gic . . . . . . . . . . . . . . . . . . . . . . . . . . 277.2 Dng tng qut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287.3 Cc trng hp n gin . . . . . . . . . . . . . . . . . . . . . . . . . . 297.4 Tch phn dng hu t i vi hm s lng gic . . . . . . . . . . . 377.5 Dng hm ph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
8 Tch phn hm v t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
iv
8.1 Biu thc c tam thc bc hai . . . . . . . . . . . . . . . . . . . . . . . 448.2 Php th Eurle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508.3 Dng c bit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
9 Tnh tnh phn bng tnh cht . . . . . . . . . . . . . . . . . . . . . . . . . . . 539.1 Tch phn c cn i nhau . . . . . . . . . . . . . . . . . . . . . . . . . 539.2 Tch phn c cn l radian . . . . . . . . . . . . . . . . . . . . . . . . . 59
10 Phng php tnh tch phn tng phn . . . . . . . . . . . . . . . . . . . . . 6410.1 Dng 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6510.2 Dng 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6810.3 Phng php hng s bt nh . . . . . . . . . . . . . . . . . . . . . . 70
11 Cc bi ton c bit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
II NGDNGCA TCH PHN 751 Tnh din tch hnh phng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
1.1 Cng thc tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 761.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
2 Th tch vt th trn xoay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 792.1 Hnh phng quay quanh Ox . . . . . . . . . . . . . . . . . . . . . . . . 79
III BI TP TNGHP 811 Cc thi tuyn sinh 2002-2014 . . . . . . . . . . . . . . . . . . . . . . . . . . 822 Bi tp tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
Nguyn Hng ip v
ITCH PHN
1. CC CNG THC Chng I. TCH PHN
1 Cc cng thc
1.1 Bng cc nguyn hm thng dng
0dx =C dx = x+Cxdx = x+1
+1 +C(ax+b)dx = 1a x
+1+1 +C 1
xdx = ln |x|+C 1
ax+bdx = 1a ln |ax+b|+Cexdx = ex +C eax+bdx = 1a eax+b +Caxdx = axlna +C
uaudx = aulna + c
cosxdx = sinx+C cos(ax+b)dx = 1a sin(ax+b)+Csinxdx =cosx+C sin(ax+b)dx = 1a cos(ax+b)+C ) 1cos2x
dx = tanx+C 1cos2ax
dx = tan(ax)+C 1sin2x
dx =cotx+C 1sin2ax
dx =cot(ax)+C
1.2 Tch phn xc nh
nh ngha
Cho y = f (x) l mt hm s lin tc trn [a,b] v y = F (x) l mt nguyn hm ca n.Tch phn xc nh t a n b c nh ngha v k hiu nh sau:
ba
f (x)dx = F (b)F (a)
Tnh cht
00
f (x)dx = 0,a
a
f (x)dx = 0
ba
k f (x)dx = kb
a
f (x)dx
ba
f (x)dx =a
b
f (x)dx
2 Nguyn Hng ip
Chng I. TCH PHN 2. PHNG PHP PHN TCH
ba
[f (x) g (x)]dx = b
a
f (x)dxba
g (x)dx
ba
f (x)dx =c
a
f (x)dx+bc
f (x)dx
Nu f (x) 0 trn [a;b] thb
a
f (x)dx 0
Nu f (x) g (x) trn [a;b] thb
a
f (x)dx b
a
g (x)dx
2 Phng php phn tch
V d 2.1. Tnh cc tch phn sau:
(a) I1 =2
1
x22xx3
dx (b) I2 =3
1
(x21)2x
dx
(c) I3 =1
0
ex +1e2x
dx (d) I4 =1
0
(pex 1
)2dx
(e) I5 =2
0
6x3x2x+5 dx
Gii
(a) Ta c: I1 =2
1
(1
x 2x2
)dx =
(ln |x|+ 2
x
)21= ln21.
(b) Ta c: I2 =3
1
x4+2x2+1x
dx =3
1
(x3+2x+ 1
x
)dx =
(1
4x4+x2+ ln |x|
)31= 28+ ln3.
(c) Ta c: I3 =1
0
(1
ex+ 1e2x
)dx =
10
(ex +e2x)dx = (ex 1
2e2x
)10= 32 1e 12e2
.
(d) Ta c: I4 =1
0
(ex 2
pex +1
)dx =
10
(ex 2e x2 +1
)dx(ex 4e x2 +x
)10= e4pe+4.
(e) Ta c: I5 = 32
0
2x1x2x+5 dx = 3
(ln |x2x+5|)20 (dng uu dx)
= 3ln 75
Nguyn Hng ip 3
2. PHNG PHP PHN TCH Chng I. TCH PHN
V d 2.2. Tnh cc tch phn sau:
(a) I1 =1
0
x(1x)2004dx (b)I2 =1
0
1px2px3 dx
Gii
(a) Ta c: I1 =1
0
[(x1)+1](x1)2004dx =1
0
[(x1)2005+ (x1)2004]dx
=1
0
(x1)2005dx+1
0
(x1)2004dx
=[(x1)2006
2006 (x1)
2005
2005
]10= 1
4022030.
(b) Nhn xt: khi trc cn thc ta s trit tiu c x mu.
Ta c: I2 =1
0
(px1px
)dx = 2
3
[(x+1) 32 x 32
]43= 43(p21)
Bi ton tng t
1.
43
1px+2px3 dx. p s:
215 (6
p65p5+1).
2.
pi2
pi2
sin7x sin2x dx. p s: 445 .
3.
pi2
pi6
1+ sin2x+cos2xsinx+cosx dx. p s: 1.
4.
pi4
0
sin2(pi4x)dx. p s: pi28 .
5.
pi2
0
sin4 x dx. p s: 3pi16
6.
pi4
0
tan2 x dx. p s: 1 pi4 .
7.
pi2
0
tan3 x dx. p s: 32 ln2.
4 Nguyn Hng ip
Chng I. TCH PHN 3. TCH PHN CHA TR TUYT I, MIN, MAX
8.
160
1px+9px dx. p s: 12.
9.
52
1px+2+px2 dx. p s:
10.
10
(e2x + 3
x+1)dx. p s: e
2
2 +3ln2 12
11.
10
x
x+px2+1
dx. p s: 23 + 23p2
3 Tch phn cha tr tuyt i, min, max
1. Tnh I =ba
| f (x)|dx ta xt du f (x) trn [a,b] kh du gi tr tuyt i.
2. Tnh I =ba
max[ f (x),g (x)]dx, I =b
a
min[ f (x),g (x)]dx ta xt du hm
h(x)= f (x) g (x)
trn [a,b] tmmin[ f (x),g (x)], max[ f (x),g (x)].
V d 3.1. Tnh I =2
0
|x2x|dx
Gii
Cho x2x = 0 x = 0 x = 1Bng xt du
x
x2+ x0 1 2
0 0 +
Khi : I =1
0
(x2+x)dx+2
1
(x2x)dx = 1
V d 3.2. Tnh I =2pi0
p1+ sinx dx
Nguyn Hng ip 5
3. TCH PHN CHA TR TUYT I, MIN, MAX Chng I. TCH PHN
Gii
Ta c: I =2pi0
p1+ sinx dx =
2pi0
(sin
x
2+cos x
2
)2dx =
2pi0
sin x2+cos x
2
dxCho sin
x
2+cos x
2= 0 tan x
2=1 x =pi
2+k2pi
Do x [0,2pi] ta c x = 3pi2
Bng xt du
x
sin x2 +cos x20 3pi2 2pi
0 + 0
Khi : I =3pi2
0
(sin
x
2+cos x
2
)dx+
2pi3pi2
(sin
x
2+cos x
2
)dx
= 2(cos x
2+ sin x
2
) 3pi20+2(cos
x
2 sin x
2
)2pi3pi2
= 4ln2.
V d 3.3. Tnh I =2
1(|x| |x1|)dx
Gii
Bng xt du chung
x
x
x 1
1 0 1 2 0 + + 0 +
Khi : I =0
1(x+x1)dx+
10
(x+x1)dx+2
1
(xx+1)dx
=0
1dx+
10
(2x1)dx+2
1
dx = 0.
V d 3.4. Tnh I =2
0
max{x2,3x+2}dx
Gii
Xt hm s h(x)= x23x+2 trn [0,2]Bng xt du
x
h(x)
0 1 2
0 + 0
6 Nguyn Hng ip
Chng I. TCH PHN 3. TCH PHN CHA TR TUYT I, MIN, MAX
Do :
Vi x [0,1] th max[x2,3x+2]= x2. Vi x [1,2] th max[x2,3x+2]= 3x2.
Khi : I =1
0
x2dx+2
1
(3x2)dx = 176.
Bi ton tng t
1.
22|x21|dx. p s: 4
2.
23|x23x+2|dx. p s: 592
3.
pi2
0
p54cosx4sinx dx. p s: 2p32 pi6
4.
55
(|x+2| |x2|)dx. p s: 8
5.
11
(|2x1| |x|)dx. p s: 32
6.
11
|x|x4x212 dx. p s:
27 ln
34
7.
41
x26x+9dx. p s: 52
8.
11
4|x|dx. p s: 2 (5p3)
9.
11
|x|x dx. p s: 2
p2
3
10.
30
|2x 4|dx. p s: 4+ 1ln2 .
11.
30
x32x2+x dx. p s: 24+
p3+8
15 .
Nguyn Hng ip 7
4. PHNG PHP I BIN S N GIN Chng I. TCH PHN
12.
pi2
pi2
|sinx|dx. p s: 2.
13.
pi0
p2+2cos2x dx. p s: 4.
14.
pi0
p1 sin2x dx. p s: 2p2.
15.
2pi0
p1+ sinx dx. p s: 4p2.
16.
20
max(x,x2)dx. p s: 556 .
17.
20
min(x,x3)dx. p s: 43 .
18.
pi2
0
min(sinx,cosx)dx
4 Phng php i bin s n gin
Thng thng khi gp:
Mt cn thc ta t t l cn thc.
Mt phn thc ta t t l mu thc.
Mt hm s ly ly tha ta t t l biu thc ly ly tha.
Mt hm s m ta t t l biu thc trn m.
4.1 Dng cn thc
Khi gp hm di du tch phn c cha biu thc dng nf (x) ni chung
trong nhiu trng hp ta t t = n f (x)
V d 4.1. Tnh
10
xx2+1dx
8 Nguyn Hng ip
Chng I. TCH PHN 4. PHNG PHP I BIN S N GIN
Gii
t t =px2+1 t2 = x2+1 x2 = t21 xdx = td t
i cn: x = 0 t = 1 ; x =p1 t =p2
Khi : I =
10
x2+1.x dx =
p2
1
t .t d t
=p2
1
t2dt = t3
3
p2
0= 13
(2p21
)
Lu :mt s hc sinh thng qun i sang cn mi theo t . Bi ny ta cn c th giitheo cch khc nh V d 5.7 trang 20.
V d 4.2. Tnh I =p3
0
x3x2+1dx
Gii
t t =px2+1 x2 = 1 t2 xdx =td t
i cn: x = 0 t = 0 ; x =p3 t = 2
Khi : I =
p3
0
x2x2+1.x dx =
20
(1 t )t (t )dx =2
0
(t3 t2)dx
=(t4
4 t
3
3
)20= 43
Nhn xt: Trc khi i sang bin t ta c bc phn tch lm xut hin kt qu vi phnxdx l x3dx = x2.xdx v ta thy cn chuyn x2 theo bin t th php i bin mi thnhcng.
Bi ton tng t
1.
10
x1p3x26x+7
dx. ps:2
p
7
3
2.
lnx0
e2xp1+ex dx. ps:2
p
2
3
3.
512
xp2x1dx. ps:
144
5
4.
62
1
2x+1+p4x+1 dx. p s: ln32 16
Nguyn Hng ip 9
4. PHNG PHP I BIN S N GIN Chng I. TCH PHN
5.
pi2
0
p1+4sinx cosx dx.
V d 4.3. Tnh
p3
1
32lnxxp1+2lnx
dx
Gii
t t =p1+2lnx t2 = 1+2lnx 2lnx = t21 td t = 1xdx
i cn: x = 1 t = 1 ; x =p2 t =p2
Khi : I =p3
1
32lnxp1+2lnx
1xdx =
p2
1
(3 t2+1)t
t d t =p2
1
(4 t2)dt = 10p2
3 11
3
Bi ton tng t
1.
pe3
1
32lnxxp1+2lnx
dx. p s: 53
2.
e1
p1+3lnx lnx
xdx (B-2004). p s: 116135
3.
ep7
1
lnx31+ ln2 xx
dx. p s: ln 32 13
V d 4.4. Tnh I =2p3
p5
1
xp4+x2
dx (A-2003)
Gii
t t =p4+x2 x2 = t24 xdx = td t
i cn: x =p5 t = 3 ; x = 2p3 t = 4
Khi : I =2p3
p5
1
xp4+x2
dx =
2p3
p5
1
x2p4+x2
x dx
=4
3
1
(t24)t t d t =4
3
1
t24 t d t =4
3
1
t24 dt
=4
3
1
(t 2)(t +2) dt =1
4
43
1
t 2 1
t +2 dt
= 14(ln |t 2| ln |t +2|)|43 =
1
4 ln 5
3
Nhn xt: khi ta phn tch lm xut hin vi phn xdx ta thy hm ban u cha c ktqu ny do ta cn nhn t vmu biu thc di du tch phn cho x. Sau ta cnchuyn x2 theo bin t th php i bin mi thnh cng.
10 Nguyn Hng ip
Chng I. TCH PHN 4. PHNG PHP I BIN S N GIN
Bi ton tng t
1.
ln8ln3
1p1+ex dx. p s: ln
32
2.
ln20
pex 1dx
4.2 Biu thc c cha cn bc khc nhau
Khi gp hm di du tch phn c cha cc biu thc dng(ax+b)cx+d
)mn
,
. . . ,
(ax+b)cx+d
) rs
ta tax+b)cx+d = t
k vi k l mu s chung nh nht ca cc s
mm
n, . . . ,
r
s.
