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PHNG PHP V K THUT IN HNH TRONG TNH PHN

Nguyn Vn Cng, THPT M c A, H Ni

T: 0127.233.45.98 - 04.33.741.526 Email: [email protected]

ng ti ti http://www.mathvn.com/2011/01/cac-phuong-phap-tinh-tich-phan-ien-hinh.html

Php tnh tch phn l mt phn quan trng ca gii tch ton hc ni ring v trong Ton hc ni chung,khng nhng nh l mt i tng nghin cu trng tm ca gii tch m cn c c lc trong nghin cu l thuyt v phng trnh, l thuyt v hm s.

Ngoi ra php tnh vi phn cn c s dng nhiu trong cc mn khoa hc khc nh Vt l Thin vn hc ,c hc ....n nh l mt gii php hu hiu ca cc m hnh ton hc c th..Hc sinh lp 12 Khi n thi tt nghip ,Thi i hc cao ng thng rt gp kh khn khi gii cc bi tp trong chuyn ny. Nhng ngi mi hc v lm quen vi Tch phn thng cha hiu r t tng cng nh phng php tip cn l thuyt , c bit l khu vn dng l thuyt vo gii cc bi ton thc t.

Bi vit ny xin nu ra mt s phng php in hnh thng c dng gii cc bi tp v tch phn trong cc k thi i hc. Ni dung bi vit cng l ni dung c bn ca ti sng kin kinh nghim ca ti trong nm hc 2010 c S gio dc v o to H Ni xp loi B.

Mc d tham kho mt s lng ln cc ti liu hin nay va vit, va i ging dy trn lp kim nghim song v nng lc v thi gian c hn ,rt mong c s ng gp ca cc bn ng nghip v nhng ngi yu thch mn ton chuyn ny c ngha thit thc hn trong nh trng ,gp phn nng cao hn na cht lng Gio dc ph thng.Gip cc em c phng php - k nng khi gii cc bi Tch phn trong cc k thi cui cp ng thi bc u trang b cho cc em kin thc v php tnh vi phn Tch phn trong nhng nm u hc i hc. Xin vui lng gii thiu vi cc bn ng nghip v nhng ngi yu ton chuyn :

Phng php v k thut in hnh tnh tch phn

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Nguyn Vn Cng, M c A, H Ni - www.MATHVN.com 2

I - K thut bin i vi phn (a v bng nguyn hm) Khi s dng k thut bng nguyn hm ta cn lu n mt s php ton vi phn n gin sau: f (x)dx=dF(x) ,Trong F(x)- l mt nguyn hm ca hm s f(x) dx= 1 ( )d ax b

a+ xkdx=d

1

( )1

kxa

k

+

++

sinxdx=d(-cosx)

2

2 2

( )dx d x x a

x a x x a

+ +=

+ + +;

2(t anx)

osdx

dc x

= ; 2

( cot x)sin

dxd

x= - ....

Mt s cng thc suy rng sau

cossin

kxkxdx c

k= - + ;

sinos

kxc kxdx c

k= + ;

kxkx ee dx c

k= + ; ,ln

kxkx aa dx c k R

k a= + " .....

V d 1( HA -2010) Tnh tch phn : 1 2 x 2 x

x0

x e 2x eI dx

1 2e+ +

=+

Li gii 1 1 12

2

0 0 0

(1 2 )1 2 1 2

x x x

x x

x e e eI dx x dx dx

e e+ +

= = ++ + ;

11 32

10 0

1;

3 3x

I x dx= = = 1

20 1 2

x

x

eI dx

e=

+ = 1

0

1 (1 2 )2 1 2

x

x

d ee

++ =

1

0

1ln(1 2 )

2xe+ =

1 1 2ln

2 3e+

Vy I = 1 1 1 2ln3 2 3

e+ +

V d 1( HA -2009) Tnh tch phan 2

3 2

0

I (cos x 1)cos xdx

p

= -

Li gii

( ) ( )

( )

2 2 2 2 223 2 5 2 4 2

10 0 0 0 0

2 2 2 2 2 2 22 4 22

0 00 0 0 0 0

cos 1 cos cos cos , cos cos 1 sin cos

8 1 cos 2 1 1 1 11 2sin sin (sinx) , cos cos 2 sin 2

15 2 2 2 2 4 4

I x xdx xdx xdx I x xdx x xdx

xx x d I xdx dx dx xdx x x

p p p p p

p p p p p p p

p

= - = - = = - =

+- + = = = = + = + =

Tnh tch phn 3

x1

dxI

e 1=

-

V d 3 HKD -09) Tnh tch phn 3

x1

dxI

e 1=

-

Li gii 3 3 3x x x

3xx x 1

1 1 1

1 e e eI dx dx dx 2 ln e 1

e 1 e 1- +

= = - + = - + -- -



3 22 ln(e 1) ln(e 1) 2 ln(e e 1)= - + - - - = - + + +

V d 1 (HKB -03) Tnh I= /4 2

0

1 2sin1 sin 2

xdx

x

p -+

Li gii: Nhn thy d(1+sin2x)= 1 os2

2c xdx , 1-2sin2x=cos2x nn ta c

I =/4 /4 /42

/40

0 0 0

1 2sin os2 1 (1 sin 2 ) 1 1ln(1 sin 2 ) ln 2

1 sin 2 1 sin 2 2 1 sin 2 2 2x c x d xdx dx x

x x x

p p pp- += = = + =

+ + +

V d 2 (H KA-06) J =

/4

2 20

sin 2

os 4sin

xdx

c x x

p

+

Li gii: Nhn thy d(cos2x+4sin2x)=sin2xdx do ta c

J=/4

2 20

sin 2

os 4sin

xdx

c x x

p

+ =

/4 2 2

2 20

1 ( os 4sin )3 os 4sin

d c x xdx

c x x

p +

+ =

12 2 /42

0

2( os 4sin )

3c x x p+ = 1 ( 10 2)

3-

V d 3 Tnh K=

3 2

1

ln 2 lne x xdx

x+

(HKB-04) Li gii: K =

3 23 2 2 1/3 2 33

1 1 1

ln 2 ln 1 32 ln ln (ln ) (2 ln ) (2 ln ) (3 3 2 2)

2 8

e e ex xdx x xd x x d x

x+

= + = + + = -

Nhn xt 1: - Cc tch phn trn c th gii c bng phng php i bin s song nu ta kho lo bin i vi phn th a c v cc tich phn c bn . -Dng php bin i vi phn a v bng nguyn hm c bn gip Li gii ngn gn,so vi Php i bin s th khng phi i cn ,Trong gii ton thm mt php ton l thm mt nguy c sai. lm r u im ca phng php ny ta xt bi ton sau

V d 4: Tnh L= ( )ln ( ) ( )( )( )

bx a x b

a

dxx a x b

x a x b+ + + + + + vi b>a>0

Li gii:



Vit li L= ( ) ln( ) ( ) ln( )( )( )

b

a

x a x a x b x bx a x b

+ + + + ++ + dx =

ln( ) ln( )b

a

x a x bdx

x b x a+ + + + + =

[ ]ln( ) ln( ) ln( ) ln( )b

a

x x d x b x b d x a+ + + + +

= [ ]ln( ) ln( ) ln( ) ln( ) ln ln( )b

ba

a

ad x b x b x a x b a b

b+ + = + + = +

Nhn xt 2 -y l mt trong nhng bi ton in hnh minh ho tnh u vit cho phng php s dng phng php bin i vi phn a v bng nguyn hm -Mt trong nhng phng php c bn nht tnh tch phn lng gic l bin i Vi phn a v bng nguyn hm c bn,khi ta cn dng cc cng thc bin i lng gic nh h bc ,nhn i ,tng thnh tch ... ta xt cc v d sau V d 5 Tnh M=

/2sin

0

( cos ) cosxe x xdxp

+ (H K D-05) Li gii: M=

/2 /2sin

0 0

1 os2(cos ) 1

2 4x c xe d x dx e

p p p++ = - +

V d 6: Tnh N=

/3

2 2/4

sin

os 1 os

xdx

c x c x

p

p +

Li gii:

N=/3 /3 /3

2 1/2 2

2 22/4 /4 /4

2

sin tan 1(2 tan ) (2 tan ) 5 3

21 os 2 tanos cos 1os

xdx xdxx d x

c x xc x xc x

p p p

p p p

-= = + + = -++

V d 7: Tnh P= 23

1

20 1

x xe dxx

+

+

Li gii: P=

2 2 2 23 3 31

1 1 2 2 1 2 1 3 2212

0 0 0

(1 ) (1 ) (1 )1

x x x xxe dx e x d x dx e d x e e ex

-+ + + += + + = + = = -

+



Mt s sai lm thng gp khi tnh tch phn bng phng php bin i vi phn

Vd 7 : Tnh I=0 1 s inx

dxp

+

Nhn xt: Hc sinh khi gii thng gp sai lm sau t x=tanx/2 dx=

22

02 2 20 0 0

2 1 1 2 2 2 2; 2 (1 ) ( 1)

1 1 s inx (1 ) 1 s inx (1 ) tan 0 1t an 1 tan 12 2

dt t dx dtt d t

xt t t

p p pp

p-+ - - -= = = + + = = -

+ + + + + ++ +

Do tan2p khng xc nh nn tch phn trn khng tn ti.

