Tich phan on thi DH

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


(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


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


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


4.3 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


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.


(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


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


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


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


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


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


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


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


=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


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


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


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


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


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.


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


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


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]


(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


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


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.


0

3sinx+5cosx+2sinx+cosx+1 dx

Gii

38 Nguyn Hng ip


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.


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


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 .


0

4cosx7sinx(2sinx+cosx)3 dx

40 Nguyn Hng ip


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


hn. Dng ny thng c p dng i vi hm s lng gic.


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


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


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


= 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


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


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


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


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


= 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


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


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


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


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


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


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


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


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


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


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


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)


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


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.


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


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


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


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.


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


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


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


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


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


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 .


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

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