V d 4.5. Tnh I =630
13px+1+px+1 dx
Gii
t x+1= t6 x = t61 dx = 6t5dti cn: x = 0 t = 1 ; x = 63 t = 2
Khi : I =2
1
6t5
t3+ t2 dt = 62
1
t3
t +1 dt =2
1
(t2 t +1 1
t +1)dt = 11+6ln 2
3
Nhn xt: do 3px+1 = (x +1) 13 ,px+1 = (x +1) 12 v mu s chung ca cc s m 1
3,1
2l 6
nn ta i bin x+1= t6.
Bi ton tng t
1.
72964
13pxpx dx
2.
32
3
x1x+1
1
x+1 dx.
Hng dn: t x+1x1 = t3 v kt hp phng php gii mc 5.3 trang 19.
Nguyn Hng ip 11
4. PHNG PHP I BIN S N GIN Chng I. TCH PHN
4.3 Dng phn thc 1
Khi gp hm di du tch phn c cha biu thc dngf (x)
g (x)ni chung
trong nhiu trng hp ta t t = g (x).
V d 4.6. Tnh I =4
0
p2x+1
1+p2x+1 dx
Gii
t t = 1+p2x+1 t 1=p2x+1 2x+1= (t 1)2 dx = (t 1)dt
i cn: x = 0 t = 2 ; x = 4 t = 4
Khi : I =4
2
t 1t
(t 1)dt =4
2
(t 1)2t
d t =4
2
t 2+ 1td t = 2+ ln2
Nhn xt: bi ny ta c th i bin dng cn thc t =p2x+1 nhng s phc tp hn,cch i bin t = 1+p2x+1 l ph hp.
V d 4.7. Tnh I =1
0
x3
x2+1 dx
Gii
t t = x2+1 x2 = t 1 xdx = dt2
i cn: x = 0 t = 1 ; x = 1 t = 2
Khi : I =1
0
x2
x2+1 x dx =2
1
t 1t
12dt = 1
2
21
(1 1
t
)dx = 1
2 ln2
2
Nhn xt: bi ny gn nht l gii bng phng php Tch phn hmhu t2 y ara hng gii khc thy nhiu cch tip cn mt bi tch phn.
Bi ton tng t
1.
pi2
0
sin3 x
1+cosx dx
1Phng php gii tng qut xemmc 6 trang 232Xemmc 6 trang 23
12 Nguyn Hng ip
Chng I. TCH PHN 4. PHNG PHP I BIN S N GIN
4.4 Dng biu thc ly tha
Thng thng ta t t l biu thc ly ly tha.
V d 4.8. Tnh I 1
0
x3(x41)5dx.
Gii
t t = x41 dt = 4x3dx x3dx = 14x3
i cn: x = 0 t =1 ; x = 1 t = 0
Khi : I = 14
01
t5dt = 124
t601 = 124.
Nhn xt: do (x4) = 4x3 nn ta kh c x3 trong bi.
Bi ton tng t
1.
10
x3(1+x4)3dx. p s: 1516 .
2.
10
x3(1x3)6dx. p s: 1168 .
3.
10
x3(1x)2014dx
4.5 Biu thc c logarit
Dng thng gp l biu thc cha1
xv lnx. Ta thng i bin t = lnx hoc
t = biu thc cha lnx.
V d 4.9. Tnh cc tch phn sau:
(a) I1 =e
1
(1+ lnx)2x
dx (b) I2 =e
1
lnx.31+ ln2 xx
dx
Gii
(a) t t = 1+ lnx dt = 1x
i cn: x = 1 t = 1 ; x = e t = 2
Khi : I1 =2
1
t2dt = t3
3
21= 73.
Nguyn Hng ip 13
5. I BIN SANG LNG GIC Chng I. TCH PHN
(b) t t = 31+ ln2 x t3 = 1+ ln2 x 3t2dt = 2 lnx
xdx lnx
xdx = 3
2 t2dt
i cn: x = 1 t = 1 ; x = e t = 3p2
Khi : I2 = 32
3p21
t3dt = 38 t43p2
1= 38(3p161)
Bi ton tng t
1.
e2e
1
x lnxdx. p s: ln2
2.
p3
0
ln(x+
px2+1
)px2+1
dx.
3.
e1
1
x9 ln2 x
dx
4.
e1
p1+ lnx2x
dx. p s: 2p213
5.
e31
1
xp1+ lnx
dx. p s: 2.
5 i bin sang lng gic
I BIN SANG LNGGIC
Hm di du tch phn i bin iu kin
1pa2x2 x = a sin t t [pi2 , pi2 ]
x = a cos t t [0,pi]2
px2a2 x = asin t t
[pi2 , pi2 ]\ {0}x = acos t t [0,pi] \ {pi2 }
3(a2+x2)k x = a tan t t (pi2 , pi2 )
x = a cot t t (0,pi)4
a+xax hoc
axa+x x = a cos2t t
[0, pi2]
5p(xa)(bx) x = a+ (ba)sin2 t t [0, pi2 ]
14 Nguyn Hng ip
Chng I. TCH PHN 5. I BIN SANG LNG GIC
5.1 Dng 1
Biu thc di du tch phn c dngpa2x2, a > 0, vi bi tp c dng ny
ta t
x = a sin t , t [pi2,pi
2
] x = a cos t , t [0,pi]
V d 5.1. Tnh I =p3
1
4x2dx
Gii
t x = 2sin t , t [pi2,pi
2
] dx = 2cos td t
i cn: x =1 t = pi6
; x =p3 t = pi3
Khi : I =pi3
pi6
44sin2 t 2cos t d t = 4
pi3
pi6
cos tcos2 t d t
Do t [pi6,pi
3
] cos t > 0pcos t = cos t
I =pi3
pi6
4cos2 t d t = 2pi3
pi6
(1+cos2t )dt
= 2pi3
pi6
dt +2pi3
pi6
cos2 t d t =pi+p3
Nhn xt:mc du hm di du tch phn c cn thc nhng nu t t =p4x2 th
s gp kh khn do:
1. T t2 = 4x2 td t =xdx nhng di du tch phn ch c dx nu lm xut hinvi phn xdx th ta phi chia cho x. Trong khi cn tch phn t 1 n p3 ccha x = 0 khi php chia khng hp l.
2. Khi i sang bin t cn tnh t theo x li xut hin du cn mi, bi ton sau phctp hn bi ton trc. (.)
y l V d chng t khng phi c thyf (x) l i bin t = f (x). C th khng
thnh cng.
V d 5.2. Tnh I =32
3p2
2
1(9x2)3 dx
Nguyn Hng ip 15
5. I BIN SANG LNG GIC Chng I. TCH PHN
Gii
t x = 3cos t vi t [0,pi] dx =3sin td t
i cn: x =3p2
2 t = 3pi
4; x = 3
2 t = pi
3
Khi I =pi3
3pi4
3sin t(9sin2 t
)3 dt =3pi4
pi3
3sin t
33 |sin3 t | dt
Do t [pi3 , 3pi4 ] sin t > 0|sin3 t | = sin3 t=
3pi4
pi3
3sin t
33 sin3 t d t =1
9
3pi4
pi3
1
sin2 td t = 1
9cot t
3pi4pi3
=p3+327
Nhn xt: trong bi ny nu t t =(
9x2)3 l khng thch hp. Bi ton tng t
1.
32
0
1(9x2)3 dx. p s: 19p3
2.
10
(1x2)3dx. p s: 3pi16
3.
p22
0
x2p1x2
dx. p s: pi8 14
4.
10
x2+1p4x2
dx. p s: pi2 p32
5.
1p22
p1x2x2
dx. p s: 1 pi4
6.
p2
0
x2p4x2
dx. p s: pi2 1
Dng tng qut
Biu thc di du tch phn c dngpa2b2x2, a > 0, vi bi tp c dng
ny ta t
16 Nguyn Hng ip
Chng I. TCH PHN 5. I BIN SANG LNG GIC
x = absin t , t
[pi2,pi
2
] x = a
bcos t , t [0,pi]
Bi tp
1.
10
x243x2dx. p s: 2
p3pi
27 + 112 .t:x=2
p
3
sint
2.
10
1px2+2x+3
dx.
p s: pi6 .Hd:I=
1
0
dx
p
4(x1)
2
.tx1=2sint
3.
p31
1
1
x2+2x+2 dx. p s:pi3
5.2 Dng 2
Biu thc di du tch phn c dngpx2a2, a > 0, vi bi tp c dng ny
ta t
x = asin t
, t (0,pi
2
) x = a
cos tcos t , t
(0,pi
2
)
V d 5.3. Tnh
63p2
1
xpx29
dx
Gii
t x = 3sin t
vi t (0,pi
2
) dx =3cos t
sin2 xdt
i cn: x = 3p2 t = pi4; x = 6 t = pi
6
Khi : I =pi6
pi4
3cos t
sin2 x 3sin t
9
sin2 t9
dt =pi4
pi3
cos t
3sin t
cos2 t
sin2 t
d t
= 13
pi4
pi6
cos t
sin t cos tsin t
d t = 13
pi4
pi6
dt = pi36
Nguyn Hng ip 17
5. I BIN SANG LNG GIC Chng I. TCH PHN
Nhn xt: bi ny ta cn c th i bin t =px2+9 s xut hin tch phn c dng
3p3
3
1
t2+9 dt ta p dng phng php gii mc 5.3 trang 19.
V d 5.4. Tnh
p22
1
1p4x21
dx
Gii
t x = 12cos t
, t (0,pi
2
) dx = sin t
2cos2 td t
i cn: x = 1 t = pi3; x =
p2
2 t = pi
4
Khi : I =pi4
pi3
1
cos td t
t u = sin t du = cos td ti cn: t = pi
3 u =
p3
2; t = pi
4 u =
p2
2
Khi : I =pi4
pi3
1
cos2 tcos t d t =
pi4
pi3
1
sin2 t 1 cos t d t
=
p22
p32
1
u21 du =1
2
p22
p32
(1
u1 1
u+1)du
= 12ln
(p2+1p3+1
)
Nhn xt: php i bin sang lng gic trong bi ny l ph hp nhng y cha phil cch lm hiu qu nht, nu ta i bin theo hng khc t = 2x+
p4x21 th bi gii
gn hn nhiu. Qua cho thy mt bi tch phn c nhiu cch gii khc nhau, tmc li gii p i hi nhiu v kinh nghim v kh nng suy lun ca mi ngi.
Bi ton tng t
1.
p22
0
1p1x2
dx. p s: 12 ln(p
2+1p3+1)
2.
p2
2p3
1
xpx21
dx. p s: pi12
18 Nguyn Hng ip
Chng I. TCH PHN 5. I BIN SANG LNG GIC
3.
4p3
2
px24x3
dx. p s: pi48 p3
32
5.3 Dng 3
Biu thc di du tch phn c dng(a2+x2)k , a > 0, vi bi tp c dng
ny ta t
x = a tan t , t (pi2,pi
2
) x = a cot t , t (0,pi)
V d 5.5. Tnh
3p3
3
1
x2+9 dx
Gii
t x = 3tan t , t (pi2,pi
2
) dx = 3
cos2 td t
i cn: x = 3 t = pi4; x = 3p3 t = pi
3
Khi : I =pi3
pi4
1
9tan2 t +9 3
cos2 td t = 1
3
pi3
pi4
1(1+ tan2)cos2 t d t
= 13
pi3
pi4
1
cos2 tcos2 t d t = 1
3
pi3
pi4
dt = pi36
V d 5.6. Tnh
20
1
x2+4 dx
Gii
t x = 2cot t , t (0,pi) dx = 2
cos2 td t
i cn: x = 0 t = pi2; x = 2 t = pi
4
Khi : I =pi4
0
1
4cot2 t +4 2
sin2 td t = 1
2
pi4
0
1(1+cot2)sin2 t d t
= 12
pi4
0
1
sin2 t sin2 t d t = 1
2
pi4
0
dt = pi8
Nguyn Hng ip 19
5. I BIN SANG LNG GIC Chng I. TCH PHN
V d 5.7. Tnh
10
x1+x2dx
Gii
t x = tan t , t (pi2,pi
2
) dx = 1
cos2 td t
i cn: x = 0 t = 0 ; x = 1 t = pi4
Khi : I =pi4
0
tan t1+ tan2 t 1
cos2 td t =
pi4
0
sin t
cos t 1cos t
1cos2 t
d t
=pi4
0
sin t
cos4 td t = = 1
3
(2p21
)
Nhn xt: y l cch gii ng v d nhin c th chp nhn c nhng ta cn ccch gii khc ngn gn hn V d 4.1 trang 8. Php i bin x = tan t c th dngc nhng khng thch hp trong trng hp ny.
Bi ton tng t
1.
10
1(1+x2)3 dx. p s: 3pi32 + 14
2.
20
1(x2+4)2 dx. p s: 132 (pi2 +1)
3.
p3
p33
1(1+x2)3 dx. p s:
p3+12
4.
p3
1
p1+x2x2
dx. p s: ln(2+p3)(p21)+ 3p32p33
Dng tng qut
Biu thc di du tch phn c dng(a2+b2x2)k , vi bi tp c dng ny ta
t
x = abtan t , t
(pi2,pi
2
) x = a
bcot t , t (0,pi)
20 Nguyn Hng ip
Chng I. TCH PHN 5. I BIN SANG LNG GIC
V d 5.8. Tnh I =1
0
1(1+3x2)2 dx
Gii
t x = 1p3,t (pi2,pi
2
) dx = 1p
3
(1+ tan2 t)dt
i cn: x = 0 t = 0 ; x = 1 t = pi3
Khi : I =pi3
0
1(1+ tan2)2 1p3 (1+ tan2 t)dt = 1p3
pi3
0
1
1+ tan2 t d t
= 1p3
pi3
0
cos2 t d t = 12p3
pi3
0
(1+cos t )dt = pi6p3+ 18
Bi ton tng t
1.