Nguyn nhn sai lm :Do tch phn l tng v hn cc hng t nn 2 0tan 1

2p-

+

vn c tha nhn. Li gii ng: I=

0 1 s inxdxp

+

=0 1 os( )

2

dx

c x

p

p+ - = 0

20

( )2 4 tan( ) tan tan( )

2 4 4 41 os ( )2 4

xd x

xc

pp

pp p p

p

-= - = - -

+ - =2

Qua bi ton trn ngi thy nn lu vi hc sinh khi i bin s trc ht phi ngh ngay ti php i bin c tn ti hay khng?( cng ging nh khi ta gii phng trnh cn t iu kin cho n s nu c) V d 8 I=

42

0

6 9x x dx- +

Nhn xt: Hc sinh thng mc sai lm sau I=

42

0

6 9x x dx- + =4 4 2

2 2 40

0 0

( 3)( 3) ( 3) ( 3) 4

2x

x dx x d x-

- = - - = = -

Nguyn nhn sai lm l php bin i 2( 3) 3x x- = - khng tng ng ng trn [ ]0, 4 v |x-3|= 3;3 4

3 ;0 3

x x

x x

- -

Li gii ng l



I=4

2

0

6 9x x dx- +

=4 4 4 3 4

2 2

0 0 0 0 3

( 3) ( 3) ( 3) | 3 | ( 3) ( 3) ( 3) ( 3) ( 3)x dx x d x x d x x d x x d x- = - - = - - = - - - + - -

=2 2

3 40 0

( 3) ( 3)| 5

2 2x x- - -

+ =

V d 9: Tnh I=2

22 ( 1)

dxx- +

Hc sinh thng mc sai lm khi bin i nh sau

I =

2

22 ( 1)

dxx- +

= 2

22

( 1)( 1)d xx-

++ =

2

2

1 4|

1 3x -- -

=+

Nguyn nhn sai lm l do hm s y=2

1( 1)x +

gin on trn on [ ]2;2- nn khng s dng c cng thc NeW ton leibnitz nh trn. Li gii ng l : hm s y=

2

1( 1)x +

khng xc nh ti x=-1 [ ]2;2 - nn gin on trn [ ]2;2- ,do vy tch phn trn khng tn ti. Tng kt: s dng c thnh tho k thut s dung bng nguyn hm hc sinh hiu c bn cht ca cc cng thc,phi hiu cng thc trong trng thi ng.khi ng trc bi ton tnh tch phn cn xem xt k biu thc di du tch phn,nu c tng s dng bng nguyn hm th nh a v cng thc no trong bng nguyn hm. lm c iu hoc sinh phi hiu k bn cht ca cng thc, c t duy trong bin i vi phn mt cch logic, tip nhn n mt cch t nhin ,khng gng p . Chng hn khi hng dn hc sinh s dung cng thc

1

1x

x dx ca

a

a

+

= ++ , hc sinh phi hiu gi tr x trong hai s x

a v dx l ging nhau, nu thay x trong hai s bi mt biu thc khc th cng thc trn vn ng v d thay



X = 2t+1 th ta c 1(2 1)

(2 1) (2 1)1

tt d t c

aa

a

+++ + = +

+ ,Nhng nu ch c dng (2 1)t dta+ mun s dng c cng thc trn phi bin i dt = 1 (2 1)2 d x + .ngha

l ta bin i vi phn. Tng t i vi cc nguyn hm khc. luyn tp k thut trn ta c th lm tng t cc bi tp sau

1/I=4

3

s inxdx

p

p ; 2/J=

4

3

cosdx

x

p

p ; 3/K= 32 31x x dx- ;4/L= tan xdx ;5/ M= 4

dxcos x

6/N=2 4 21 os 1

x

x c x+ + ; 7/ P= 2

1

ln(ln 1)

e xx x + ; 8/Q=

2001

2 1002(1 )x dx

x+ ;

9/y=2

2 20

sin x cos3sin 4 os

xdxx c x

p

+ ; 10/T=3

3 5

6

sin os

dx

xc x

p

p ; 11/H=

4

6 60

sin 4sin os

xdxx c x

p

+

II-Tnh tch phn bng cch a biu thc di du tch phn v do hm ca mt hm s khi s dng k thut ny ta ch n cc tnh cht quan trng sau

( UV)=UV+UV

' ' '

2

U U V UVV V

- =

( ) ( )' 'U V UV dx d UV+ =

' '

2

U V UV Udx d

V V- =

V d 1 I= 2

12 ln

ln

e

e

x dxx

+

(H NT-00) Li gii: Ta c ' ' '12 ln 2 ln .( ) (2 ln ) (2 ln )

lnx x x x x x x

x+ = + =



Do I= 2 2

2 212 ln (2 ln )= 2 ln 2 2 2ln

e eee

e e

x dx d x x x x e ex

+ = = -

V d 2 J=2

0

1 s inx1 cos

dxx

p

++ (H -Dc -00)

Li gii:

J=

2 2 2 2

2 20 0 0 0

2 20

1 2sin os1 s inx 12 2 tan tan1 cos 2 22 os 2 os

2 2

tan2

x x x x x

x

x xc x x

e dx e dx e e dx d ex xx c c

xe e

p p p p

p p

+ + = = + = = +

=

Nhn xt :Ngoi cch gii trn ta cn c th gii nh sau

Cch 2 Phn tch K=2 2 2

1 20 0 0

1 s inx 1 s inx1 cos 1 cos 1 cos

x x xe dx e dx e dx K Kx x x

p p p

+= + = +

+ + +

2 2 2 22

1 020 0 0 0

2 22 2 2 2

2 2 220 0

1 1(tan ) tan tan

1 cos 2 2 22 os2

sin sin1 cos2 os

2

x x x x x

x x

x x xK e dx e dx e d e e dx

xx c

x xe e dx e e dx K K e K K e

x xc

p p p pp

p pp p p p

= = = = -+

= - = - - = + - =+

Cch 3: C th t 2 2(1 cos ) s inx1 s inx(1 cos ) (1 cos )1 cos

x x

xduu

x xxdv e dx v e

++ = -= + ++ = =

dx

T ta c K= 2 2

'2 22 2

0

2 2 2

1 s inx (1 cos ) s inx 1 e2 ( )

1 cos (1 cos ) (1 cos ) 2 (1 cos )

12 ( )

2 1 cos

x x xx

oo

x

o

x e ee dx e dx

x x x x

ee e

x

p pp p

p p p

+ +- - = - - = + + + +

- - =+

V d 3 K =2

2

.( 2)

xx ex + dx

Li gii:



K =

2'

2 2 2

'

. 4 4 2 1 1 14 4 ( ) ( )

( 2) ( 2) ( 2) 2 2

4 ( ) 4( )2 2

xx x x x x x

x xx x

x e x xdx e e dx e dx e e e dx

x x x x x

e ee dx e C

x x

+ + - = - = - = - + + + + + +

- = - ++ +

luyn tp ta tnh cc tch phn sau

I=4

2 2

0

4 tan (1 tan )2 2x x

x x dx

p

+ + HD: I=2

tan8 8p p

J=1 2

20

( 1)( 1)

xx edx

x++ HD: J=1

K=2

sinx

0

(1 cos )e x x dx

p

+ HD: K= 2ep

III-K thut i bin s 1/i bin s dng 1: i bin s l mt trong nhng phng php quan trng nht tnh nguyn hm v tch Phn .C s ca phng php i bin s dng 1 l cng thc sau ,[ ( )] ( )

b

a

f u x u x dx = ( )f u dub

a

Trong f(x) l hm s lin tc v hm s u(x) c o hm lin tc trn K sao cho f[u(x) ] xc nh trn K v ( ), ( )u a u ba b= = . p dng tnh cht trn ta c quy tc i bin sau Xt tch phn ( )

b

a

f x dx . t t=V(x) khi ta bin i f(x)dx=g(t)dt do

( )b

a

f x dx = ( )g t dtb

a v ( ), ( )u a u ba b= =



Khi i bin s iu quan trng l chn c hm V(x) thch hp sao cho tch phn vi bin mi phi n gin hn so vi tch phn ban u ,v gn lin vi vic i bin l phi i cn , ta xt mt s bi ton sau trc khi rt ra nhng kinh nghim trong vic la trn hm V(x). V d 0(HKB-2010): Tnh tch phn I =

21

ln(2 ln )

e xdx

x x+

( )21ln

2 ln

e xI dx

x x=

+ ; 1

lnu x du dxx

= =( ) ( )

1 1

2 20 0

1 222 2

uI du du

uu u

= = - ++ +

1

0

2ln 2

2u

u = + + +

( )2ln3 ln 2 13

= + - +

3 1

ln2 3 = -

V d 1: Tnh I= 2 3

25 4

dx

x x + (HKA-03)

Li gii: t t= 2 4x + khi x= 5 ,t=3 x= 2 3 ,t=4. t2=x2+4 suy ra x2=t2-4,tdt=xdx

I=2 3

25 4

dx

x x + =

2 3

2 25 4

xdx

x x + =

4 4 4 4 4

2 23 3 3 3 3

1 ( 2) ( 2) 1 ( 2) 1 ( 2)( 4) 4 4 ( 2)( 2) 4 2 4 2

tdt dt t t d t d tdt

t t t t t t t+ - - - +

= = = - =- - + - - +

4

3

1 2 1 5ln ln

4 2 4 3tt-

=+

.

Nhn xt 1: -Dng tng qut ca tch phn trn l

2( )

b

a

dx

mx n px qx c+ + + ngoi cch gii nh

trn l t t= 2px qx c+ + ta cn c th gii nh sau: t mx+n=1

t. Sau chuyn tch phn trn v bin mi t ta cng thu c kt qu

trn -i vi cc tch phn c cha biu thc ( )n f x ta thng ngh ti vic la chon t= ( )n f x ( tr mt s trng hp s c du hiu i bin s dng 2 s trnh by sau ).Ta xt thm mt s v d lm sng t V d 2 : Tnh (HKA-04)



J=2

1 1 1

dx

x+ -

Li gii: Thc hin php bin i t= 1x - ,x=1 th t=o,x=2 th t=1,t2=x-1 suy ra x=t2+1 2tdt = dx t ta c

2

1 1 1

dx

x+ - =1 2

0

( 1)21

t tdtt

++ =

1 1 1 12 2

0 0 0 0

4 ( 1) 11[2 2 4 ] 2 2 4 4ln 2

1 1 3d t

t t dt t dt dtt t

+- + - = - - = -

+ +

V d 3:( HKB-04) K=

1

1 3ln lne x xdx

x+

Li gii: Nhn thy K=

1

1 3ln ln (ln )e

x xd x+ do vy ta chn t= 1 3ln x+ , x=1,t=1,x=2,t=2

lnx=2 12

t - v 2tdt= 3dxx

.Do

K=1

1 3ln lne x xdx

x+

=2 2 22

4 2

1 1 1

( 1)2 2 116[ ]

3.3 9 135t t

tdt t dt t dt-

= - =

V d 4: Tnh L=3 2

1

ln

ln 1

e xdx

x x + ( thi d b KD-2005) Li gii:

t t=2 2

3

ln 1 ln 1 2 , 1 ln

2; 1 1

dxx t x tdt t x

xx e t x t

+ = + = - =

= = = =

L=3 22

4 2

1 1

ln 762 ( 2 1) ...