3p3
30
x2(4x2+9)2 dx. p s: 148
(pi3
p34
)
2.
0 12
1
2x2+2x+1 dx. p s:pi4
3.
10
x
x4+x2+1 dx. p s:pip3
18
5.4 Dng 4
Biu thc di du tch phn c dng
a+xax hoc
axa+x , vi bi tp c
dng ny ta t x = a cos2t , t [0,pi
2
]
V d 5.9. Tnh I =1
1
1+x1x dx
t x = cos2t , t [0,pi
2
] dx =2sin2t
i cn: x =1 t = pi2; x = 0 t = pi
4
Nguyn Hng ip 21
5. I BIN SANG LNG GIC Chng I. TCH PHN
Khi : I =pi2
pi4
1+cos2t1cos2t (2sin2t )dt =
pi2
pi4
cot2 t (2sin2t )dt
=pi2
pi4
cot t (2sin2t )dt =pi2
pi4
(4cos2 t)dt =2pi2
pi4
(1+cos2t )dt =2 pi2
Bi ton tng t
1.
10
1x1+x dx. p s:
pi2 1
2.
p2
0
2+x2x dx. p s:
pi2 +2
p2
5.5 Dng 5
Biu thc di du tch phn c dngp(xa)(bx), vi bi tp c dng ny
ta t x = a+ (ba)sin2 t , t [0,pi
2
]
V d 5.10. Tnh I =32
54
(x1)(x2)dx
Gii
t x = 1+ sin2 t , t [0,pi
2
] dx = 2sin td t
i cn: x = 54 t = pi
6; x = 3
2 t = pi
4
Khi : I = 12
pi4
pi6
sin22t d t = 14
pi4
pi6
1cos4t d t = pi48+p3
32
Bi tp tng hp
1.
p2
2p3
1
xpx21
dx. p s: pi12
22 Nguyn Hng ip
Chng I. TCH PHN 6. TCH PHNHMHU T
6 Tch phn hm hu t
6.1 Tch phn cha nh thc
Dng I =
1
(ax+b)n dx ta i bin t = ax+b
6.2 Tch phn cha tam thc
3 Dng 1
I =
1
ax2+bx+ c dx, xt cc trng hp ca = b24ac
1. > 0Khi : I =
1
a(xx1)(xx2)dx
= 1a(x1x2)
(1
xx1 1xx2
)dx
2. = 0Khi : I = 1
a
1
(xx0)2dx (tch phn hm cha nh thc).
3. < 0Khi : I = 1
a
1
(x+ A)2+B2 dx (i bin sang lng gic3 xemmc 5 trang 14).
V d 6.1. Tnh I =p31
1
1
x2+2x+2 dx
Gii
Ta c: I =p31
1
1
(x+1)2+1 dx
t x+1= tan t , t (pi2,pi
2
) dx = (1+ t2)dt
i cn: x =1 t = 0 ; x =p31 t = pi3
Khi : I =pi3
0
(1+ t2)(1+ t2) dt =
pi3
0
dt = pi3.
3Dng 3
Nguyn Hng ip 23
6. TCH PHNHMHU T Chng I. TCH PHN
Bi ton tng t
(a)
10
1
2x2+5x+2 dx (b)12
1
x2+2x+5 dx
(c)
254
1
2x25x+7 dx (d)4
3
1
x27x+10 dx
(e)
20
1
4x224x+36 dx (f)1
0
x
x42x2+2 dx
(g)
42
2x
x43x2+2 dx (h)pi2
pi6
cosx
sin2 x6sinx+2 dx
(i)
pi6
0
1
3sin2 x6sinx cosx+5cos2 x dx
3 Dng 2
Tch phn c dng I =
mx+nax2+bx+ c dx ta phn tch
mx+n = A(ax2+bx+ c)+B
t ta a c v cc dng tch phn bit cch gii.
V d 6.2. Tnh I =3
4
2x+3x23x+2 dx
Gii
Phn tch: 2x+3= A(2x3)+B = 2Ax3A+Bng nht h s hai v ta c:{
2A = 23A+B = 3
{A = 1B = 6
Khi : I =3
4
(2x3)+6x23x+2 dx =
34
2x3x23x+2 dx+6
34
1
x23x+2 dx
= I1+ I2
I1 =3
4
2x3x23x+2 dx = ln |x
23x+2|34 = ln 5624 Nguyn Hng ip
Chng I. TCH PHN 6. TCH PHNHMHU T
6.3 Dng tng qut
3 Phn tch phn thc
Cho f (x) l a thc bc b hn n khi ta c phn tch
f (x)(xa)n =
A1(xa)n +
A2(xa)n1 + +
An+1aa
f (x)(xa)m(xb)n =
A1(xa)m +
A2(xa)m1 +
Am+1xa +
A1(xb)n + +
An+1xb
Ta qui ng, khmu v xc nh cc h s Ai bng phng php ng nht thc hoctr s ring.
V d 6.3.
1.x+2
(x2)2 =A
(x2)2 +B
x2 =A+B(x2)(x2)2
Cho x+2= A+B(x2). Ln lt cho x = 0,2 ta c h phng trnh:{A2B = 2A = 4
{A = 4B = 1
2. f (x)= x24x
x34x2+5x2Ta c: x34x2+5x2= (x1)2(x2)Do : f (x)= x
24x(x1)2(x2) =
A
(x1)2 =B
x1 =C
x2Qui ng mu s v khmu hai v ta c:
x24x = A(x2)+B(x1)(x2)+C (x1)2
Ln lt cho x = 1,2,0 ta c h phng trnh:B +C = 1A3B 2C = 42A+2B +C = 0
A = 3B = 5C =4
3 Dng tng qut
tnh bi ton tch phn c dng phn thc 4 I =
f (x)
g (x)dx ta thc hin theo cc bc:
1. Xt xemf (x)
g (x) l phn thc thc s cha. C th:
(a) Nu bc f (x) nh hn bc ca g (x) ta c phn thc thc s.
4Ta ch xt trng hpmu c nghim
Nguyn Hng ip 25
6. TCH PHNHMHU T Chng I. TCH PHN
(b) Nu bc ca f (x) ln hn hoc bng bc ca g (x) ta chia f (x) cho g (x) lmxut hin phn thc tht s.
2. Cn c vo dng tch camu thcm ta phn tch thnh tng cc phn thc ngin.
V d 6.4. Tnh I = 34 x24x
x34x2+5x2Gii
Ta c:x24x
x34x2+5x2 l phn thc tht s. Cch phn tch xt trong V d 6.3 trang62.
Khi : I =4
3
(3
(x1)2 +5
x1 4
x2)dx
=( 3x1 +5ln |x1|4ln |x2|
)43= 12+5ln 3
24ln2.
Nhn xt:
1) Biu thc x4+a4(a > 0) c phn tch thnh
x4a4 =(x2+p2ax+a2
)(x2p2ax+a2
)2) Phng php trn gii quyt bi ton tch phn hmhu t nhng ni chung cn di
dng. Khi ta kt hp vi cc phng php khc th bi ton c gii quyt gnhn.
Bi ton tng t
1.
21
1
x3+x dx. p s: 18 ln
516 + ln2 12 ln 52 .
2.
32
x32x3x dx. p s: 12ln
32 + 32 ln2+ 12 43 .
3.
10
x+4x3+6x2+11x+6 dx. p s:
32 ln22ln 32 + 12 ln 43
4.
p32
0
1
(1x2)2 dx. p s: 12 ln(2
p3)+p3+ 32 .
5.
21
1x5x+x6 dx.
26 Nguyn Hng ip
Chng I. TCH PHN 7. TCH PHNHM LNG GIC
6.
10
x4
x416 dx. p s:112 ln3arctan 12
7.
10
1916x
2+x+1(x2+4)(x2+2x+5) dx. p s:
14 ln
2532 5234 arctan 12 7pi128 .
8.
10
x21x4+1 dx. p s:
12p2ln(2p22+p2
)
7 Tch phn hm lng gic
7.1 Cc cng thc lng gic
(a). Cng thc cng
sin(a+b)= sina cosb+ sinb cosa cos(a+b)= cosa cosb sina sinbsin(ab)= sina cosb sinb cosa cos(ab)= cosa cosb+ sina sinbtan(a+b)= tana+ tanb
1 tana tanb tan(pi4+x)= 1+ tanx1 tanx
tan(ab)= tana tanb1+ tana tanb tan
(pi4x)= 1 tanx1+ tanx
(b). Cng thc nhn Cng thc nhn i
sin2x = 2sinx cosx cos2x = cos2 x sin2 x=2cos2 x1= 12sin2 x
tan2x = 2tanx1 tan2 x cot2x =
cot2 x12cotx
Cng thc h bc
sin2 x = 1cos2x2
cos2 x = 1+cos2x2
tan2 x = 1cos2x1+cos2x
Cng thc nhn ba
sin3x = 3sinx4sin3 x tan3x = 3tanx tan3 x
1 tan2 xcos3x = 4cos3 x3cosx
(c). Cng thc theo tan x2t t = tan x2 th
Nguyn Hng ip 27
7. TCH PHNHM LNG GIC Chng I. TCH PHN
sinx = 2t1+ t2 tanx =
2t
1 t2cosx = 1 t
2
1+ t2
(d). Cng thc bin i tng thnh tch
sin(a+b)= 2sin(a+b2
)cos
(ab2
)cos(a+b)= 2cos
(a+b2
)cos
(ab2
)sin(ab)= 2cos
(a+b2
)sin
(ab2
)cos(ab)=2sin
(a+b2
)sin
(ab2
)tan(a+b)= sin(a+b)
cosa cosbcot(a+b)= sin(a+b)
sina sinb
tan(ab)= sin(ab)cosa cosb
cot(ab)= sin(ab)sina sinb
sina+cosa =p2sin(a+ pi
4
)sinacosa =p2sin
(a pi
4
)=p2cos
(a pi
4
)=p2cos
(a pi
4
)(e). Cng thc bin i tch thnh tng
sina. sinb = 12[cos(ab)cos(a+b)] cosa.cosb = 1
2[cos(ab)+cos(a+b)]
sina.cosb = 12[sin(ab)+ sin(a+b)]
7.2 Dng tng qut
Khi gp tch phn hm lng gic trong trng hp tng qut ta c th i bin
t = tan x2. Khi :
dx = 2dt1+ t2 ; sinx =
2t
1+ t2 ;cosx =1 t21+ t2
V d 7.1. Tnh I =pi2
0
1
4+5sinx dx
Gii
t t = tan x2 sinx = 2t
1+ t2 dx = 2td t
1+ t2i cn: x = 0 t = 0 ; x = pi
2 t = 1
Khi : I =1
0
1
4+5 2t1+ t2
21+ t2 dt =
10
1
2t2+5t +2 dt
28 Nguyn Hng ip
Chng I. TCH PHN 7. TCH PHNHM LNG GIC
=1
0
1
2(t +2)(t + 12) dt =1
3
10
(1
t + 12 1t +2
)dt
= 13
(lnt + 12 ln |t +2|)10 = 13 ln2.
V d 7.2. Tnh I =pi2
0
1
sin2 x+2sinx cosxcos2 x dx
Gii
Ta c: I =pi2
0
11cos2x
2+ sin2x 1+cos2x
2
dx =pi2
0
1
sin2xcos2x dx
t t = tanx sin2x = 2t1+ t2 v cos2t =
1 t21+ t2
dx = dt1+ t2
i cn: x = 0 t = 0 ; x = pi4 t = 1
Khi : I =1
0
1
2t
1+ t2 1 t21+ t2
11+ t2 dt =
10
1
t2+2t +1 dt
=1
0
1
(t +p21)(t p21) dt =1
2p2ln
(p2+ t 1p2 t 1
)1
0
= (.).
7.3 Cc trng hp n gin
Phng php tng qut gii bi ton tch phn hm lng gic l i bin t = tan x2
nhng trong mt s trng hp phng php ny tr nn phc tp, di dng. Ta ccch gii ring i vi mt s dng c bit.
3 Dng 1
Hm s l l i vi sin ta t t = cosx. Hm l i vi cos ta t t = sinx.
V d 7.3.
1. Xt hm f (sinx,cosx)= cos3 x sin2 xTa c: f (sinx,cosx)= (cosx)3 sin2 x =cos3 x sin2 x
= f (sinx,cosx)y l trng hp hm l i vi cos
Nguyn Hng ip 29
7. TCH PHNHM LNG GIC Chng I. TCH PHN
2. Xt hm s f (sinx,cosx)= (sinx+ sin3 x)cosxTa c: f (sinx,cosx)= [sinx+ (sinx)3]cosx
= (sinx sin3 x)cosx =(sinx+ sin3 x)cosx= f (sinx,cosx)
y l trng hp hm l i vi sin
V d 7.4. Tnh tch phn sau:
(a) I =pi2
0
cos3 x sin2 x dx
(b) I =pi2
0
sin2x cosx
1+cosx dx
Gii
(a) t t = sinx cos2 x = 1 t2 dt = cosxdx
i cn: x = 0 t = 0 ; x = pi2 t = 1
Khi : I =pi2
0
cos2 x. sin2 x.cosx dx =1
0
(1 t2) t2dt
=1
0
(t2 t4)dt = ( t3
3 t
5
5
)10= 215
(b) Nhn xt:sin2x cosx
1+cosx =2sinx cos2 x
1+cosx y l trng hp hm l i vi sin.t t = cosx sin2 x = cosx
sinxdx =dti cn: x = 0 t = 1 ; x = pi
2 t = 0
Khi : I =pi2
0
2sinx cos2 x
1+cosx dx =1
0
2t2
1+ t d t
= 21
0
(t 1+ 1
t +1)dt = 2
[t2
2 t + ln(1+ t )
]10
= 2ln21
Bi ton tng t
1.
pi2
0
sin3 x cos5 x dx. p s: 124
30 Nguyn Hng ip
Chng I. TCH PHN 7. TCH PHNHM LNG GIC
2.
pi2
0
cos5 x dx. p s: 815
3.
pi2
0
cosx sin3 x dx
4.
pi2
0
cosxp7+cos2x dx. p s:
pip2
12
3Dng 2
Hm bc chn i vi sin v cos.
t t = tanx dx = dt1+ t2
Cng thc thng s dng: cos2 x = 11+ t2 ; sinx =
t2
1+ t2
V d 7.5. Tnh tch phn sau
(a) I =pi4
0
tan5 x dx
(b) I =pi3
pi4
1
sin2 x cos4 xdx
Gii
(a) Nhn xt: hm tan5 x = sin5 x
cos5 xl hm chn i vi sinx,cosx
t t = tanx dx = 11+ t2dt
i cn: x = 0 t = 0 ; x = pi4 t = 1
Khi : I =1
0
t5
1+ t2 dt =1
0
(t3 t + t
t2+1)dt =
[t4
4 t
2
2+ 12ln(t2+1)]1
0=1
4+ 12ln2.
(b) t t = tanx dt = (1+ tan2 x)dx dt = (1+ t2)dx dx = 1
1+ t2dt
Ta c: sin2 x = t2
1+ t2 v cos2 x = 1
1+ t2
Nguyn Hng ip 31
7. TCH PHNHM LNG GIC Chng I. TCH PHN
Khi I =p3
1
1
t2
1+ t2 (
1
1+ t2)2 11+ t2 dt =
p3
0
(1+ t2)2t2
dt
=p3
0
t4+2t2+1t2
dt =p3
0
(t2+2+ 1
t2
)dt
=(t3
3+2t 1
t
)p3
1= 4+4
p3
3
Bi ton tng t
1.
pi2
pi4
cotx+1sin4 x
dx. p s: 2512
2.
pi2
pi6
cos3 xpsinx
dx. p s: 85 1910p2
3.
pi4
0
1
cosxdx. p s: ln
(2+p22p2
) 12
4.
pi2
0
cos5 x dx. p s: 815
5.
pi4
pi6
1
sin3 c cosxdx. p s: 1+ ln3
6.
pi2
pi4
cot6 x dx. p s: 1315 pi4
7.
pi4
0
1
sinx+2sinx cosx8cos2 x dx. p s:16 ln
25
3Dng 3
Dng I =b
a
sinm cosn dx trong m,n l cc s nguyn dng chn. Loi ny l trng
hp c bit ca Dng 2 nhng ta c th gii gn hn bng cch:
32 Nguyn Hng ip
Chng I. TCH PHN 7. TCH PHNHM LNG GIC
1. Nhm ly tha chung ca sinx v cosx s dng cng thc sinx cosx = 12sin2x.
2. Phn cn li dng cos2 x = 1+cos2x2
v sin2 x = 1cos2x2
gim dn bc casinx,cosx.
V d 7.6. Tnh I =pi2
0
cos4 x sin2 x dx
Gii
Ta c: I =pi2
0
cos2 x sin2 x.cos2 x dx = 14
pi2
0
sin22x cos2 x dx
= 14
pi2
0
sin22x
(1+cos2x
2
)dx = 1
8
pi2
0
(sin22x+ sin22x cos2x)dx
= 18
pi2
0
sin22x dx+ 18
pi2
0
sin22x cos2x dx = I1+ I2
I1 = 18
pi2
0
sin22x dx = 116
pi2
0
(1cos4x)dx = 116
(x sin4x
4
)pi20= pi32
I2 = 18
pi2
0
sin22x cos2x dx
t t = sin2x cos2xdx = 12dt
i cn: x = 0 t = 0 ; x = pi2 t = 0 I2 = 0Vy I = pi
32
Bi ton tng t
1)
pi2
0
sin2 x cos4 x dx 2)
pi4
0
sin2 x cos2 x dx
3)
pi2
0
cos4 x dx 4)
pi6
0
sin6 x dx
3Dng 4
Nguyn Hng ip 33
7. TCH PHNHM LNG GIC Chng I. TCH PHN
1. I =b
a
1
sin2m x cos2n xdx . i bin t = tanx v p dng cng thc
cos2 x = 11+ tan2 x , sin
2 x = tan2 x
1+ tan2 x
2. I =b
a
tanm x
cos2nxdx. i bin t = tanx.
3. I =b
a
cotm x
sin2n xdx. i bin t = cotx.
4. I =b
a
sinm x
cos2n xdx. Ta a v tanx sau ty trng hp m i bin t = tanx hoc
t = cotx. Dng I =b
a
cosm x
sin2n xdx c cch lm tng t.
5. I =b
a
tanm x dx. Ta s dng cng thc tan2 x = 1cos2 x
1 sau i bin ty tng
bi c th. Dng I =b
a
cotm x dx c cch gii tng t.
6. Dng I =b
a
1
sinn xdx.
Bin i: I = 12n
ba
1
sinn x2 cosn x2
dx = 12n
ba
1
tann x2 cos2n x
2
dx
i bin t = tan x2 (xemmc 7.2 trang 28).
7. Dng I =b
a
1
cosn xdx
Ta a v dng trn bng cch i bin t = pi2x.
V d 7.7. Tnh cc tch phn sau:
a)
pi4
0
tan2 x dx b)
pi3
0
tan3 x dx
Gii
a) Ta c: I =pi4
0
(1
cos2 x1)dx = (tanxx)|
pi40 = 1
pi
4
34 Nguyn Hng ip
Chng I. TCH PHN 7. TCH PHNHM LNG GIC
b) Ta c: I =pi3
0
tanx. tan2 x dx =pi3
0
tanx
(1
cos2 x1)dx
=pi3
0
tanx 1cos2 x
dxpi3
0
tanx dx = I1 I2
I1 =pi3
0
tanx 1cos2 x
dx
t t = tanx dt = 1cos2 x
dx
i cn: x = 0 t = 0 ; x = pi3 t =p3
Khi : I =p3
0
t d t = t2
2
10= 32.
I2 =pi3
0
tanx dx
t t = cosx dt =sinxdxi cn: x = 0 t = 1 ; x = pi3 t = 12
Khi : I =12
1
1
tdx = ln |t ||
121 = ln
1
2= ln2
Vy I = 32 ln2.
Bi ton tng t
1)
pi4
0
tan2 x
cos6 xdx 2)
pi2
0
1
cos4 xdx
3)
pi4
0
tan5 x dx 4)
pi2
pi4
cot6 x dx
5)
pi4
0
tan4 x dx 6)
pipi4
cot3 x dx
7)
pipi4
1
sin4 xdx 8)
pi4
pi6
1
sin4 x cos4 xdx
9)
pi4
0
1
cos4 x+ sin4 x dx 10)pi4
pi6
1
a sin2 x+b cos2 x dx
11)
pi0
(tan2
x
3+ tan4 x
4
)dx 12)
pi2
0
sin2 xcos2 xsin4 x+cos4 x dx
Nguyn Hng ip 35
7. TCH PHNHM LNG GIC Chng I. TCH PHN
13)
pi2
pi3
1
sinxdx 14)
pi6
0
1
cosxdx
15)
pi2
pi3
1
sin3 xdx
3 Dng 5
tnhsina cosbdx;
sina sinbdx;
cosa cosbdx ta p dng cng thc bin i tch
thnh tng
sina cosb = 12[sin(a+b)x+ sin(ab)x]
sina sinb = 12[cos(ab)xcos(a+b)x]
cosa cosb = 12[cos(a+b)x+cos(ab)x]
V d 7.8. Tnh cc tch phn sau:
(a) I1 =pi2
pi2
sin7x cos2x dx
(b) I2 =pi2
pi6
1+ sin2x+cos2xsinx+cosx dx
Gii
(a) Ta c: I1 = 12
pi2
pi2
(cos5xcos9x)dx = 445
(b) Ta c: I2 =pi2
pi6
(1+ sin2x
sinx+cosx cos2x
sinx+cosx)dx
=pi2
pi6
[(sinx+cosx)2sinx+cosx +
cos2 x sin2 xsinx+cosx
]dx
=pi2
pi6
(sinx+cosx+cosx sinx)dx = 2pi2
pi6
cosx dx = 1.
36 Nguyn Hng ip
Chng I. TCH PHN 7. TCH PHNHM LNG GIC
Bi ton tng t
1)
pi8
0
sin3x cos5x dx 2)
pi0
sinx
3cos
x
4dx
3)
pi0
cosx cos23x dx 4)
pi6
0
sinx cos2x cos3x dx
7.4 Tch phn dng hu t i vi hm s lng gic
3 Dng 1
Dng I =
1
a sinx+b cosx+ c dx ta p dng cch gii tng qut:
t t = tan x2 dx = 2td t
1+ t2 .
Khi : sinx = 2t1+ t2 , cosx =
1 t21+ t2
V d 7.9. Tnh I =pi2
0
1
sinx+cosx+1 dx
Gii
t t = tan x2 dx = 2
1+ t2dti cn: x = 0 t = 0 ; x = pi
2 t = 1
Khi : sinx+cosx+1= 2t1+ t2 +
1 t21+ t2 +1=
2t +21+ t2
Do : I =1
0
(1+ t22t +2
2
1+ t2)dt =
10
1
t +1 dt = ln |t +1|10 = ln2.
Bi ton tng t
1.
pi0
1
sinx+1 dx. p s: pi
2.
pipi2
1
3sinx2cosx+3 dx. p s:12 ln
53
3.
pi2
0
1
4sinx+3cosx+5 dx. p s: 16
Nguyn Hng ip 37
7. TCH PHNHM LNG GIC Chng I. TCH PHN
3 Dng 2
Tch phn c dng I =
m sinx+n cosxa sinx+b cosx dx
Ta tm hai s A,B tha mn T s= A(Mu s)+B(Mu s)
V d 7.10. Tnh I =pi4
0
2sinx+16cosx2sinx+3cosx dx
Gii
Ta tm hai s A,B tha mn:2sinx+16cosx = A(2sinx+3cosx)+B(2sinx+3cosx)
= A(2cosx3sinx)+B(2sinx+3cosx)Ln lt cho x = 0, pi
2ta c h phng trnh:
{2A+3B = 163A+2B = 2
{A = 2B = 4
Khi : I = 2pi4
0
2cosx3sinx2sinx+3cosx dx+4
pi4
0
dx
= 2ln(2sinx+3cosx)|pi40 +pi= 2ln
5
3p2+pi.
Bi ton tng tpi
pi2
11sinx+10cosx4sinxcosx dx. p s: 3ln4+pi.
3 Dng 3
Tch phn dng I =
m sinx+n cosx+ka sinx+b cosx+ c dx.
Ta tm 3 s A,B ,C tha : T s= A(Mu s)+B(Mu s)+C . Khi a v c cc dngtch phn bit cch gii.
V d 7.11. Tnh I =pi2
0
3sinx+5cosx+2sinx+cosx+1 dx
Gii
38 Nguyn Hng ip
Chng I. TCH PHN 7. TCH PHNHM LNG GIC
Ta tm ba s A,B ,C tha mn:3sinx+5cosx+1= A(3sinx+5cosx+1)+B(3sinx+5cosx+1)+C
= A(3cosx5sinx)+B(3sinx+5cosx+1)+CLn lt cho x = 0, pi
2,pi ta c h phng trnh:
A+2B +C = 7A+2B +C = 5A+B +C =5
A = 1B = 4C =2
Khi : I =pi2
0
cosx sinxsinx+cosx+1 dx+4
pi2
0
dx2pi2
0
1
sin+cosx+1 dx
= I1+4I22I3
I1 =pi2
0
cosx sinxsinx+cosx+1 dx = ln |sinx+cosx+1||
pi20 = 0.
I2 =pi2
0
dx = x|pi20 =
pi
2.
I3 =pi2
0
1
sin+cosx+1 dx = ln2 (tch phn ny c trnh by V d 7.9)
Vy: I = 2pi2ln2.
3 Dng 4
Tch phn dng I =
m sinx+n cosx+k(a sinx+b cosx+ c)2 dx.
Ta tm 2 s A,B tha : T s= A(Mu s)+B(Mu s). Khi a v c cc dng tchphn bit cch gii.
V d 7.12. Tnh I =pi2
0
5cosx+ sinx(sinx+cosx)2 dx
Gii
Ta tm 2 s A,B tha
sinx+5cosx = A(cosx sinx)+B(sinx+cosx)
Ln lt cho x = 0, pi2ta c h phng trnh:
{A+B = 5A+B = 1
{A = 2B = 3
Nguyn Hng ip 39
7. TCH PHNHM LNG GIC Chng I. TCH PHN
Khi : I = 2pi2
0
cosx sinx(sinx+cosx)2 dx+3
pi2
0
1
sinx+cosx dx = 2I1+3I2.
I1 =pi2
0
cosx sinx(sinx+cosx)2 dx =
1
sinx+cosxpi20= 0 (hoc ta c th i bin t = sinx+cosx.)
I2 =pi2
0
1
sinx+cosx dx. i bin t = tanx
2(V d 7.9 ) ta tnh c I2 =
p2ln(
p2+1)
Vy I = 3p2ln(p2+1).
Bi ton tng tpi2
0
3sinx+29cosx(3sinx+4cosx)2 dx. p s:
14 + ln6.
3 Dng 5
Tch phn dng I =b
a
m sinx+n cosx(a sinx+b cosx)2 dx
Cch gii
1. Ta tm hai s A,B tha
T s= A(Mu s)+B(Mu s)
2. Khi : I = Ab
a
a cosxb sinx(a sinx+b cosx)3 dx+B
ba
1
(a sinx+b cosx)3 dx
= AI1+BI2.3. Tnh I1 : i bin t = a sinx+b cosx.