15ln 1

e xdx t t dt

x x= - + = =

+

V d 5: Tnh M=

4 7 3

3 40 1 1

x dx

x+ +

Li gii: t t= 3 4 1x + x=0,t=1,x= 4 7 ,t=2.Ta c t3=x4+1 suy ra 3t2dt=4x3dx do M=

4 7 3

3 40 1 1

x dx

x+ + =

2 2 2 22 2

1 1 1 1

3 3 ( 1) 1 3 3 ( 1) 3 3 3( 1) ( 1) ln

4 1 4 1 4 4 1 8 4 2t dt t d t

t d tt t t

+ - += = - - + = +

+ + +

Nhn xt 2: Do c th mt s tch phn phc tp ,trc khi i bin s dng 1 i khi ta phi bin



i d nhn thy bin mi r hn V d 6: Tnh N=

2

11 41 1

dx

x x+

Li gii:

Bin i N=2

11 41 1

dx

x x+ =

1

1304

11

dx

xx

+ t t=

52

4 4

4

211 1

1 1,1

1

dxxsuyra t dt

x x

x

-+ = - =

+

V t 2

4 52 2

13 4 4

1 1( ) 1

( 1)21 1

dxdx x x t dtx x x

= = - -+ +

,x=1,t= 172, 2, 2,16

x x t= = =

N=1

11 40 1

dx

x x+ =

1716

4 2

2

1( 2 1)

2t t dt- - + ta a v tch phn quen thuc

Nhn xt 3 -Cc tch phn cha cc hm s lng gic trc khi nhn din c bin mi cn c hng bin i lng gic nh vo cc cng thc quen thuc nh:cng thc nhn i , h bc,tng thnh tch ,... V d 7: Tnh L=

/4

0

sin( / 4)sin 2 2(1 s inx cos )

xdx

x x

p p-+ + + (HKB-08)

Li gii: Nhn xt :tch phn trn mi nhn ta thy kh nhn din c bin mi ta th xem mu v t s c mi qua h g ? sin2x+2(1+sinx+cosx)=1+2sinxcosx+2(sinx+cosx)+1=(sinx+cosx)2+ 2(sinx+cosx)+1 =(sinx+cosx+1)2, d(sinx+cosx)=(cosx-sinx)dx= - 2 sin( / 4)x p- dx Do t t=sinx+cosx khi i cn ta c:

L=/4

0

sin( / 4)sin 2 2(1 s inx cos )

xdx

x x

p p-+ + + =

2

21

( 1)2 4 3 22 ( 1) 4

d tdt

t

+- -=

+

V d 8 Tnh P=

/6 4

0

tanos2

xdx

c x

p

(HKA-08) Li gii:



Nhn xt P=/6 4

0

tanos2

xdx

c x

p

=/6 4

2 20

tanos sin

xdx

c x x

p

- =/6 4

2 20

tan(1 tan ) os

xdx

x c x

p

- =/6 4

20

tan (t anx)(1 tan )

xdx

p

-

t tanx=t ,x=0th t=0,x=6p th t= 1

3.

Do P=/6 4

0

tanos2

xdx

c x

p

=1

3 4

20 1

tdt

t- =

1 1 1

3 3 342

2 20 0 0

1 (1 ) 10 1(1 ) ln(2 3)

1 1 29 3

t dtdt t dt

t t- -

= - + = - + +- -

Nhn xt 4 -khi tnh tch phn dng (tan )

os2

b

a

f xdx

c x hoc (tan )sin 2

b

a

f xdx

x ta vit nh sau Cos2x=cos2x(1-tan2x); sin2x=2cos2xtan2x sau t t= tanx th dt=

2osdx

c t sau a v tch phn c bn.

-Bi ton tng qut ca bi trn l P=4tan

; ,os2

a xdx a b R

bc x

b

a

- Vi cnh khai thc trn ta c th gii quyt bi ton tng qut hn nh sau

P1= 4

2 2

tan; , , ,

sin sin x cos osa x

dx a b c d Rb x c x dc x

b

a

+ + vi ch l

(bsin2x+csinx cosx+dcos2x)=(btan2x+ctanx+d)cos2x do ta chn t =tanx - i vi cc tch phn lng gic (s inx,cos )

b

a

R x dx cha hai hm lng gic sinx,cosx ta c my iu quan trng sau + Nu l theo bc ca sinx th nn chn t=cosx +Nu l theo bc ca cosx th nn t t=sinx +chn theo sinx v cosx th t t=tanx

V d 9: Tnh Q=/2

0

sin 2 cos1 os

x xdx

c x

p

+ (HKB-05) Li gii: Bin i Q=

/2

0

sin 2 cos1 os

x xdx

c x

p

+ =/2 2

0

sin cos2

1 osx x

dxc x

p

+



t t=1+cosx (v bc ca sinx l) suy ra dt=-sinxdx ,x=0th t= / 2p ,x= / 2p th t=1

Q=/2

0

sin 2 cos1 os

x xdx

c x

p

+ = Q=2 2

1

( 1)2ln 2 1

tdt

t-

= -

Nhn xt 5: -Tch phn trn c dng tng qut Q= sin 2 cos

osa x x

dxb Cc x

b

a + c hai cch t

C1: t=b+ ccosx C2: t=cosx V d 10: R=

/3

2/4 sin 2sin x cos

dxx x

p

p -

Li gii: Nhn xt bc ca sinx chn nn ta ngh ti cch t t= tanx R=

/3

2/4 sin 2sin x cos

dxx x

p

p - = R=

/3

2 2/4

1(tan 2 t anx) os

dxx c x

p

p - =

/3

2/4

(t anx)(tan 2 t anx)

dx

p

p -

t t=tanx ta c R=/3

2/4

(t anx)(tan 2 t anx)

dx

p

p - =

3 3

21 1

1 1 1 1 2( ) ln(1 )

2 2 2 2 3

dtdt

t t t t= - = -

- -

Nhn xt 6 - Tch phn tng qut ca tch phn trn l R=

2 2sin sin x cos osdx

a x b x cc x

b

a + +

Ta bin i R= 2 2( tan t anx ) osdx

a x b c c x

b

a - + = 2

(t anx)( tan t anx )

da x b c

b

a - +

sau t t=tanx - Tng t i vi tch phn lng gic c dng R= 2 2 2 2sin x cos ,( sin os )n

xdxn N

a x b c x

b

a

+

Nhn xt 7: i vi mt s tch phn khng c du hu c bit nh cha ( )n f x hay cha cc hm s lng gic nh xt trn khi ta phi quan st k v kho lo phn tch c th nhn din c bin mi.Ta xt thm mt s cc v d sau V d 11 Tnh G=

2 2

41

11

xdx

x-+

Li gii: Nhn xt ( 2 2

2 2

1 1 1 1) ( ) 2, ( ) (1 )x x d x dx

x x x x+ = + - + = - t ta bin i nh sau



G=2 2

41

11

xdx

x-+ =

2 2

212

11

1x dx

xx

-

+ = G=

2 2

21

11

1( ) 2

x dxx

x

-

+ - =

2

21

1( )

1( ) 2

d xx

xx

+

+ -

t t=x 1x

+ ,x=1 th t=2,x=2 th t=5/2

Khi ta c G=

2 2

41

11

xdx

x-+ =

5/2 5/25/222

2 2

(5 2 2)(2 2)1 ( 2) ( 2) 1 2 1ln ln

2 2 2 ( 2)( 2) 2 2 2 2 2 6 2

dt t t tdt dt

u t t t

- +- - - -= = =

- - + + -

V d 12

Y=ln5

ln3 2 3x x

dxe e-+ - (HKB-06)

Li gii: Nhn xt

2

( )2 3 3 2 ( 1)( 2)

x x

x x x x x x

dx e dx d ee e e e e e-

= =+ - - + - -

t t=e x,x=ln3 th t=3,x=ln5 th t=5 Y=

ln5

ln3 2 3x x

dxe e-+ - =

5

3 ( 1)( 2)dt

t t=

- -

5 5 55

33 3 3

( 1) ( 2) ( 2) ( 1) 2 3ln ln

( 1)( 2) 2 1 1 2t t dt d t d t t

t t t t t- - - - - -

= - = =- - - - -

V d 13 H=

2 7

71

1(1 )

xdx

x x-+

Li gii: Nhn thy

7 7 6 77

7) 7 7 7

1 (1 ) 1 (1 )( )

(1 (1 ) 7 (1 )x x x x

dx dx d xx x x x x- - -

= =+ + +

do ta ngh ti t t=x7

H=2 7

71

1(1 )

xdx

x x-+ =

128

1

1 (1 )7 ( 1)

t dtt t-+ =

128

1

1 (1 ) 27 ( 1)

t tdt

t t+ -

=+

128 128

1 1

1 1 512[ 2 ] ln

7 1 7 16641dt dtt t- =

+

V d 14 K=

15 3 6

0

(1 )x x dx-



Li gii: Nhn xt x5(1-x3)6=x3(1-x3)6x2=- 3 3 6 31 (1 ) (1 )

3x x d x- - = 3 3 6 31[1 (1 )](1 ) (1 )

3x x d x- - - - -

Do ta t t=1-x3 v ta c K=

15 3 6

0

(1 )x x dx- =0

6

1

1(1 )

3t dt- - =

06

1

1 1(1 ) (1 )

3 168t d t- - =

Nhn xt 8: Cc v d trn c gii nh vo vic bit phn tch mi quan h gia cc biu thc di du tch phn.ta gi chung l i bin nh Phn tch Nhn xt chung: i bin s dng 1 l mt trong nhng phng php rt c bn, hc sinh thng gp trong Cc k thi tt nghip v thi vo cc trng i hc,bi n c th pht huy ti a t duy Linh hot ca hc sinh ,Hc sinh khng th dng mt cng thc i bin tng qut no p dng Cho cc bi ton khc nhau.Chnh v l trong ging dy hc sinh dng phng php i bin s dng 1 ,ngi thy khng qu sa vo vic dy hc sinh nhng dng ton c tnh cht cng thc,my mc. iu quan trng l pht trin hc sinh t duy logc,s sng to ,cc em t mnh chim lnh kin thc ,t rt ra nhng bi hc b ch t vic gii c hay khng gii c nhng bi tch phn,c nh vy khi ng trc nhng bi ton mi hay nhng bi ton c ngy trang th cc em vn c c sc khng vt qua.Ti coi l t tng ch yu ca dy hc tch phn ni ring v mn ton ni chung. 2-i bin s dng hai: T tng ca k thut ny l :Gi s ta cn tnh tch phn I= ( )

b

a

f x dx th ta chn X=u(t),vi u(t) l hm s ta chn thch hp Biu din dx=u(t)dt, u( ) , ( )a u ba b= = Biu th f(x)dx theo t v dt,gi s f(x)dx=g(t)dt I= ( )

b

a

f x dx = ( )g t dtb

a l tch phn d tm hn tch phn ban u.