4. Tnh I2 =b
a
1
(a sinx+b cosx)3 dx
Ta bin i mu s
a sinx+b cosx =a2+b2.cos(x)
vi sin= apa2+b2
, cosbp
a2+b2v dxcos2 x
= tanx+C .
V d 7.13. Tnh I =pi2
0
4cosx7sinx(2sinx+cosx)3 dx
40 Nguyn Hng ip
Chng I. TCH PHN 7. TCH PHNHM LNG GIC
Gii
Tm hai s A,B tha
4cosx7sinx = A(2cosa sinx)+B(2sinx+ cosx)
Ln lt cho x = 0, pi2ta c h phng trnh{
2A+B = 4A+2B =7
{A = 3B =2
Khi : I = 3pi2
0
2cosx sinx(2sinx+cosx)3 dx2
pi2
0
1
(2sinx+cosx)3 dx
= I1+ I2.
I1 =pi2
0
2cosx sinx(2sinx+cosx)3 dx
t t = 2sinx+cosx ta tnh c I1 = 12t2
21= 38
I2 =pi2
0
1
(2sinx+cosx)3 dx
Ta bin i:
2sinx+cosx =p5(2p5sinx+ 1p
5cosx
)(t
pa2+b2 =p5 lm nhn t chung)
t sin= 2p5, cos= 1p
5Ta c: 2sinx+cosx =p5(sinx sin+coscosx)=p5cos(x)
Khi : I2 =pi2
0
1
5cos2(x) dx =1
5tan(x)|
pi20 =
1
5(cot+ tan)= 1
2.
Vy : I = 3I12I2 = 18.
Bi ton tng tpi
pi2
18cosx sinx(3sinx2cosx)3 dx. p s:
79 .
7.5 Dng hm ph
i khi thay v tnh trc tip tch phn ca hm s f (x), ta c th kt hpvi mt hm s khc g (x) bng cch tnh tch phn ca hm a f (x)+bg (x) vc f a(x)+dg (x). Da vo s lin kt nh vy ta tnh c tch phn d dng
Nguyn Hng ip 41
7. TCH PHNHM LNG GIC Chng I. TCH PHN
hn. Dng ny thng c p dng i vi hm s lng gic.
V d 7.14. Tnh I =pi2
0
cos2 x.cos2x dx
Gii
Xt thm J =pi4
0
sin2 x. sin2x dx
Ta c: I + J =pi4
0
(cos2 x+ sin2 x)cos2x dx =pi4
0
cos2x dx
= 12sin2x|
pi40 =
1
2
I J =pi4
0
(cos2 x sin2x)cos2x dx =pi4
0
cos22x dx
=pi4
0
1+cos4x2
dx = 12
(x+ 1
4sin4x
)pi40= pi8
Ta c h phng trnh: {I + J = 12I J = pi8
{I = 14 + pi16J = 14 pi16
Bi ton tng t
1.
pi2
0
cos4 x
sin4 x+cos4 x dx
Bi tp tng hp
1.
pi2
pi4
cos2x.(cot+2)sin2 x
dx. p s: 52 ln2pi
2.
pi4
0
sin2 x
cos6 xdx. p s: 815
3.
pi2
pi6
cos3 xpsinx
dx. p s: ln(p2+1)
42 Nguyn Hng ip
Chng I. TCH PHN 7. TCH PHNHM LNG GIC
4.
pi4
0
cosx sinxp2+ sin2x dx. p s: ln
(p3+p2p2+1)
5.
pi2
pi3
cotx.3sin3sinx
sin3 xdx. p s:
3p924
6.
pi0
(xcos4 x. sin3 x)dx. p s: 4pi35
7.
pi2
0
(pcosx
psinx
)dx. p s: 0
8.
pi2
0
cosx.cos(sinx)dx. p s: sin1
9.
pi2
0
sinx+2cosx3sinx2cosx+3 dx.
10.
pi2
0
sinx cosx
sinx+cosx dx. p s: 1+p22 ln(
p21).
Hd:sinxcosx=
1
2
[
(sinx+cosx)
2
1
]
11.
pi2
0
sin2
sin2 x+2cos2 x dx.
Hng dn:
sin
2
x
sin
2
x+2cos
2
x
=
1cos2x
1cos2x+2(1+cos2x)
12.
pi4
pi4
1ptanx
dx.
Hng dn: tt=p
tanxavdng
dx
1+t
4
13.
pi4
pi3
1
sin3 xdx. p s: 14 ln3+ 13 .
14.
pi4
0
cosx sinxp2+ sin2x dx. p s: ln
(p2+p31+p2
)
2+sin2x=1+(sinx+cosx)
2
.tt=sinx+cosx.
Nguyn Hng ip 43
8. TCH PHNHM V T Chng I. TCH PHN
15.
pi4
0
sin32x cos23x dx p s: 14 .
1
4
(3sin2xsin6x)
1+cos6x
2
vpdngkhaitrintchthnhtng.
8 Tch phn hm v t
Mt s dng tch phn v t c gii quyt cc phn trc:
1. Biu thc cha cn (xemmc 4.1 trang 8 ).
2. Biu thc cha cn bc khc nhau (xemmc 4.2 trang 11).
3. i bin sang lng gic (xemmc 5 trang 14).
8.1 Biu thc c tam thc bc hai
3 Dng 1
Dng I =
1pax2+bx+ c
dx. Ta phn tch biu thc trong cn thnh tng hoc hiu cc
bnh phng. Sau a v cc dng tch phn bit, ta c th p dng i bin sanglng gic (xemmc 5 trang 14) hoc da vo ch sau:Ch :
1. Ta chng minh c cng thc :1p
x2a2dx = ln |x+
x2a2|+C .
2. Ring dng
1p(x+a)(x+b) dx ta cn c th i bin
t =px+a+px+b nu x+a > 0 v x+b > 0. t =pxa+pxb nu x+a < 0 v x+b < 0.
Tam thc bc hai f (x)= ax2+bx+ c c hai nghim x1,x2 th
f (x)= a(xx1)(xx2).
V d 8.1. Tnh I =4
3
1px22x
dx
44 Nguyn Hng ip
Chng I. TCH PHN 8. TCH PHNHM V T
Gii
Ta c: I =4
3
1px(x2) dx
t t =px+px2 dt =(1px+ 1p
x2
)dx
dt =(p
x2+pxpx(x2)
)dx dt
t= dxp
x(x2)i cn: x = 3 t =p3+1 ; x = 4 t = 2p2
Khi : I =2p2p3+1
1
td t = ln |t ||
p3+1
2p2 = ln(p
3+12p2
).
V d 8.2. Tnh I =14
1px22x
dx
Gii
Ta c: I =14
1px(x2) dx
t t =px+px+2 dt =(
1px +1px+2
)dx
dt =(px+2+pxp
x(x2)
)dx dt
t= dxp
x(x2)i cn: x =4 t =p6+2 ; x =1 t = 1+p3
Khi : I =p6+2
1+p3
1
td t = ln |t ||1+
p3p
6+2 = ln(1+p3p6+2
).
V d 8.3. Tnh I =1
0
1px2+1
dx
Gii
Chng minh
1px2a2
dx = ln |x+x2a2|+C .
Ta c:(lnx+px2+a2) =
(x+
px2+a2
)x+
px2+a2
=1+ xp
x2+a2x+
px2+a2
= 1px2+a2
p dng kt qu trn ta c:
I =(lnx+x2+1)1
0= ln2.
Nguyn Hng ip 45
8. TCH PHNHM V T Chng I. TCH PHN
Bi ton tng t
(a)
52
1px2+4x+5
dx (b)
53
x26x+13dx
(c)
p7+34
34
1p2x2+3x+2
dx (d)
75
1p3x218x+15
dx
(e)
232
1p2x2+6x4
dx (f)
10
1px2+2x
dx
(g)
10
xp1+ex +e2x
dx (h)
10
xp2x43x2+1
dx
3 Dng 2
Tnh tch phn c dng I =
Ax+Bpax2+bx+ c
dx ta phn tch
Ax+B =C (ax2+bx+ c)+D
V d 8.4. Tnh I =3
2
x+4px2+2x3
dx
Gii
Ta c: x+4=C (x2+2x3)+D =C (2x+2)+D = 2Cx+2C +Dng nht h s {
2C = 12C +D = 4
{C = 12D = 3
Khi : I = 12
32
2x+2px2+2x3
dx+33
2
1px2+2x3
dx = I1+ I2
I1 = 12
32
2x+2px2+2x3
dx =3
2
x+1px2+2x3
dx =x2+2x3
32= 2p3p5.
I2 = 33
2
1px2+2x3
dx (y l tch phn Dng 1)
= 33
2
1(x+1)24
dx = 3(lnx+1+x2+2x3)3
2
46 Nguyn Hng ip
Chng I. TCH PHN 8. TCH PHNHM V T
= 3ln(4+2p33+2p5
)
Vy I = 2p3p5+3ln(4+2p33+2p5
)
Bi ton tng t
(a)
01
x1px22x+3
dx (b)
2+p62
5x+3px2+4x+10
dx
(c)
21
x1px2+2x+3
dx (d)
2+p52
1
2x1px2+3x1
dx
(e)
ep52
1e2
lnx
x14lnx ln2 x
dx
3 Dng 3
Tnh tch phn dng I =
1
(Ax+B)pax2+bx+ c
dx ta i bin
Ax+B = 1t
s a c v Dng 1.
V d 8.5. Tnh I =1
0
1
(x+1)px2+2x+3
dx
Gii
t x+1= 1t dx =dt
t2
i cn: x = 0 t = 1 ; x = 1 t = 12
Khi : I =12
1
1
1
t(1
t1)2+2(1
t1)+3
1t2dx
=12
1
11
t 1|t |
p4t 1
1t2dx =
12
1
1p4t 1 dx
= 12
(p4t 1)11
2= 12
p3 1
2.
Nguyn Hng ip 47
8. TCH PHNHM V T Chng I. TCH PHN
V d 8.6. Tnh I =0
1
1
(x+2)p2x2+4x+4
dx
Gii
t x+2= 1t dx = 1
t2
i cn: x =1 t = 1 ; x = 0 t = 12
Khi : I =12
1
1
1
t2
(1
t2)2+4(1
t2)+4
1t2dx
=1
12
1p4t24t +2
dx =1
12
(2t 1)2+1dx
= 12
(ln2t 1+p4t24t +2)11
2
= ln 1+p2 Bi ton tng t
(a)
21
1
xpx2+2x+3
dx (b)
0 12
1
(x+1)px2+1
dx
(c)
112
1
xpx2+1
dx (d)
107
1
(3x6)px24x+1
dx
(e)
2 94
1
(x+1)px2+3x+2
dx (f)
12
1
1
xpx2x
dx
3 Dng 4
Tch phn c dng I =
Ax+B(x+)
pax2+bx+ c
dx ta bin i
Ax+B =C (x+)+D
s a c v tch phn c Dng 1 v Dng 3.
V d 8.7. Tnh I =0
2
2x1(x+1)
px2+3x+3
dx
Gii
48 Nguyn Hng ip
Chng I. TCH PHN 8. TCH PHNHM V T
Ta c: 2x1= 2(x1)3
Khi : I =0
2
2(x1)3(x+1)
px2+3x+3
dx
= 20
2
1px2+3x+3
dx30
2
1
(x1)px2+3x+3
dx
= I1+ I2.Cc tch phn I1, I2 bit cch gii.
Ta tnh c: I =2ln(3
4+p3
2
)+3ln
(1
4+ 34+p3
2
).
3 Dng 5
Dng tng qut ca Dng 2, tch phn dng
I =
Pn(x)pax2+bx+ c
dx trong Pn(x) l a thc bc n
ta lm nh sau Phn tch:I =
Pn(x)pax2+bx+ c
dx =Qn1(x)ax2+bx+ c+
1p
ax2+bx+ cdx viQn1(x) l a thc
bc n1 v l s thc. Cc h s ca a thcQn1 v c xc nh bng cch:
1. o hm 2 v bc phn tch trn.
2. Cn bng h s.
V d 8.8. Tnh I =21
x2+4xpx2+2x+2
dx
Gii
Phn tch:
I =21
x2+4xpx2+2x+2
dx = (ax+b)x2+2x+2+
21
1px2+2x+2
dx
Xc nh cc h s:Ly o hm hai v v thu gn ta c:
x2+4x 2ax2+ (2s+ s+b)x+2a+b
ng nht h s hai v ta c:2a = 13a+b = 42a+b+ = 0
a = 12b = 52 =72
Nguyn Hng ip 49
8. TCH PHNHM V T Chng I. TCH PHN
Khi : I = 12(x+5)
px2+2x+3
21 7221
1px2+2x+2
dx
= 2 3p2
2 72
21
1(x+1)2+1
dx
= 2 3p2
2 72ln |x+1+
px2+2x+2|
21= 2 3
p2
2 72ln
(2p2
2
).
Bi ton tng t
1.
10
x2+1px2+2x+3
dx. Hd: a = 12 ,b =32 ,= 1.
2.
10
2x3+1px2+x+2
dx. Hd: a = 23 ,b =56 ,c =1712 ,= 8124 .
8.2 Php th Eurle
Trong trng hp tng qut khi tnh tch phn dng
I =
f (x,ax2+bx+ c)dx,a 6= 0
ta dng php th Eurle.
1. Nu a > 0, tpax2+bx+ c = t pax hoc t +pax
2. Nu c > 0, tpax2+bx+ c = xt +pc hoc xt pc
3. Nu ax2+bx+ c c hai nghim x1,x2 th ta tax2+bx+ c = t (xx1)
Ch : Nhng trng hp xt trn (a > 0,c > 0) c th a trng hp ny v trnghp kia bng cch t x = 1
z.
V d 8.9. Tnh I =0
1
1
x+px2+x+1
dx
Gii
50 Nguyn Hng ip
Chng I. TCH PHN 8. TCH PHNHM V T
y a = 1> 0 nn ta dng php th th nht.t
px2+x+1= t x x = t
211+2t dx =
2t2+2t +2(1+2t )2 dt
i cn: x =1 t = 2 ; x = 0 t = 1
Khi : I =1
2
2t2+2t +2t (1+2t )2 dt (dng tch phn hm hu t)
Phn tch:2t2+2t +2t (1+2t )2 =
A
(1+2t )2 +B
1+2t +C
tTa tm c A =3,B =3,C = 2.
Khi : I =31
2
1
(1+2t )2 dt 31
2
1
1+2t d t +21
2
1
td t
=[
3
2(1+2t ) +1
2ln
t4
(1+2t )3]1
2= 12+ 12ln
125
432
V d 8.10. Tnh I =
p615
p312
1
1+p12xx2
dx
Gii
Do c = 1> 0 nn theo php th th 2 ta tp12xx2 = xt x = 2 t 1
t2+1 dx = 2 t2+2t +1(
t2+1)2 dti cn: x =
p311 t = 0 ; x =
p615
t = 2
Khi : I =2
2
t2+2t +1t (t 1)(t2+1) dt =
10
(1
t 1 1
t 2t2+1
)dt
=2ln2+ pi42arctan2.
V d 8.11. Tnh I =1
0
xpx2+3x+2
x+px2+3x+2
dx
Gii
Ta c: x2+3x+2= (x+1)(x+2) nn ta dng php th th 3.t
px2+3x+2= t (x+1) x = t
221 t2 dx =
2t(1+ t2)3dt
i cn: x = 0 t =p2 ; x = 1 t =p6
Khi : I =p6
p2
2t2+2t
(1 t )(t 2)(1+ t )3 dt
=
p6
p2
[1
3(t +1)3 +5
18(t +1)2 17
108(t +1) +3
4(t 1) 16
27(t 2)]dt
Nguyn Hng ip 51
8. TCH PHNHM V T Chng I. TCH PHN
= 21466600p333
225+ 17108
ln
(p2+1p6+1
)+ 34ln
(p61p21
)
+1627
ln
(p2+2p
62
)
8.3 Dng c bit
Tch phn c dng
I =xr(a+bxp)q dx
trong r,p,q l cc s hu t.
1. Nu q l s nguyn t x = t s vi s l bi s chung nh nht ca mu s cc phns r v p.
2. Nur +1p
l s nguyn t a+bxp = t s vi s l mu s ca phn s p.
3. Nur +1p
+q l s nguyn t axp = t s vi s l mu s ca phn s q.
V d 8.12. Tnh I =25616
1p2( 4px1) dx
Gii
V q = 3 nguyn nn ta t x = t4, t > 0 dx = 4t3dti cn: x = 16 t = 2 ; x = 256 t = 4
Khi : I =4
2
4t3
t2(t 1)3 dt = 44
2
t
(t 1)3 dt =2[
2
t 1 +1
(t 1)2]4
2= 40
9
V d 8.13. Tnh I =p2
1
x5
(2x2)p2x2
dx
Gii
Ta c:x5
(2x2)p2x2
= x5(ax2) 32 , r = 5,p = 2,q =32Do
r +1p
= 3 nguyn nn ta t 2x2 = t2, t > 0 xdx =td ti cn: x = 1 t = 1 ; x =p2 t = 0
Khi : I =0
1
t44t2+4t2
dt =( t
3
3+4t + 4
t2
)01=
52 Nguyn Hng ip
Chng I. TCH PHN 9. TNH TNH PHN BNG TNH CHT
Bi ton tng t
(a)
21
1
x2 3(2+x3)5
dx (b)
161
31+ 4pxpx
dx
(c)
10
x1+ 3
px2
dx (d)
321
1
x3p1+x5
dx
(f)
81
1px3 31+ 4
px3
dx (g)
10
x6p1+x2
dx
9 Tnh tnh phn bng tnh cht
9.1 Tch phn c cn i nhau
Khi gp tch phn c dng I =a
af (x)dx ta c th dng phng php sau:
1. Ta c: I =0
af (x)dx+
a0
f (x)dx = I1+ I2
2. Xt I =0
af (x)dx
t x =t dx =dti cn: x =a t = a ; x = 0 t = 0
I1 =0
a
f (t )dt =a0
f (t )dt
Sau ta ty tng hm f (t )m c hng gii c th.
V d 9.1. Tnh I =1
1x2014 sinx dx
Gii
Ta c: I =1
1x2014 sinx dx =
01
x2014 sinx dx =1
0
x2014 sinx dx
= I1+ I2
Xt I1 =0
1x2014 sinx dx
t x =t dx =dti cn: x =1 t = 1 ; x = 0 t = 0
Nguyn Hng ip 53
9. TNH TNH PHN BNG TNH CHT Chng I. TCH PHN
Khi : I1 =0
1
(t )2014 sin(t )dt =0
1
t2014 sin t d t
=1
0
t2014 sin t d t =1
0
x2014 sinx dx =I2
T ta c: I = I1 I1 = 0Nhn xt: vi bi ton trn a s hc sinh suy ngh theo hai hng:
Hng 1: s dng phng php Tch phn tng phn v c dngf (x)sinxdx, nhng
trong trng hp ny cn thc hin 2014 ln tch phn tng phn, iu ny lkhng thc t.
Hng 2: tm cng thc tng qut ca bi ton tch phn c dng11
xn sinxdx, t
rt ra kt qu ca11
xn sinxdx. y l hng suy ngh haymang tnh khi qut cao
nhng cha hn l phng php ngn gn.
Qua cho thy tm quan trng ca vic nhn xt tnh cht cn tch phn v tnh chthm s di du tch phn nh hng phng php gii. T V d trn ta rt rac tnh cht sau
Tnh cht 9.2. Hm s f (x) lin tc trn [a,a]
1. Nu f (x) l hm s l trn [a,a] th I =a
af (x)dx = 0
2. Nu f (x) l hm s chn trn [a,a] th I = 2a0
f (x)dx = 0
Chngminh
Ta c: I =0
af (x)dx+
a0
f (x)dx = I1+ I2
Xt I =0
af (x)dx
t x =t dx =dti cn: x =a t = a ; x = 0 t = 0
I1 =0
a
f (t )dt =a0
f (t )dt
1. Nu f (x) l hm l th f (x)= f (x) f (t )= f (t )
Do : I1 =a0
f (t )dt =a0
f (x)dx =I2
T ta c: I = I1+ I2 =I2+ I2 = 0.
54 Nguyn Hng ip
Chng I. TCH PHN 9. TNH TNH PHN BNG TNH CHT
2. Nu f (x) l hm chn th f (x)= f (x) f (t )= f (t )
Do : I1 =a0
f (t )dt =a0
f (x)dx = I2
T ta c: I = I1+ I2 = I2+ I2 = 2a0
f (x)dx
Nhn xt: nu p dng kt qu Tnh cht 9.2 ta c ngay kt qu I = 0 , nhng trongkhun kh chng trnh ton ph thng khng c tnh cht ny, khi trnh by trongbi thi ta phi chngminh li nh trong V d 9.1.
V d 9.3. Tnh
11
x4+ sinxx2+1 dx
Gii
Ta c: I =1
1
x4
x2+1 dx+1
1
sinx
x2+1 dx = I1+ I2
Nhn xt: I1 l dng tch phn hm hu t ta gii c; hm s trong I2 l hm s l ,theo Tnh cht 9.2 ta c I2 = 0.
Tnh I1 =1
1
x4
x2+1 dx
Ta c: I1 =1
1
(x21+ 1
x2+1)dx =
11
(x21)dx+ 1
1
1
x2+1 dx
=43+ I12
Vi I12 =1
1
1
x2+1 dx ta s dng phng php i bin sang lng gic Dng 3 (mc 5.3
trang 19)
t x = tan t , t (pi2,pi
2
) dx = dt
cos2 ti cn: x =1 t =pi
4; x = 1 t = pi
4
Khi : I =pi4
pi4
1
tan2 t +1 1
cos2 td t =
pi4
pi4
cos2 t 1cos2 t
d t =pi4
pi4
dt
= pi2
Vy: I1 = pi4 43
Tnh I2 =1
1
sinx
x2+1 dx =0
1
sinx
x2+1 dx+1
0
sinx
x2+1 dx = I21+ I22
Xt I21 =0
1
sinx
x2+1 dx
Nguyn Hng ip 55
9. TNH TNH PHN BNG TNH CHT Chng I. TCH PHN
t x =t dx =dti cn: x =1 t = 1 ; x = 0 t = 0
Khi : I21 =0
1
sin(t )(t )2+1 dt (1)=
10
sin t
t2+1 dt
=1
0
sinx
x2+1 dx =I22
Do : I2 = I21 I22 = 0Vy: I = I1+ I2 = pi
2 43
Nhn xt: hm s ban u di du tch phn khng chn khng l, khi ta tch I = I1+I2th hm s l xut hin v ta bit c kt qu ca I2 = 0 nhng vn phi chng minhkt qu. Bi ny c th gii theo phng php Tch phn tng phn nhng rc ri hnnhiu.
Bi ton tng t
1.
11
(x2+2)sinx dx. p s: 0
2.
12
12
cosx ln
(1x1+x
)dx. p s: 0
3.
11
ln3(x+x2+1
)dx. p s: 0
4.
pi2
pi2
cosx ln(
x+x2+1)dx. p s: 0
5.
12
12
cosx ln(1x1+x
)dx. p s: 0
Nhn xt: qua cc v d trn ta thy hiu qu ca Tnh cht 9.2 i vi hm s l,cn i vi hm s chn t c s dng hn. Trong mt s bi ton tch phn c cni nhau ta cng p dng phng php nh phn chng minh Tnh cht 9.2.
V d 9.4. Cho hm s f (x) lin tc trn R v
f (x)+ f (x)=p22cosx,x R
Tnh I =3pi2
3pi2
f (x)dx
56 Nguyn Hng ip
Chng I. TCH PHN 9. TNH TNH PHN BNG TNH CHT
Gii
Ta c: I =0
3pi2
f (x)dx+3pi2
0
f (x)dx = I1+ I2
Xt I1 =0
3pi2
f (x)dx
t x =t dx =dti cn: x =3pi
2 t = 3pi
2; x = 0 t = 0
Khi : I1 =0
3pi2
f (t ) (1)dt =3pi00
f (t )dt =3pi20
f (x)dx
Ta c: I =3pi2
0
f (x)dx+3pi2
0
f (x)dx =3pi2
0
[f (x)+ f (x)]dx
=3pi2
0
p22cosx dx =
3pi2
0
2|sinx|dx
= 2
pi0
sinx dx3pi2
pi
sinx dx
= 6.Tnh cht sau cho ta mt kt qu p, thu gn ng k hm di du tch phn.
Tnh cht 9.5. Nu f (x) l hm chn v lin tc trn R th
I =
f (x)
ax +1 dx =0
f (x)dx vi R+ v a > 0
Chngminh
Ta c: I =
f (x)
ax +1 dx =0
f (x)
ax +1 dx+0
f (x)
ax +1 dx = I1+ I2
Xt: I1 =0
f (x)
ax +1 dx
t x =t dx =dti cn: x = t = ; x = 0 t = 0
Khi : I1 =0
f (t )at +1 dt =
0
at f (t )
at +1 dt =0
ax f (x)
ax +1 dx
Ta c: I =0
ax f (x)
ax +1 dx+0
f (x)
ax +1 dx
=0
(ax +1) f (x)ax +1 dx =
0
f (x)dx
Nguyn Hng ip 57
9. TNH TNH PHN BNG TNH CHT Chng I. TCH PHN
V d 9.6. Tnh I =1
1
cosx
ex +1 dx
Gii
Ta c: I =1
1
cosx
ex +1 dx =0
1
cosx
ex +1 dx+1
0
cosx
ex +1 dx = I1+ I2
Xt I1 =0
1
cosx
ex +1 dx
t x =t dx =dti cn: x =1 t = 1 ; x = 0 t = 0
Khi : I1 =0
1
cos(t )et +1 dt =
10
e t cos t
e t +1 dt =1
0
ex cos t
ex +1 dx
Ta c: I =1
0
ex cos t
ex +1 dx+1
0
cosx
ex +1 dx =1
0
(ex +1)cosxex +1 dx
=1
0
cosx dx = sin1
Nhn xt: da vo Tnh cht 9.5 ta d on kt qu I =10cosxdx, mt kt qu p trong
bi tch phn tng chng rc ri. Nhng ta khng c p dng trc tip kt qu nym ch dng nh hng phng php gii.
Bi ton tng t
1.
11
x4
1+2x dx. p s:15
2.
pi2
pi2
sinx sin2x cos5xex +1 dx. p s: 0
3.
11
p1x21+2x dx. p s:
pi4
4.
pi2
pi2
x2
1+2x |sinx|dx. p s: pi2
5.
pipi
sin2
32+1 dx. p s:pi2
58 Nguyn Hng ip
Chng I. TCH PHN 9. TNH TNH PHN BNG TNH CHT
6.