V d 1: Tnh I=2

22

20 1

x dx

x-

Li gii : Nx: ta c sin2t+cos2t =1 nn 1-sin2t=cos2t, 21 sin cost t- = do ta ngh ti t x=sint t ;

2 2p p -

x=0,t=0,x= 2 ,2 4

tp

= ,dx=costdt



I=

222

20 1

x dx

x- = I=

/4 /4 /42 2

20 0 0

sin cos sin cos 1 os2 1cos 2 8 41 sin

x tdt t t c tdt dt

tt

p p p p-= = = -

-

Nhn xt 1 : - C th t x=cost t [ ]0;p -i vi nhng tch phn c cha cc biu thc 2 2a x- ta c th t x=acost , t [ ]0;p

hoc x= asint , t ;2 2p p -

V d 2: Tnh J=6

23 2 9

dx

x x -

Li gii: t x= 3 ,

sin t (0; / 2)t p

dx=2

3cossin

tdtt

- , 1 13 2,sin , 6,sin4 2 62

x t t x t tp p

= = = = = =

J=6

23 2 9

dx

x x - =

/6

2/42

3cos

3 9sin 9

sin sin

tdt

tt t

p

p

-

- =

/4 4

/6 /6

1 cos 1cos3 3 36sinsin

tdtdt

tt

t

p p

p p

p= =

Nhn xt 2: - c th t x= 3 ,

osc t

- i vi nhng tch phn c cha biu thc 2 2x a- (a>0) ta c th t x=osa

c t

hoc X= ,sin

at

V d 3 Tnh K=

3 2

21

1 xdx

x+

Li gii; t x=tant,t ( ; )

2 2p p

-

1 , 34 3

x t x xp p

= = = = , 2 221

; 1 1 tanos cosdt

dx x xc t t

= + = + =

K=3 2

21

1 xdx

x+

= /3 /3 /3

2 2 2 2 2/4 /4 4

2

1 (sin )sin os cos sin sin (1 sin )

cosos

dt dt d tt c t t t t t

tc t

p p p

p p p

= = =-

32

2 22

2

(1 )du

u u=

-



32

22

2

duu

+ 3

2

22

2

1du

u- =3 2 2 3

ln(2 3)( 2 1)3-

+ - +

Nhn xt 3: -i vi nhng tch phn c cha biu thc (a2+x2)k (a>0)ta thng t x=atant hoc x=acott -Mt s tch phn sau khi bin i mi a v dng c cha biu thc (a2+x2)k .ta xt v d sau V d 4 Tnh L=

1

4 20 1

xdxx x+ +

Li gii:

L=1

4 20 1

xdxx x+ + =

1 2

2 2 20

1 ( )2 ( ) 1

d xx x+ + =

1

20

1 ( )2 ( ) 1

d tt t+ + =

1

2 20

1( )1 2

2 1 3( ) ( )

2 2

d t

t

+

+ + =

32

1 2 22

1 ( )2 3

( ) ( )2

d u

u +

t 3 tan , ( ; )2 2 2

tp pa - ,u= 3 1tan 3 ,,

2 3 2 6u

p pa a a = = = =

L =

32

1 2 22

1 ( )2 3

( ) ( )2

d u

u + =

3 3

2 2

6 6

31 3 32

32 3 18os . (1 tan )4

dd

c

p p

p p

a pa

a a= =

+ ,

Nhn xt 5 Mt s tch phn c cha cc biu thc ( )( )x a b x- - ,b>a>0 Khi ta t X=a+(b-a)sin2t , t 0;

2p

.ta xt v d sau

V d 5: Tnh M=32

54

( 1)(2 )x x dx- -

Li gii : Nhn xt a=1,b=2 t x=1+sin2t t 0;

2p

,dx=2sintcostdt,x= 5 3;4 6 2 4

t x tp p

= = =

M=

32

54

( 1)(2 )x x dx- - =32

54

( 1)(2 )x x dx- - =24

2 2

6

sin (1 sin ) sin cost x t tdt

p

p

- =



24

2 2

6

sin cost tdx

p

p

=4

6

1 1 3(1 os2 ) ( )

2 8 12 8c t d

p

p

p- = - .

Nhn xt 6: Bng cch khai thc tng t ta s rt ra c cc cch bin s dng 2 i vi nhng tch phn c cha nhng biu thc c thng k qua bng sau:

Du hiu Cch chn 2 2a x- (a>0) X=asint t ;

2 2p p -

hoc x=acost t [ ]0;p

2 2x x- (a>0) X=sin

at t ;

2 2p p -

\0

X=osa

c t t [ ]0;p \ / 2p

2 2a x+ (a>0) X=atant t ;2 2p p -

ho c

X=acott t ( )0;p

a xa x+-

hoc a xa x-+

X=acos2t

( )( )x a b x- -

X=a+(b-a)sin2t

Nhn xt 7: -i khi s dng i bin s dng 2 la bt u t dng 1

V d 6: Tnh K=24

4 2

4

sinos (tan 2 t anx 5)

xdxc x x

p

p- - + ( thi d b 2008-B)

Li gii:



Bin i K=2 24 4

4 2 2

4 4

sin tan (t anx)os (tan 2 t anx 5) (tan 2 t anx 5)

xdx xdc x x x

p p

p p- -

=- + - +

t tanx=t i cn a K v dng K=

1 1 1 12 2

2 2 21 1 1 1

( 2 5)3

( 2 t 5) 2 5 ( 1) 4t dt d t t dt

dtt t t t- - - -

- += + -

- + - + - +

Li t t-1=2tant i cn tnh ton ta c K=2-ln2 38p

-Mt trong nhng php i bin hay dng na l php thay bin x=a-t i vi nhng tch phn c cn trn l a v hm di du tch phn cha cc biu thc lng gic v cc biu thc ny c lin quan n cn trn l a (Theo ngha chng c mi quan h n cc gc lin quan c bit).V l cc tch phn ny thng c cn trn l ; ;2 ,...

2p p p

Khi tnh cc tch phn ny thng dn ti gii mt phng trnh n gin vi n s l t V d 7:

Tnh H=/2 4

4 40

sinos sin

xdxc x x

p

+

Li gii: t x=2

t dx dtp- = - v ta c

I=0 4

4 42

sinos sin

xdxc x xp

-+ =

/2 4

4 40

osos sin

c xdxc x x

p

+ suy ra

2I=/2 /24 4

4 40 0

sin os/ 2

os sin 4x c xdx

dx xc x x

p p pp+ = = =+

V d 8 Tnh F=

23

0

osxc xdxp

Li gii: t x= 2 t dx dtp - = - v ta c I=

0 2 2 23 3 3 3

2 0 0 0

(2 ) os (2 ) (2 ) os 2 os ost c t dt t c tdt c tdt tc tdtp p p

p

p p p p- - - = - = - 2

3

0

2 osI c tdtp

= =2

0

os3 3cos0

4c t t

dtp +

=

V d 9:



Tnh M=2

0

s inx1 os

xdx

c x

p

+

Li gii: t x= t dx dtp - = - M=

20

s inx1 os

xdx

c x

p

+

=0

2 2 2 2 20 0 0 0

( )sin sin sin sin sin2

1 os 1 os 1 os 1 os 2 1 ost tdt tdt t tdt tdt tdt

M Mc t c t c t c t c t

p p p p

p

p pp p- = - = =+ + + + +

Li t u=cost suy ra du=sintdt M=

1 1 2

2 2 20 1 0

s int2 1 os 2 1 1 4

dt dtdt

c t t t

pp p pp-

= = =+ + +

Nhn xt 8: Li gii ca cc bi ton trn da vo tnh cht : Nu hm s f(x) lin tc trn [ ];a b tho mn f(x)=f(a+b-x) th

( ) ( )2

b b

a a

a bxf x dx f x dx

+=

c bit hn : Nu f(x) l hm s lin tc trn [ ]0;1 th (s inx) (s inx)

2xf dx f dx

p a p a

a a

p- -=

Nu f(x) l hm s lin tc trn [ ]0;1 th 2 2

( osx) ( osx)xf c dx f c dxp a p a

a a

p- -

=

Cc tnh cht ny s c chng minh v ng dng trong k thut s dng lp cc Tch phn c bit .

IV-K thut s dng Tch phn tng phn C s l thuyt :Theo cng thc v php tnh vi phn ta c

d(uv)=udv+vdu Hay udv=uv-vdu



b b

ba

a a

udv uv vdu = - (I)

Cng thc trn gi l cng thc tnh tch phn tng phn ,Phng php s dng cng thc Trn tnh gi l phng php tch phn tng phn. Nhn dng : Hm s di du tch phn thng l hm hai bin s khc nhau ngha: a mt tch phn phc tp v mt tch phn n gin hn .Trong nhiu trng hp khi s dng tch tng phn s gim bt hm s di du tch phn v cui cng ch cn mt hm s di du tch phn. Nh vy tnh

b

a

udv ta chuyn v tnh b

a

vdu ,Nh vy iu quan trng nht khi tnh tch phn tng phn l phi chn u,v thch hp m bo hai nguyn tc c bn sau -Chon u,v sao cho du n gin dv d tnh -Tch phn

b

a

vdu d tnh hn so vi b

a

udv

Sau y l mt s dng Tch phn thng c s dng k thut Tch phn tng phn 1-Dng I ( ) ln

bk

a

P x xdx : ( )K Z

Thng chn: 1lnln

( )( )

kk du k xdxu x

v p x dxdv p x dx

- = = ==

Chn u nh vy kh lnx di du Tch phn , ng thi d tm V V D 0: (HKD-2010)Tnh tch phn

1

32 ln

e

I x xdxx

= -

1 2

1 1 1

3 12 ln 2 ln 3 ln .

e e e

I I

I x xdx x xdx x dxx x

= - = - 1424314243

11

lne

I x xdx= ;t lndx

u x dux

= = ;

2

2x

dv xdx v= =2 2 2 2

111 1

1 1 1ln

2 2 2 2 2 4

e eex e x eI x xdx

+= - = - =



Tnh I2 : t t = lnx dxdtx

= x = 1 ; t = 0; x = e ; t = 1.11 2

20 0

12 2t

I tdt

= = =

. Vy 2 22

eI

-=

V D 1: (HKD-2009): Tnh tch phn 3

21

3 ln xI dx

(x 1)+

=+ 33 3 3 3 3

1 22 2 2 2 21 1 1 1 11

3 ln x dx ln x dx 3 3 ln xI dx 3 dxI 3 I dx

(x 1) (x 1) (x 1) (x 1) (x 1) 4 (x 1)+ -

= = + = = = =+ + + + + +

t u = lnx dxdux

=2

dxdv .

(x 1)=

+ Chn 1v

x 1-

=+

3 3 3 3

21 1 1 1

ln x dx ln 3 dx dx ln 3 3I ln

x 1 x(x 1) 4 x x 1 4 2= - + = - + - = - +

+ + + Vy : 3

I (1 ln 3) ln 24

= + -

V D 2: (HKD-08) I=2

31

ln xdx

x Li gii:

t 3

3 2

ln

12

dxduu x

xdx dxdv vx x x

== -= = = : I=

2 22

3 2 311 1

ln ln 3 2ln 22 2 16

x x dxdx

x x x- -

= + =

Nhn xt: Mt s tch phn mun a v dng trn cn thng qua i bin s dng 1 V d 3:

J= 2

2

6

cos ln(s inx)sinx

dxx

p

p

Li gii:

Vit li J= 2

2

6

ln(s inx)(s inx)

sind

x

p

p t t=sinx , i cn ta i n tch phn sau

J= 1

212

ln tdt

t t 2

2

ln

1

dtduu t

tdt dtdv v dtt t t

== -= = = M=

1112 1

2

ln1 2ln 2

t dtt t

-= + = -



V d 4: K=3

2

4

ln(tan x)os

dxc x

p

p

Li gii: Ngoi cch trnh by bng i bin sau dng tch phn tng phn nh trn ta c th trnh by trc tip nh sau

K=3

2

4

ln(tan x)os

dxc x

p

p =

3

4

ln(t anx) (t anx)d

p

p t ln (t anx) sin x cos(t anx) t anx

dxduu

xdv d v

== = =

K=3

3 3

4 44

t anx 3 ln 3 3 ln 3tan x ln(t anx) t anx 3 1

sin cos 2 2dx

x x

pp p

p pp

- = - = - +

Nhn xt 2 :Do khng c cng thc tnh nguyn hm ca biu thc cha lnx nn mc ch ca ta khi tnh tch phn trn l kh lnx ,v vy s ln s dung cng thc Tnh tch phn tng phn ph thuc vo s K trong tch phn ( ) ln

bk

a

P x xdx .C th l k=1 (nh v d trn) dng mt ln,k=2 s dng 2 ln ......ta xt thm v d sau m ta iu

V d 5: L= 3 21

lne

x xdx (HKD-07) Li gii:

t 2

433

2 lnln

4

xdxdu

u x xxdv x dx

v x dx

= = = = =

L=

4 2 43 '

11

ln 1ln

4 2 4

eex x e

x x L- = -

Li t 43

3

ln

4

dxdu

u x xxdv x dx

v x dx

= = = = =

L=

'4 43

11

ln 1 3 14 4 16 16

eex x e

x dx- = -

T L=45 132

e -

Nhn xt 3: M s tch phn cha p(x) phc tp ,ta vn da vo cch t trn kh lnf(x)



V d 6: N=3 2

21

ln( 1)

1

x x x

x

+ +

+

Li gii: t

232

2 2 31

2 022

ln( 1)1

1ln( 1) 2ln 3 311 1

dxduu x x

xN x x x dxx xdv v dx xx x

= = + + + = + + + - = - = = = ++ +

Dng 2: ( ) cosb

a

P x xdx ( ( ) cosb

a

P x xdx )

t cos( )

u x

dv P x dx

= =

Hoc cos( )

dv x

u P x dx

= =

V d 1: I=/4

2

0

(2cos 1)x x dxp

-

Li gii: vit li I=

/42

0

(2cos 1)x x dxp

- =/4

0

os2xc xdxp

t /4

/4 /400

0

sin 2cos 2 sin 2 1 sin 2 os2

sin 2 ( )22 2 2 4

18 2

xdv xdx v dx x x x x c x

I xdxu x

du dx

pp p

p

= = = + = - = =

= +

V d 2 J=

1

cos(ln )e

x dxp

Li gii

t sin(ln )

cos(ln )x dx

u x dux

dv dxv x

= = = =

J=xcos(lnx)| 1ep - 1

1

sin(ln )e

x dx e Jp

p= - -



t os(ln )

sin(ln )c x dx

u x dux

dv dxv x

= = = =

J1 = xsin(lnx)| 1ep -

1

1os(ln ) 1

2

e ec x dx J J e J J

p pp += - = - - - = -

V d 3: K=3

2

3

s inxcosx

dxx

p

p- (H Vinh-2001)

Li gii:

Bin i K=3

2

3

s inxcosx

dxx

p

p- =

3 3

33 00

0 0

1 1 4 4 52 ( ) 2 2 tan 2 ln(tan )

cos cos cos 3 2 4 3 12dx x

xd xx x x

p ppp p p p p

== - = - + -

Nhn xt : Mt s tch phn trc khi s dung tch phn tng phn cn bin i a v c dng trn, ta xt mt v d sau m t iu

V d 4: L=2

2

0

(2 1)cosx xdx

p

- ( thi d b KB-2005) Li gii Bin i L nh sau

2 2 2 22

0 0 0 0

222 2

1 00

2

20

1 os2 1 1(2 1) cos (2 1)( ) (2 1) (2 1) os2

2 2 2

1 1(2 1)

2 2 8 4

1(2 1) os2

2

c xx xdx x dx x dx x c xdx

L x dx x x

L x c xdx

p p p p

pp

p

p p

+- = - = - + -

= - = - = -

= -

2

22 0

0

1 1(2 1) ; cos 2 sin 2

2 2

1 1 1(2 1)sin 2 sin 2

4 2 2

u x du dx dv xdx v x

L x x xdx

pp

= - = = =

= - - = -



Do L=2 1

8 4 2p p

- -

V d 5:M=4

sinx

0

(t anx cos )e x dx

p

+ ( thi d b KB-2005) Li gii: Bin i M nh sau

[ ]

4 4 4sinx sinx

0 0 0

14 4inx sinx 24 4

000 0

(t anx cos ) t anx cos

(cos )(s inx) ln(cos ) ln 2

coss

e x dx dx e xdx

d xdx e d x e e

x

p p p

p pp p

+ = + =

- + = - + = +

Dng 3: cosb

kx

a

e mxdx ( hoc sinb

kx

a

e mxdx )

t 1

1coss inmx

kxkx du e dxu e k

dv mxdxv

m

= = = =

Hoc t 1

1coss inmx

kxkx v e dxdv e k

u mxdxdv

m

= = = =

V d 1: K= 2

2

0

cos3xe xdx

p

t 2

2 2

1cos3 s in3x

3

xx du e dxu e

dv xdx v

= = = =

K= 2

2

0

cos3xe xdx

p

=2

2 220 1

0

1 2 1 2sin 3 sin 3

3 3 3 3x xe x e xdx e K

pp

p- = - -

li t 2

2 2

1sin 3 cos3x

3

xx du e dxu e

dv xdx v

= = = = -

K1 = 2 201 2

os33 3

xe c x Kp

+ t suy ra

K= 1 2 4 3 23 9 9 13

ee K K

pp- - -- - = .

Nhn xt :

-Cng c th t 22 1

2cos3 3s in3x

xx v e dxdv e

u xdx d

== = = -

v thu c kt qu nh trn .



-Tch phn ( hoc sinb

kx

a

e mxdx ) dng phng php tng phn Bao gi cng phi dng hai ln song iu ch l trong hai ln dng phi thng nht trong hai cch t u,v cng mt kiu nu khng th sau hai ln s dung tch phn li tr li ban u.