11
1
(ex +1)(x2+1) dx. p s: pi49.2 Tch phn c cn l radian
Trng hp tng qut dng I =b
a
f (x)dx ta i bin x = a+b t
V d 9.7. Tnh I =pi2
0
sin6 x
sin6 x+cos6 x dx
Gii
t x = pi2 t dx =dt
i cn: x = 0 t = pi2; x = pi
2 t = 0
Khi : I =0
pi2
sin6(pi2 t)
sin6(pi2 t)+cos6
(pi2 t) (1)dt
=pi2
0
cos6 t
cos6 t + sin6 t d t =pi2
0
cos6 x
cos6 x+ sin6 x dx = I
Do : 2I = I + I =pi2
0
cos6 x
cos6 x+ sin6 x dx+pi2
0
sin6 x
sin6 x+cos6 x dx
=pi2
0
sin6 x+cos6 xsin6 x+cos6 x dx =
pi2
0
dx = pi2
I = pi4
Nhn xt: V d trn cn c th gii bng phng php hm ph, V d trn minh hacho tnh cht:
Tnh cht 9.8. Nu f (x) l hm s lin tc trn [0,1] th
I =pi2
0
f (sinx)dx =pi2
0
f (cosx)dx
Hng dn chng minh: t t = pi2 t
Nguyn Hng ip 59
9. TNH TNH PHN BNG TNH CHT Chng I. TCH PHN
Bi ton tng t
1.
pi2
0
sinn x
sinn x+cosn x dx. p s:pi4
2.
pi2
0
ln
(1+ sinx1 sinx
)dx. p s: 0
3.
pi2
0
(1
cos2(cosx) tan2(sinx)
)dx. p s: pi2
Tnh cht sau cho ta thu gn hm di du tch phn
Tnh cht 9.9. Nu f (x) lin tc trn [0,1] th
I =pi0
x f (x)dx = pi2
pi0
f (x)dx
Hng dn chng minh: t t =pi t
V d 9.10. Tnh I =pi0
x. sinx.cos2 x dx
Gii
t x =pi t dx =dti cn: x = 0 t =pi ; x =pi t = 0
Khi : I =0
pi
(pi t ).sin(pi t ).cos2(pi t ).(t )dt
=pi0
(pi t ).sin t .cos2 t d t
=pipi0
sin t .cos2 t d t pi0
t . sin t cos2 t d t
=pipi0
sinx cos2 x dx I
2I =pipi0
sinx cos2 x dx I = pi2
pi0
sinx cos2 x dx
t t = cosx dt =sinxdxi cn: x = 0 t = 1 ; x =pi t =1
60 Nguyn Hng ip
Chng I. TCH PHN 9. TNH TNH PHN BNG TNH CHT
Khi : I =pi2
11
t2dt = pi2 t
3
3
11 = pi3Nhn xt: V d trn c th gii bng phng php Tch phn tng phn nhng bi giidi dng hn. y ta c nhn xt
sinx.cos2 x = sinx(1 sin2 x)= f (sinx)v theo Tnh cht 9.9 ta thu gn c bi ton.
Bi ton tng t
1.
pi0
x sinx
4cos2 x dx. p s:pi ln98
2.
pi0
x sinx
9+4cos2 x dx. p s:pi6 arctan
23
Tng t ta c Tnh cht i vi hm cosin
Tnh cht 9.11. Nu f (x) lin tc trn [0,1] th
I =2pi
x f (cosx)dx =pi2pi0
f (cosx)dx
Hng dn chng minh: t x = 2pi t
V d 9.12. Tnh I =2pi0
x cos3 x dx
Gii
t x = 2pi t dx =dti cn: x = 0 t = 2pi ; x = 2pi t = 0
Khi : I =0
2pi
(2pi t )cos3(2pi t ).(t )dt =2pi0
(2pi t )cos3 t d t
= 2pi2pi0
cos3 t d t 2pi0
t cos3 t d t
= pi2
2pi0
(cos3t +3cos t )dt 2pi0
t cos3 x dx
(do cos3x = 4cos3 x3cosx)= pi2
(1
3sin3t +3sin t
)2pi0 I
= 0 I 2I = 0 I = 0
Nguyn Hng ip 61
9. TNH TNH PHN BNG TNH CHT Chng I. TCH PHN
Tnh cht sau l dng tng qut ca hai Tnh cht trc
Tnh cht 9.13. Nu f (x) lin tc v f (a+bx)= f (x) th
I =b
a
x f (x)dx = a+b2
ba
f (x)dx
Hng dn chng minh: t x = a+b tTng t ta chng minh c Tnh cht sau:
Tnh cht 9.14. Nu f (x) lin tc trn [a,b] th
I =b
a
f (x)dx =ba
f (a+bx)dx
Hng dn chng minh: t t = a+b t
V d 9.15. Tnh I =pi4
0
ln(1+ tanx)dx
Gii
t x = pi4x dx =dt
i cn: x = 0 t = pi4; x = pi
4 t = 0
Khi : I =0
pi4
ln[1+ tan(1+ tan t )]dt =pi4
0
ln
(1+ 1 tan t
1+ tan t)dt
=pi4
0
ln2
1+ tan t d t =pi4
0
[ln2 ln(1+ tan t )]dt
= ln2pi4
0
dt pi4
0
ln(1+ tan t )dt = ln2|pi40
pi4
0
ln(1+ tanx)dx
= pi ln24
I 2I = pi ln2
4 I = pi ln2
8
Tnh cht 9.16. Nu f (x) lin tc trn R v tun hon vi chu k T th
I =a+Ta
f (x)dx =Ta
f (x)dx
62 Nguyn Hng ip
Chng I. TCH PHN 9. TNH TNH PHN BNG TNH CHT
Chngminh tnh cht
Ta c: I =T0
f (x)dx =a0
f (x)dx+a+Ta
f (x)dx+T
a+Tf (x)dx
= I1+ I2+ I3
Xt I3 =T
a+Tf (x)dx
t x = T + t dx = dti cn: x = a+T t = a ; x =T t = 0
Khi : I3 =0
a
f (t +T)dt =a0
f (t )dt =a0
f (x)dx =I1
Ta c: I = I1+ I2+ I3 = I1+ I2 I1I = I2
V d 9.17. Tnh I =2014pi0
p1cos2x dx
Gii
Ta c: I =2014pi0
p1cosx dx =
2014pi0
2sin2 x dx =p2
2014pi0
|sinx|dx
=p2 2pi0
|sinx|dx+4pi
2pi
|sinx|dx+ +2014pi
2012pi
|sinx|dx
Theo Tnh cht 9.16 ta c:2pi0
|sinx|dx =4pi
2pi
|sinx|dx = =2014pi
2012pi
|sinx|dx
Ta c: I = 1007p22pi0
|sinx|dx = 1007p2 pi0
sinx dx2pipi
sinx dx
= 4028p2
Nhn xt: vic xt du hm sinx trn [0,2014pi] l kh khn, V d trn cho ta thy hiuqu ca Tnh cht 9.16, . Do Tnh cht ny khng c trong sch gio khoa, v vic chngminh Tnh cht i vi hm c th l di dng nn trc khi lm bi cc em hc sinhcn chng minh li dng tng qut ca n nh phn Chng minh, sau p dng ktqu lm bi tp.
Bi tp tng hp
1.
21
ln(1+x)x2
dx. p s: 32 ln3+3ln2
2.
pi2
0
cosx. ln(1+cosx)dx. p s: pi2 1
Nguyn Hng ip 63
10. PHNG PHP TNH TCH PHN TNG PHN Chng I. TCH PHN
3.
pi0
x.cos4 x. sin3 x dx. p s: 2pi35
4.
22
ln(x+x2+1
)dx. p s: 0
5.
pi2
pi2
sinx. sin2x. sin5x
ex +1 dx. p s: 0
6.
3pi2
0
sinx. sin2x. sin3x.cos5x dx. p s: 0
7.
pi2
pi2
x+cosx4 sin2 x dx.
8.
11
x2014
2007x +1 dx.
9.
pi4
pi4
sin6 x+cos6 x6x +1 dx
10.
pi4
0
log2008(1+ tanx)dx
11.
pi2
0
sinx cosxa2 cos2+b2 sin2
dx. p s: 1|a|+|b|
12.
pi2
0
cosxp2+cos2x dx. p s:
1p2arcsin
p63
10 Phng php tnh tch phn tng phn
Mt s iu lu khi tch tch phn tng phn
1. Hm no kh ly nguyn hm ta t l l u.
2. Trong trng hp c hm a thc ta t u l hm a thc gim dn bc ca athc.
3. Trong tch phn cn tnh c cha hm logarit th ta t u l hm cha logarit.
64 Nguyn Hng ip
Chng I. TCH PHN 10. PHNG PHP TNH TCH PHN TNG PHN
10.1 Dng 1
Tnh tch phn c dng I =P (x). f (x)dx trong P (x) l a thc bc n v f (x) l mt
trong cc hm sinx,cosx,ex ,ax .Khi ta t {
u = P (x)dv = f (x)
V d 10.1. Tnh cc tch phn sau:
1. I1 =pi2
0
x cosx dx
2. I2 =1
0
(x2)e2x dx
3. I3 =1
0
(x+1)2e2x dx
Gii
1. Tch phn tng phn dng 1
t u = t du = dxdv = cosx dx ; v = sinx
Khi : I1 = x. sinx|pi20
pi2
0
sinx dx = pi21.
2. I2 =1
0
(x1)e2x dx
t u = x1 du = dxdv = e2x dx ; v =
e2x
2
Khi : I2 = (x1) e2x
2
10 12
10
e2x dx = 13e2
4.
3. a thc bc 2 ta tnh tch phn tng phn hai ln.
Nguyn Hng ip 65
10. PHNG PHP TNH TCH PHN TNG PHN Chng I. TCH PHN
t u = (x+1)2 du = 2(x+1) dxdv = e2x dx ; v =
1
2e2x
Khi : I3 = 12e2x(x+1)2
10
10
(x+1)e2x dx
= 2e2 12
10
(x+1)e2x dx
Tch phn I2 =1
0
(x+1)e2x dx c tnh trn.
Vy: I = 5e214
Nhn xt: a thc P (n) c bc n ta tnh n n ln.
V d 10.2. Tnh I =pi2
0
e2x cos3x dx
Gii
t u = e2x du = 2e2x dxdv = cos3x dx ; v =
1
3sin3x
Khi : I = 13e2x sin3x
pi20 23
pi2
0
e2x sin3x dx =13epi 2
3J
Xt J =pi2
0
e2x sin3x dx
t u = e2x du = 2e2x dxdv = sin3x dx ; v = 1
3cos3x
Khi : J = 13e2x cos3x
pi20+ 23
pi2
0
e2x cos3x dx = 13+ 23I
T ta c:
I =13epi 2
9 49I I = 3e
pi213
.
Nhn xt:
1) Trong tch phn J nu ta t u = cos3x (khc kiu vi cch t ca tch phn I ) th tathy xut hin I nhng khi thay vo s c iu hin nhin ng, bi ton i vong ct.
66 Nguyn Hng ip
Chng I. TCH PHN 10. PHNG PHP TNH TCH PHN TNG PHN
2) Trong bi ton tch phn cha ex .cosx (hoc ex . sinx) ta tnh tch phn tng phn hailn. Ln u ta t kiu no cng c nhng ln 2 phi cng kiu vi ln 1.
V d 10.3. Tnh I =1
0
ex+exdx
Gii
Ta c: I =1
0
ex .eexdx
i bin:t t = ex dt = exdxi cn: x = 0 t = 1 ; x = 1 t = e
Khi : I =e
1
te t dt =e
1
xex dx
Tch phn tng phn
t u = t du = dxdv = e t dx ; v = e t
Khi : I = xex |e1e
1
ex dx = ee(e1).
Nhn xt: V d trn cho thy cc bi tp tch phn khng ch l cc dng n l m ltng hp nhiu phng php, gii quyt mt bi ton tch phn ta cn nm vngcc dng ton v bit cch kt hp chng.
Bi ton tng t
1.
pi2
0
(x+ sin2 x)cosx dx. p s: 3pi46 .
2.
pi4
0
x
1+cos2x dx. p s:pi8 14 ln2.
3.
pi4
0
x tan2 x dx. p s: pi4 12 ln2 pi32 .
4.
20
x2ex
(x+2)2 dx. p s: e2+1, hd u=x
2
e
x
5.
pi2
0
(2x1)cos2 x dx. p s: pi28 pi4 12 .
Nguyn Hng ip 67
10. PHNG PHP TNH TCH PHN TNG PHN Chng I. TCH PHN
6.
pi2
0
(x+1)sin2x dx. p s: pi4 +1
7.
pi3
pi4
x
sin2 xdx. p s: (94
p3)pi
36 + 12 ln 34
8.
pi2
0
e2x sin3x dx. p s: 32epi
13
10.2 Dng 2
Tnh tch phn c dng I =P (x) f (x)dx trong f (x) l mt trong cc hm lnx, loga x.
Khi ta t {u = f (x)dv = P (x)
V d 10.4. Tnh I =3
2
ln(x3x)dx
Gii
t u = ln(xx2) du = 2x1x(x1) dx
dv = dx ; v = (x1)
Khi : I = (x1)ln(x2x)323
2
2x1x
dx = 3ln32.
Nhn xt: ta c chn la v , thng thng vi bi ny ta chn v = x nhng yv = x1 th bi ton gn hn.
V d 10.5. Tnh I =3
1
3+ lnx(x+1)2 dx
Gii
t u = 3+ lnx du = 1xdx
dv =1
(x+1)2 dx ; v = 1
x+1
68 Nguyn Hng ip
Chng I. TCH PHN 10. PHNG PHP TNH TCH PHN TNG PHN
Khi : I =(3+ lnx
x+1)31+
31
1
x(x+1) dx
=(3+ lnx
x+1)31+
31
(1
x 1x1
)dx
=(3+ lnx
x+1)31+(ln
x
x+1)31= 34 ln2+ 3
4ln3.
V d 10.6. Tnh I =pi2
0
cosx ln(1+cosx)dx
Gii
t u = ln(1+cosx) du = sinx1+cosx dx
dv = cosx dx ; v = sinx
Khi : I = sinx ln(1+cosx)|pi20
=0+
pi2
0
sin2 x
1+cosx dx
=pi2
0
1cos2 x1+cosx dx =
pi2
0
(1cosx)dx = pi21.
V d 10.7. Tnh I =epi1
cos(lnx)dx
Gii
t u = cos(lnx) du = 1xsin(lnx dx
dv = ) dx ; v =
x Khi : I = x cos(lnx)|epi1 +epi1
sin(lnx)dx
J
=epi1+ J
t u = sin(lnx) du = 1xcos(lnx) dx
dv = dx ; v = x
Khi : J = x. sin(lnx)|epi1 0
epi1
cos(lnx)dx =I
Do : I =epi1 I I =epi+12
Nguyn Hng ip 69
10. PHNG PHP TNH TCH PHN TNG PHN Chng I. TCH PHN
Bi ton tng t
1.