Tht vy sau khi t 2

2 2

1cos3 s in3x

3

xx du e dxu e

dv xdx v

= = = =

K= 2

2

0

cos3xe xdx

p

=2

2 220 1

0

1 2 1 2sin 3 sin 3

3 3 3 3x xe x e xdx e K

pp

p- = -

ta khng t nh trn m

t 22 1

2sin 3 3cos3x

xx v e dxdv e

u xdx du

== = =

K1= 2 201 3 1 3

sin 32 2 2 2

xe x K e Kp

p- = -

t ta c K= 1 2 1 3( )3 3 2 2

e e K K Kp p- - = nh vy ta khng tnh c tch phn trn . Hin tng gi l hin tng xoay vng trong tch phn. l iu m ta phi trnh khi tnh tch phan bng phng php tng phn , Dng 4: ( )

bkx

a

P x e dx

t , ( )( )

1ekx kx

du P x dxu P x

dv e dx v dxk

== = =

V d 1: I=

12

0

( 2) xx e dx- (HKD-06)

t 1 2

2 1 2022

0

( 2) 1 1 5 3( 2)1

2 2 4e2

x xxx

du dxu x eI x e e dx

v dxdv e dx

== - - = - - = ==

Nhn xt : Mt s tch phn c dng trn ,nhng trc khi dng tng phn ta phi i bin s vic s dng tch phn tng phn c thun li hn V d 2: I= 2

13

0

xx e dx

Li gii:



Nhn thy I= 21

3

0

xx e dx = I=2

12 2

0

12

xx e dx v vy

t x2=t ,dt=2xdx 23 12

x tx e dx te dt= x=0 th t=0,x=1 th t=1

I=2

13

0

xx e dx = I=1 1 1

1 1 10 0

0 0 0

1 1 1 1 1 12 2 2 2 2 2 2

t t t tete dt tde te e dt e= = - = - =

V d 3: Tnh I=2 2

20 ( 2)

xx edx

x +

Li gii: t u=x2e x suy ra du=xe x(x+2)dx; dv=

2

1( 2) 2

dxv

x x-

=+ +

I=2 22

2 20

0 02

xx xx e xe dx e xe dx

x-

+ = - ++ . li t u=x,dv=e

x ta c

2 22 20

0 0

1x x xx e dx xe e dx e= - = +

Nn I=-e2+e 2+1=1 Nhn xt 1 : nu t

2

2( 2)x

ux

=+

th vic tnh tch phn trn li tr v tnh tch phn 2

0 2

xxex + r dng l phc tp hn rt nhiu so vi tch phn ba u.v vy vic chon

u,v thch hp l rt quan trng trong tnh tch phn tng phn. Nhn xt 2:

Mt s tch phn c dng ( )b

kx

a

P x e dx mun s dng c phng php nh trn phi qua i bin s:

V d 4: Tnh K=2

0

1 s inx1 cos

xe dxx

p

++

Li gii:

Phn tch K=2 2 2

1 20 0 0

1 s inx 1 s inx1 cos 1 cos 1 cos

x x xe dx e dx e dx K Kx x x

p p p

+= + = +

+ + +

Xt K1=



2 2 2 220

20 0 0 0

2 22 2 2 2

2 2 220 0

1 1(tan ) tan tan

1 cos 2 2 22 os2

sin sin1 cos2 os

2

x x x x x

x x

x x xe dx e dx e d e e dx

xx c

x xe e dx e e dx K K e K K e

x xc

p p p pp

p pp p p p

= = = -+

= - = - - = + - =+

Cch 2: C th t 2 2(1 cos ) s inx1 s inx(1 cos ) (1 cos )1 cos

x x

xduu

x xxdv e dx v e

++ = -= + ++ = =

dx

T ta c K= 2 2

'2 22 2

0

2 2 2

1 s inx (1 cos ) s inx 1 e2 ( )

1 cos (1 cos ) (1 cos ) 2 (1 cos )

12 ( )

2 1 cos

x x xx

oo

x

o

x e ee dx e dx

x x x x

ee e

x

p pp p

p p p

+ +- - = - - = + + + +

- - =+

lyn tp ta c th tnh cc tch phn sau

I=2

51

ln xdxx ;J=

2

2

1 1ln ln

e

e

dxx x

- ;K=

4

0

tan xx dx

p

; L= 21

ln(1 )

e xdx

x+ ;M=2

2

1

( 2 ) xx x e dx+

V-Tch phn lin kt i khi vic tnh mt s tch phn I= ( )

b

a

f x dx gp nhiu kh khn ta thng

Hay ngh n mt tch phn J= ( )b

a

g x dx sao cho vic tnh cc tch phn mI+nJ v pI-qJ thun li .Khi vic tnh I,J thng a v gii h phng trnh 1

2

mI nJ C

pI qJ C

+ = - =

Khi ta ni I v J l cc tch phn lin kt vi nhau. Vic la chn tch phn lin kt vi mt tch phn cho trc ph thuc vo c im ca hm di du tch phn v cn ca chng .Do c th ca cc hm lng gic nn ta thng dng phng php lin kt cc tch phn i vi cc tch phn cha cc hm s lng gic .



V d 1 I=4

30

4sin x(s inx cos )

dxx

p

+

Li gii:

Xt J=4

30

4 osx(s inx cos )

c dxx

p

+ ta c

I+J=4 4

402

20 0

42 2 tan( ) 2

(s inx cos ) 4os ( )4

dx dxx

x c x

p ppp

p+ = - =

+ - (1)

Mt khc J-I =4 4

403 2 2

0 0

4( osx-sinx) d(sinx+cosx) 22 2

(s inx cos ) (s inx cos ) (s inx cos )c dx dx

x x x

p pp

= = = -+ + + (2)

T (1) v (2) ta c I= 22

V d 2: Tnh I=23

0

sin x

sinx 3 cos

dx

x

p

+

Li gii

Xt tch phn J=23

0

os x

sinx 3 cos

c dx

x

p

+ ta c

I-3J=2 23 3 3

0 0 0

sin 3 os x (sin 3 osx)(sin 3 osx)(s inx 3 cos )

s inx 3 cos s inx 3 cos

x c dx x c x c dxx dx

x x

p p p

- - += = - =

+ +

3(cos 3 sinx) 1oxp

- + = -

I+J=2 23 3 3 3

0 0 0 0

(sin os x) 1 12 2s inx 3 cos s inx 3 cos 1 3 sin( )(s inx cos ) 32 2

x c dx dx dx dx

x x xx

p p p p

p+

= = =+ + ++

=3 3

30

20 0

tan( )1 1 1 ln 32 6 ln tan( )2 2 2 2 6 2os ( ) tan( ) tan( )

2 6 2 6 2 6

xd

dx xx x x

c

p pp

pp

p p p

+ = = + =+ + +

Vy ta c h phng trnh 3 1

3ln 3 2ln 3

82

I JI

I J

- = -- =

+ =



Tng qut: 2sin x

sinx cosdx

A B x

b

a + thng lin kt vi J=

2os xs inx cos

c dxA B x

b

a +

V d 3: I=2

30

(5cos 4sinx)(s inx cos )

x dxx

p

-+

Li gii

Xt J=2

30

(5sin 4 osx)(s inx cos )

x c dxx

p

++ ta c

I+J=2 2 2

203 2

20 0 0

(sin osx) 1 1tan( ) 1

(s inx cos ) (s inx cos ) 2 2 4os4

x c dx dx dxx

x x c x

p p ppp

p+

= = = - =+ + -

(1)

Mt khc t x=2

tp- ta c

I=2 2 2

3 3 30 0 0

(5cos 4s inx) (5sin 4cos ) (5sin 4cos )(s inx cos ) (s int cos ) (s inx cos )

x dx t t dx x x dxJ

x t x

p p p

- - -= = =

+ + + (2)

T (1) v (2) ta c I= 12

V d 4: I=2

2 2

0

( os3 ) ( os6 )c x c x dx

p

Li gii

Xt tch phn J=2

2 2

0

(sin 3 ) ( os6 )x c x dx

p

Ta c I+J=2 2 2

2 2 2 2

0 0 0

1[(sin 3 ) ( osx) ]( os6 ) ( os6 ) (1 os12 )

2 4x c c x dx c x dx c x dx

p p p

p+ = = + =

Mt khc I - J= 2 2 2

2 2 2 2 2

0 0 0

1[(sin 3 ) ( osx) ]( os6 ) os6 ( os6 ) 1 sin (sin 6 )

6 8x c c x dx c x c x dx x d x

p p p

p - = = - =

Vy ta c I = J = 8p

Nhn xt :Ta c th chn tch phn lin kt khc l J=2

2 2

0

( os3 ) (sin 6 )c x x dx

p



V d 5: Tnh I=20092

2009 20090

( os )(s inx) (cos )

c xdx

x

p

+

Li gii:

Xt J= 20092

2009 20090

(sin )(s inx) (cos )

xdx

x

p

+

t x= ; 0; 02 2 2

t dx dt x t x tp p p- = - = = = =

Ta c J =

2009020092

2009 20092009 20090

2

2009 20092 2

2009 2009 2009 20090 0

(sin( ))(sin ) 2 ( )(s inx) (cos ) 2(s in( )) (cos( ))

2 2

(cos ) ( os )(s int) (cos ) (s inx) (cos )

txdx d t

x t t

t c xdx dx

t x

p

p

p p

pp

p p

-= - =

+ - + -

=+ +

M I+J= 2009 20092 2

2009 20090 0

(s inx) ( os )(s inx) (cos ) 2 4

c xdx dx I J

x

p p

p p+= = = =

+

V d 6: Tnh I=36

0

sins inx cos

xdxx

p

+

Li gii:

Xt tch phn lin kt J=36

0

oss inx cos

c xdxx

p

+

Ta c I+J = 3 36 6

60

0 0

(sin os ) 1 1 1(1 sin 2 ) ( os2 )

s inx cos 2 4 6 8x c x dx

x dx x c xx

p pp p+

= - = + = -+

Li c I-J = 23 36 6

0 0

660

0

1 (s inx cos ) (s inx cos )(sin os ) 1s inx cos 2 s inx cos

1 1 s inx cos 1 3 2(s inx cos ) (s inx cos ) [ ln s inx cos ] ln

2 sinx cos 2 2 2

x x dxx c x dxx x

xx d x x

x

p p

pp

+ + -- = =+ +

+ + + + + = + + = - +

t ta c I= 1 1 3 2 1ln12 2 2 4 16p ++ - -

Nhn xt : i vi tch phn ta c th trnh by theo cch khc nh sau



2I=3 3 3 3 3 3 3 36 6 6

0 0 0

(sin os ) (sin os ) (sin os ) (sin os )s inx cos s inx cos s inx cos

x c x x c x x c x dx x c x dxdx

x x x

p p p

+ + - + -= +

+ + +

Sau trnh by nh trn . Nhn xt chung : Do c im ca cc hm s lng gic nn cc tch phn Lin kt thng c dng rt nhiu trong tch phn lng gic .V vy khi tnh tch phn lng gic hc sinh cn nhn k vo biu thc lng gic nm ngay di du tch phn c th la chn cch gii nhanh v thun li luyn tp ta xt cc tch phn sau

I=4

0

sin x1 sin 2

dxx

p

+ :J=4

0

sin x cossin 2 os2

xdxx c x

p

+ ;K=4

0

sin 59 os5 7sin 5

xdxc x x

p

- ; L=4 5

5 50

os2

os2 os2

c xdx

c x c x

p

+

M=4

0 1 t anxdx

p

+ ; K=4

2 2

0

( os 2 )( os 4 )c x c x dx

p

; P=42

4 40

( os )(s inx) (cos )

c xdx

x

p

+

VI-Tch phn cc hm s phn thc hu t A L thuyt :

Phn thc hu t : ( )( )

P xQ x

vi P(x) ,Q(x) l cc a thc vi cc h s thc

Phn thc thc s :l phn thc hu t ( )( )

P xQ x

v degP(x )< degQ(x) Phn thc n gin l phn thc c mt trong 4 dng sau

2 2; ; ;( ) ( ) ( )k kA A Bx c Bx c

x a x a x px q x px q+ +

- - + + + +

nh l tng quan v phn tch cc a thc Mi a thc Q(x) khc khng vi h s thc , u c th phn c mt cch phn Tch duy nht thnh cc nhn t ( khng tnh theo th sp xp cc nhn t) gm cc nh thc bc nht hoc cc tam thc bc hai vi bit thc en ta m.

B . phng php tnh tch phn hu t



Xt tch phn ( )( )

p xdx

Q x vi P(x),Q(x) l cc a thc vi h s thc + Nu bc ca p(x) ln hn bc ca Q(x) ta thc hin php chia a thc cho a thc 1 1( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )P x P xP x P x

G x G xQ x Q x Q x Q x

= + = + vi bc ca P1(x) nh hn bc ca Q(x)

v tch phn ( )G x dx tnh d dng .v vy vic tnh tch phn ( )( )p x

dxQ x a v tnh

tch phn 1( )( )

p xdx

Q x c bc ca t s nh hn bc ca mu s. tnh cc tch phn hu t ta xt cc php bin i c bn sau

I=2

22 2

( )ln | |

dx d x ax a C

x a x a+

= = + ++ +

J=2 2

( ) 1( ) ( )

dx d x aC

x a x a x a- -

= = +- - -

K=2 2

1 1ln

2 2dx dx dx x a

Cx a a x a x a a x a

- = - = + - - + +

L=2 2

dxx a+ , t x=atant dx=ad(tant)=a(1+tan

2t)dt, 22 2 21dx dt dx

dtx a a ax a

= =+ +

Vic tnh cc tch phn cc hm hu t cui cng dn n cc tch phn c dng nh trn V vy trong qu trnh tnh cc tch phn hu t ta thng hng n cc tch phn trn. thun tin ta chia cc tch phn hm hu t theo mt s dng sau:

1.Dng 2

dxax bx c

b

a + + (mu s cha tam thc bc hai)

Trng hp1: ax2+bx+c=0 c hai nghim phn bit x1,x2 I=

2

dxax bx c

b

a + + = 1

1 2 1 2 1 2 1 2 2

1 1 1 1( ) ln | |

( )( ) ( ) ( )x xdx

dxa x x x x a x x x x x x a x x x x

b bba

a a

-= - =

- - - - - - -

Trng hp 2: ax2+bx+c=0 c nghim kp x=x1=x2 I= 12 2 2

1 1 1

( )1 1( ) ( ) ( )

d x xdx dxax bx c a x x a x x a x x

b b bba

a a a

-= = -

+ + - - -

Trng hp 3: ax2+bx+c=0 v nghim



Phn tch ax2+bx+c = 2 2 2( ) ( )

2 4b

a x h mx n ka a

D + - = + + sau t mx+n=ktant

Ta dn n I= 2 dxax bx cb

a + + =A

qqp

p

dt At=

V d 1: I=

1

20 5 6

dxx x- +

Li gii: I=

1

20 5 6

dxx x- + =

1 110

0 0

1 1 3 4( ) ln | | ln

( 2)( 3) 3 2 2 3dx x

dxx x x x x

-= - = =

- - - - -

V d 2: K=1

20 4 12 9

dxx x- +

Li gii: K=

1

20 4 12 9

dxx x- + =

1102

0

1 1(2 3) 2(3 2 ) 3

dxx x

= =- -

V d 3: Tnh L= 2

21 5 6

dxx x- +

Li gii: L=

2

21 5 6

dxx x- + =

2

21 ( 1) 1

dxx - + t x-1= tant dx=(1+tan

2t)dt

44

0

2

41 0

x tL dt

x t

pp p= = = =

= =

2-Dng 2

( )mx n dxax bx c

b

a

++ +

ng li chung gii l ta phn tch mx+n=p(ax2+bx+c) +q Khi K=

2 '

2 2

( ) ( )ax

mx n dx ax bx cp q

ax bx c bx c

b b

a a

+ + += +

+ + + + 2dx

ax bx c

b

a + +

=p 2ln ax bx c ba+ + +q 2dx

ax bx c

b

a + +

c bit hn nu ax2+bx+c=0 c hai nghim phn bit th Li gii trn c th thu ngn li nh sau



21 2

( )( )

mx n dx p qM dx

ax bx c x x x x

b b

a a

+= +

+ + - -

V d 1: Tnh I =

2

21

5( 1)6

x dxx x

-- -

Li gii: I=

2

21

5( 1)6

x dxx x

-- - =

221

1

2 3( ) (2 ln | 3 | 3ln | 2 |) 6 ln 3

3 2dx x x

x x+ = - + + =

- +

Nhn xt: i n

2

5( 1) 2 36 3 2

xx x x x

-= +

- - - + ta phn tch

2

5( 1) 5( 1) ( 2) ( 3)5( 1) ( 2) ( 3)

6 ( 3)( 2) 3 2 ( 3)( 2)x x a b a x b x

x a x b xx x x x x x x x

- - + + += = + = - = + + -

- - - + - + - +(1)

tm a,b ta c mt trong hai cch sau Cch 1 (tr s ring): (1) ng vi mi x suy ra ng vi cc gi tr c bit ca x.ta thng chn cc gi tri ca x lm trit tiu mt trong cc tham s a hoc b chng hn cho x=3 th a=2 ;x=2 ta c b=3 Cch 2 ng nht cc h s ca cc s hng t s 5(x-1)=a(x+2)+b(x-3)=(a+b)x+2a-3b dn n h phng trnh

5 2

2 3 5 3

a b a

a b b

+ = = - = - =

V d 2: Tnh K=

1

20

(2 1)2 2 1

x dxx x

++ +

Li gii: K=

1 1 22 1

02 20 0

(2 1) 1 (2 2 1) 1 1ln(2 2 1) ln 5

2 2 1 2 2 2 1 2 2x dx d x x

x xx x x x

+ + += = + + =

+ + + +

3-Dng ( )( )

P x dxQ x

b

a

Khi tnh cc tch phn dng trn ta thng gp cc phn thc ng ( )( )

P xQ x



(ngha l bc ca t s nh hn bc ca mu s) sau khi ta phn tch s a c v dng c bn nh xt. Khi tnh cc tch phn dng ny cn phi ch n cc trng hp mu thc c nghim n ,nghin kp,v nghim . ta tm thi chia ra mt s dng c bn sau a/ Mu thc c nghim n ,cha nhn t l tam thc bc hai v nghim V d 1: I=

1 2

30

2 51

x xdx

x+ ++

Li gii: Nhn xt

2 2 2 '

3 2 2 2

2 5 2 5 ( 1)1 ( 1)( 1) 1 1 1

x x x x a b x x cx x x x x x x x x+ + + + - +

= = + ++ + - + + - + - +

=

22 2

3

( 2 ) ( )2 5 ( 2 ) ( )

12 2 2

5 0

1 3

a b x b c a x a b cx x a b x b c a x a b c

xa b a

a b c b

b c a c

+ + + - + - + + + = + + + - + - +

++ = =

- + = = + - = =

Vy I=1 1 1

2 20 0 0

2 3( ) 2 3 2ln 2 3

1 1 1 1dx dx

dx Jx x x x x x

+ = + = ++ - + + - + (*)

li c 1 1

220 0

1 31 ( )2 4

dx dxx x x

=- + - +

t

21 3 3tan (1 tan )2 2 2

16

06

x t dx t dt

x t

x t

p

p

- = = +

= = - = =

J=6

6

2 393

dt

p

p

p

-

= th vo (*) ta c I=2ln2+ 39p

Nhn xt : Nu quan st k ta thy : t s h s ca x2 l 2 iu chng t 2x2+x+5=2(x2-x+1)+b(x+1) c h s ca x bng 1 ta phi c b=3 v th ta c ngay

2

3 2

2 5 2 31 1 1

x xx x x x+ +

= ++ + - +

v d 2: Tnh J=

6 2

32

3 2 62

x xdx

x x+ ++ :



Li gii: J=

6 2

32

3 2 62

x xdx

x x+ ++ =

6 2

22

3( 2) 2( 2)

x xdx

x x+ +

=+

6

22

3 2( )