32
ln(x2x)dx. p s: 3ln32
2.
e1
x3 ln2 x dx. p s: 5e4132
3.
e1
x2 lnx dx. p s: 5e3227
4.
e1
x3 ln2 x dx. p s: 5x4132
5.
21
lnx
x3dx. p s: 316 18 ln2.
6.
e1
(2x 3
x
)lnx dx. p s: e
2
2 1.
7.
e1
x2+1x
lnx dx. p s: 3e2+54
8.
31
x2 ln(x+1)dx. p s: 18ln2 209
9.
pi4
pi3
sinx. ln(tanx)dx. p s: 34 ln3+ ln(p21)
10.3 Phng php hng s bt nh
3 Dng 1
Tnh tch phn c dng I =
f (x)ex dx trong f (x) l a thc bc n,1 n Z.
Ta thc hin theo cc bc:
1. Ta c:I = g (x)ex +C (1)
trong g (x) l a thc cng bc vi f (x).
2. Ly o hm hai v ca (1) v p dng phng php tr s ring hoc ng nhtthc xc nh cc a thc g (x).
70 Nguyn Hng ip
Chng I. TCH PHN 10. PHNG PHP TNH TCH PHN TNG PHN
V d 10.8. Tnh I =1
0
(2x3+5x22x+4)e2x dx
Gii
Gi s
I1 =(2x3+5x22x+4)e2x dx = (ax3+bx2+ cx+d)e2x +C (1)
Ly o hm hai v ca (1) ta c:
2x3+5x22x+4)e2x = [2ax3+ (3a+2b)x2+ (2b+2c)x+2d ]e2x
ng nht ng thc trn ta c2a = 23a+2b = 52b+2c =2c+2d = 4
a = 1b = 1c =2d = 3
Do I1 = (x3+x22x+3)e2x +CVy: I = (x3+x22x+3)e2x10 = e23.Nhn xt: phng php trn hu hiu trong trng hp a thc bc ln hn hoc bng3, trong bi ton trn nu p dng cch thng thng ta phi ly tch phn tng phn3 ln.
Bi ton tng t1
0
x5ex dx. p s: 12044e.
3 Dng 2
Tnh tch phn c dng I =
f (x)sinx dx hoc I =
f (x)cosx dx trong
f (x) l a thc bc n,1 n Z.Ta thc hin theo cc bc:
1. Ta c:I = g (x)sinx+h(x)cosx+C (1)
trong g (x),h(x) l a thc cng bc vi f (x).
2. Ly o hm hai v ca (1) v p dng phng php tr s ring hoc ng nhtthc xc nh cc a thc g (x),h(x).
Nguyn Hng ip 71
10. PHNG PHP TNH TCH PHN TNG PHN Chng I. TCH PHN
3 Dng 3
Tnh tch phn c dng I =ex cosx dx hoc I =
ex sinx dx ta thc hin
theo cc bc:
1. Ta c:I = (A cosx+B sinx)ex +C (1)
2. Ly o hm hai v ca (1) v p dng phng php tr s ring hoc ng nhtthc xc nh A,B .
V d 10.9. Tnh I =pi2
0
ex .cosx dx
Gii
Ta c:
I1 =ex .cosx dx = (A cosx+B sinx)ex +C (1)
Ly o hm hai v ca (1) ta c
ex cosx = [(A+B)cosx+ (B A)sinx]ex
ng nht thc ng thc trn ta c{A+B = 1B A = 0
{A = 12B = 12
Khi : I1 = 12(sinx+cosx)ex +C
Vy: I = 12(sinx+cosx)ex
pi20= 12epi2 1
2.
V d 10.10. Tnh I =pi2
0
ex . sin2 x dx
Gii
Ta c: I = 12
pi2
0
(1cos2x)ex dx
Mt khc
I1 = 12
(1cos2x)ex dx = (A+B cos2x+C sin2x)ex +D
72 Nguyn Hng ip
Chng I. TCH PHN 11. CC BI TON C BIT
Ly o hm hai v ng thc trn ta c
1
2(1cos2x)ex = [a+ (2C +B)cos2x+ (C 2B)sin2x]ex
ng nht ng thc ta c2A = 12(2C +B) =12(C 2B) = 0
A = 12B = 110C =15
Do : I1 = 110
(5cos2x2sin2x)ex +D
Khi : I = 110
(5cos2x2sin2x)expi20= 35epi2 2
5.
11 Cc bi ton c bit
C nhng bi ton tch phn c th gii c bng phng php tch phn tng phn.i khi gii quyt mt bi tch phn m i bin mi khng c ta ngh n phngphp ny.
V d 11.1. Tnh I =1
0
x3px2+1
dx
Gii
Cch 1 i bin s t t =px2+1 x2 = t21 xdx = td t
i cn: x = 0 t = 1 ; x = 1 t =p2
Ta c: I =p2
1
(t21)dt = 23p2
3.
Cch 2 Tch phn tng phn
t u = x2 du = 2x dxdv =
xpx2+1
dx ; v =px2+1
Vy I =(x2px2+1
)10=
10
2xx2+1dx 2
3p2
3
V d 11.2. Tnh I =p3
1
px2+1x2
dx
Nguyn Hng ip 73
11. CC BI TON C BIT Chng I. TCH PHN
Gii
Cch 1 i bint x = tan t , t
(pi2,pi
2
)px2+1= 1
cos2 t
dx = dtcos2 t
i cn: x = 0 t = pi4; x =p3 t = pi
3
Khi : I =pi3
pi4
1
cos t sin2 tdx =
pi3
pi4
cos t
cos2 sin2 tdx
i bin s u = sin t ta c
I =
p32
p22
1
(1u2)u2 dx =
p32
p22
(1
1u2 +1
u2
)dx
=p2 23
p3+ ln(2+p3) ln(1+p2)
Cch 2 Tch phn tng phn
t u =px2+1 du = xp
x2+1dx
dv =1
x2dx ; v = 1
x
Khi : I =(1x
px2+1
)p3
1+
p3
1
1px2+1
dx
=(1x
px2+1
)p3
1+ ln(x+
px2+1)
p31
=p2 23
p3+ ln(2+p3) ln(1+p2)
Nhn xt: bi ny dng phng php tch phn tng phn l hp l.
Bi tp
1.
10
x3
x8+1 dx. Hd: t = x4
2.
10
2x
4x +1 dx. Hd: t = 2x
3.
pi2
0
cosx
1310sinxcos2x dx. Hd: t = sinx
74 Nguyn Hng ip
IING DNG CA TCH PHN
1. TNH DIN TCH HNH PHNG Chng II. NG DNG CA TCH PHN
1 Tnh din tch hnh phng
1.1 Cng thc tnh
Mt hnh phng gii hn bi (C1) : y = f (x), (C2) : y = g (x), v x = a,x = b, khi din tch hnh phng tnh bi cng thc:
S =b
a
f (x) g (x)dx
Mt s lu
1) Trong trng hp bi khng cho sn cn a,b ta tm honh giao im (C1) v(C2) l nghim phng trnh f (x) g (x)= 0.
2) b du gi tr tuyt i ta c 3 cch
(a) Da vo th: nu nhn vo th ta thy (C1)nm trn (C2) th f (x)g (x) 0khi
f (x) g (x)= f (x) g (x).(b) Lp bng xt du ca f (x)g (x) (xem li Tch phn hm cha du gi tr tuyt
i)
(c) Nu phng trnh f (x) g (x) = 0 ch c hai nghim l x = a,x = b v v hms h(x)= f (x) g (x) lin tc nn f (x) g (x) khng i du trn [a,b] khi tac em du tr tuyt i ra ngoi du tch phn:
S =b
a
f (x) g (x)dx =b
a
[f (x) g (x)]dx
1.2 Cc v d
V d 1.1. Tnh din tch hnh phng c gii hn bi ng cong (C ) : f (x)= 3x1x1
v hai trc ta .
Gii
Ta tm cn ca tch phn l honh giao im ca (C ) vi cc trc ta . Honh
giao im ca (C ) v trc honh l x =13, vi trc tung l x = 0
Khi : S =0
13
3 4x1dx =
0 13
(3 4
x1)dx =1+ ln 4
3.
76 Nguyn Hng ip
Chng II. NG DNG CA TCH PHN 1. TNH DIN TCH HNH PHNG
1
1
1
013
V d 1.2. Tnh din tch hnh phng gii hn bi cc ng (C1) : f (x) = (e + 1)x v(C2) : g (x)= (1+ex)x
Gii
Honh giao im ca hai ng cong l nghim ca phng trnh
(e+1)x = (1+ex)x x = 0 x = 1
Do (C1) ct (C2) ti hai im phn bit nn ta c
S =1
0
(e+1)x+ (1+ex)xdx = 10
exxexdx =1
0
(exxex)dx= |I |
Ta c: I =1
0
(exxex)dx =1
0
ex dx1
0
xex dx
=(e.x2
2
)10 (xex ex)|10 =
e
21.
Vy: S =e21= e
21.
12 1 2 3
1
3
0
V d 1.3. Tnh din tch hnh phng gii hn bi ng cong (C ) : y = |x2 4x + 3| vd : y = 3
Gii
Phng trnh honh giao im ca (C ) v d
x24x+3= 3 [ x24x+3 = 3x24x+3 =3
[x = 0x = 4
Nguyn Hng ip 77
1. TNH DIN TCH HNH PHNG Chng II. NG DNG CA TCH PHN
Ta c S =4
0
(|x24x+3|3)dx =4
0
(3|x24x+3|)dx
Xt du f (x)= x24x+3 b du gi tr tuyt i ca |x24x+3|Cho x24x+3= 0 x = 1 x = 3Bng xt du:
x
f (x)
1 3 ++ 0 0 +
Khi :
2 21 3 4 6 8
2
2
10
0
V d 1.4. Tnh din tch hnh phng gii hn bi (P ) : y2 = 4x v d : y = 2x4
Gii
Ta c: y2 = 4x x = y2
4
y = 2x4 x = y +42
Tung giao im (P ) v d l nghim phng trnh
y2
4= y +4
2 y =2 y = 4
Khi din tch hnh phng c tnh bi
S =4
2
( y +42 y2
4
)dy =
42
(y +42
y2
4
)dy
= 9.78 Nguyn Hng ip
Chng II. NG DNG CA TCH PHN 2. TH TCH VT TH TRN XOAY
2 2 4 6 8
4
2
2
4
0
a
f
g
2 Th tch vt th trn xoay
2.1 Hnh phng quay quanh Ox
3 Cng thc
Hnh phng (H) gii hn bi (C1) : y = f (x), (C2) : y = g (x), x = a,x = b khi quay (H) quanhtrcOx ta c mt vt th trn xoay c th tch c tnh theo cng thc
V =pib
a
f 2(x) g 2(x)dxc bit khi (C2) l trc honh th cng thc trn tr thnh
V =pib
a
f 2(x)dx
3 Cc v d
V d 2.1. Tnh th tch vt th trn xoay khi quay min (D) gii hn bi (C ) : y = lnx,y = 0,x = 2 quanh trcOx.
Nguyn Hng ip 79
2. TH TCH VT TH TRN XOAY Chng II. NG DNG CA TCH PHN
Gii
Honh giao im ca (C ) v trc honh y = 0 l nghim phng trnh
lnx = 0 x = 1
Khi : V =pi2
1
ln2 x dx
Sau khi tnh tch phn tng phn 2 ln ta thu c kt qu
V = 2ln22+4ln2+2
.
80 Nguyn Hng ip
IIIBI TP TNG HP
1. CC THI TUYN SINH 2002-2014 Chng III. BI TP TNGHP
1 Cc thi tuyn sinh 2002-2014
1. Tnh din tch hnh phng gii hn bi y = |x24x+3|, y = x+3. (2002-A). p s:1096
2. Tnh din tch hnh phng gii hn bi y =4 x
2
4, y = x
2
4p2. (2002-B).
p s: 2pi+ 43
3. Tnh din tch hnh phng gii hn bi y = 3x1x1 ,Ox,Oy . (2002-D).
p s: 1+4ln 43
4. Tnh
pi2
0
61cos5 x sinx dx (D b 2002-A). p s: 1291
5. Tnh
01
x(e2x + 3px+1
)dx (D b 2002-A). p s: 3
4e2 47 .
6. Tnh I =ln30
ex(ex +1)3
dx (D b 2002-B). p s:p21
7. Tnh din tch hnh phng gii hn bi (C ) : y = 13x3 2x2+ 3x v trc Ox (D b
2002-D). p s: 94
8. Tnh I =1
0
x3
1+x2 dx (D b 2002-D). p s:12 (1 ln2).
9. Tnh I =2p3
p5
1
xpx2+4
dx (2003-A). p s: 14 ln53
10. Tnh I =pi4
0
12sin2 x1+ sin2x dx (2003-B). p s:
12 ln2
11. Tnh I =2
0
|x2x|dx (2003-D). p s: 1
12. Tnh I =pi4
0
x
1+cos2x dx (D b 2003-A). p s:pi8 14 ln2.
13. Tnh I =1
0
x31x2dx (D b 2003-A). p s: 215 .
82 Nguyn Hng ip
Chng III. BI TP TNGHP 1. CC THI TUYN SINH 2002-2014
14. Tnh I = tpln2ln5 e2x
pex 1 (D b 2003-B). p s:
203
15. Cho f (x)= a(x+1)3 +bx.e
x . Tm a,b bit f (0)=22 v I =1
0
f (x)dx = 5 (D b 2003-
B). p s:a = 8,b = 2.
16. Tnh I =1
0
x3ex2dx (D b 2003-D). p s: 12
17. Tnh I =e
0
x2+1x
lnx dx (D b 2003-D). p s: e2
4 + 34
18. Tnh I =2
1
Recommended