2dx

x x+

+ =6 6

122 2

33 2 ln 3 2

2 2dx dx

Jx x+ = +

+

li t

I=18ln 3 212p+

b/Mu thc c nghim n v nghim kp ,cha nhn t l tam thc bc hai v nghim V d 1: I=

0 2

31

4 7 33 2

x xdx

x x-

- -- +

Li gii: nhn xt

2 2 2

3 2 2 2

4 7 3 4 7 3 ( 1)( 2) ( 2) ( 1)3 2 ( 1) ( 2) 1 ( 1) 2 ( 1) ( 2)

x x x x a b c a x x b x c xx x x x x x x x x

- - - - - + + + + -= = + + =

- + - + - - + - +

ng nht thc ta c 1 2

2 3

0 1

x b

x c

x a

= = - = - = = =

I=0 02 3

013 2

1 1

4 7 3 2 3 ( 2)( ) ln | | ln16

3 2 1 ( 1) 2 1x x dx dx dx x

dxx x x x x x -- -

- - += - + = =

- + - - + -

V d 2: Tnh J=

2 2

31`

3 3 2x xdx

x x+ ++

Li gii: Ta c s phn tch

2 2 2 ' 2

3 2 2 2 2

3 3 2 3 3 2 ( 1) (1 ) 2( 1) 1 1 (1 )

x x x x a b x c a x bx cxx x x x x x x x x+ + + + + + + +

= = + + =+ + + + +

ng nht thc ta c

2 2 2

3

1

4

2 tan 2(1 tan ) , 2 2(1 tan )

623 2 2 2( )

3 4 122

4

x t dx t dt x t

x tJ dt

x t

p

p

pp p p

p

= = + + = +

= = = = - = = =



22

11 8 2

21 2 2 2 3

ax o a

x a b c b

x a b c c

== = = = + + = = - + - = =

J=3 3 2 32 2

3 2 21` 1 1 1

3 3 2 1 (1 )2 3

2 1 1x x dx d x dx

dxx x x x x+ + +

= + ++ + +

=ln3- ln 22

+3J1

li t x=tant ta c J1= 4p thay vo th J= ln 2ln 3

2 4p

+ +

Nhn xt : Nhiu khi s thnh tho ca ng nht thc gip ta b qua nhanh bc phn tch bng Phng php h s bt nh . chng hn ta c ng thc sau * 1 1 ( ) ( ) 1 1 1( )

( )( ) ( )( )x b x a

x a x b a b x a x b a b x b x a- - -

= = -- - - - - - - -

*

22

2 2 2 2 2

2 2 2

1 1 [( ) ( )] 1 1 1( )

( ) ( ) ( ) ( ) ( ) ( )

1 1 1 2 1 1[ ( )]

( ) ( ) ( )

x b x ax a x b a b x a x b a b x b x a

b a x a x b a b x a x b

- - -= = -

- - - - - - - -

= + - -- - - - - -

V d 3 (Hc vin mt m -98) Tnh K=

1

2 20 ( 1) ( 2)

dxx x+ +

Li gii: Ta phn tch theo 2

2 2 2 2

1 1 1 1 1 1 1( ) =[ 2( )]

( 1) ( 2) 1 2 ( 1) ( 2) 1 2x x x x x x x x= - + + -

+ + + + + + + +

K=1 1

2 2 2 20 0

1 1 1 1 11 4[ 2( )] .... 2 ln

( 1) ( 2) ( 1) ( 2) 1 2 12 3dx

dxx x x x x x

= + + - = = ++ + + + + +

luyn tp ta xt cc bi ton sau 1/

3 2

20`

3 2xdx

x x++ 2/

1

31`

1dx

x x+ 3/1 2

30`

2 3 7x xdx

x x+ ++ 4/

1 2

20`

53 2

xdx

x x+

+ - 2 2

41`

11

xdx

x-+

5/2 4

31`

2xdx

x x-- 6/

2 2

31`

1( 1) ( 3)

xdx

x x+

- + 7/2

2 21`

1( 3 2)

dxx x+ + 8/

2

21`

1(1 )

dxx x-

9/2 2

21` 7 5

xdx

x x- + 10/2

4 21` 1

xdx

x x+ + 11/2 3

20`

32 1x

dxx x+ + ; 12/

2

31` ( 1)

dxx x +



VII- Tch Phn cc hm s v T 1-Dng 1:

2

dx

ax bx c

b

a + +

Phng php : .Bin i a v tch phn

2

du

u k

b

a +

.s dng b : 22

ln | |du

u u ku k

bba

a

= + ++

CM: Tht vy: ( )2 ''

2

2 2

( ) 1ln | |

u u Ku u K C

u u K u k

+ ++ + + = =

+ + +

V d 1: K=

0 02 0

1221 1 2 4

4 4

1 1 1 1 1 2 6ln | 2 3 | ln

41 23 2 2 232 3 ( )4 16

dx dxx x x

x x x-

- -

+= = + + + + =

+ + + +

2- Dng 2: 2

( )mx n dx

ax bx c

b

a

+

+ +

Phng php: phn tch

2

2 2 2 2 2

( ) (2ax ) ( )2 2 2 2

mx n dx m b dx mb dx m d ax bx c mb dxa aax bx c ax bx c ax bx c ax bx c ax bx c

b b b b b

a a a a a

+ + + += - = -

+ + + + + + + + + + V D 2: J=

1

20

( 4)

4 5

x dx

x x

+

+ +

Li gii: J=

1 1 1 1 12

2 2 2 2 20 0 0 0 0

( 4) ( 2) 1 ( 4 5) ( 2)2 2

24 5 4 5 4 5 4 5 ( 2) 1

x dx x dx dx d x x dx d x

x x x x x x x x x

+ + + + += + = +

+ + + + + + + + + +

= 3 1010 5 2ln2 5

+- +

+



V d 3: K=1

20

( 2)

2 2

x dx

x x

+

+ + :

Li gii: K=

0 1 0 1 02

2 2 2 2 21 0 1 0 1

( 2) 1 (2 2) 1 ( 2 2)2 22 2 2 2 2 2 2 2 ( 1) 1

x dx x dx dx d x x dx dx

x x x x x x x x x- - -

+ + + += + = +

+ + + + + + + + + +

= 2 2 012 2 ln | 1 2 2 2 1 ln(1 2)x x x x x - + + + + + + + = - + +

3/Dng 3 2( )

dx

px q ax bx c

b

a + + +

Phng php : t px+q=2

1 1 1; ( )

dtpdx x q

t t p t-

= = -

Khi 2

2 22

2

/

1 1 1( ) ( ) ( )

dx dt t dt

a bpx q ax bx c mt nt eq q ct p t q t

b b b

a a a

-= =

+ + + + +- + - +

V d 4: M=3

22 ( 1) 2 2

dx

x x x- - +

Li gii: t

2

2 1

1 1 11 3

2

x t

tx x x t

t tdt

dxt

= =+ - = = = = = -

Do M= 3 1 2

2 22 1/2

12 1

1/221/2

/

( 1) 2 2 1 1 12 2

2 2 2ln 1 ln

1 51

dx dt t

x x x t tt t t

dtt t

t

=- - + + + - +

+= = + + =

++

Dng 4: 2

( )

( )

mx n dx

px q ax bx c

b

a

+

+ + +

Phng php : Phn tch

2

( )

( )

mx n dx

px q ax bx c

b

a

+

+ + + =



2 2

2 2

( ) ( )( )

( ) ( )

( )( )

( )

m mqpx q n dx

mx n dx p p

px q ax bx c px q ax bx c

m dx mq mx n dxn

p pax bx c px q ax bx c

b b

a a

b b

a a

+ + -+

= =+ + + + + +

++ -

+ + + + +

y l cc tch phn xt trn.

VIII-Lp tch phn c bit : Trong khi tnh tch phn nhiu bi ton tm c nguyn hm ca n khng h n gin .Khi chng ta nn khai thc ti a cc tnh cht c bit ca hm s di du tch phn v cn ca n ta s c cc kt qu p trong tch phn vo gii Ton .Trong phn ny Ti mun trao i vi cc bn ng nghip hng dn hc Sinh cch tm n nhng kta qu p .trong cc th d sau c s dung bi vit ca thy :Trnh tun-Ging vin i hc Thu Li ,ng trn bo Ton hoc & tui tr s 367 thng 1/2008 . 1/ Nu f(x) l hm s lin tc trn [-a;a] vi a>0 ta c

0

( ) ( ( ) ( ))a a

a

f x dx f x f x dx-

= + -

V d 1: Cho f(x) l hm s lin tc trn [-a;a] ,a>0 tho mn

f(x)+f (-x)= 2 2 os2c x- hy tnh I=

32

32

( )f x dx

p

p-

Li gii: t x=-t ta c 0 0

0

0

0 0 0 0

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ( ) ( ))

a

a a

a a a a a

a a

f x dx f t d t f x dx

f x dx f x dx f x dx f x hay f x dx f x f x dx

-

- -

= - - = -

+ = - + = - +



p dng vi a= 3 ; ( ) ( ) 2 2 os22

f x f x c xp

+ - = - ta c

I=

3 3 32 2 2

0 0 0

| s inx |( ) 2 2 os2 2

dxf x dx c xdx

p p p-

= - = = 2(32

0

sin x sin xdx dx

pp

p

- )=6

Nhn Xt : -Nu khng bit tnh cht trn ta rt kh khn tnh c tch phn trn v gi thit

Cha gip ta xc nh c hm s f(x). Ta tnh I=32

32

( )f x dx

p

p- m khng cn

Quan tm n hm f(x) bng bao nhiu. -Do tnh cht trn khng c cp n trong sch gio khoa nn hc sinh phi trnh by Li gii nh trn .xem tnh cht nh l nh hng Li gii.

2/Nu f(x) lin tc trn [-a;a] ,a>0 th 02 ( )

( )

0

aa

a

f x dxf x dx

-

=

nu f(x)- chn hoc l

Trn [-a;a]

V d 2: Tnh J=1

2

1

ln( 1)x x dx-

+ +

Li gii: Nhn thy ( )( )

( )( )( )2 2

2 2 2 2

0 12 2

1 0

1` 1 1

ln 1` 1 0 ln( 1) ln( 1)

ln( 1) ln( 1)

t t t t t

t t t t t t t t

t t t t-

- + + + + = "

- + + + + = - + + = + +

- + + = - + +

Hay 0

2

1

ln( 1)x x dx-

+ + =-1

2

0

ln( 1)x x dx+ + t J=1

2

1

ln( 1)x x dx-

+ + =0

V d 3: Tnh K=

12

1

2ln

2x

x dxx-

-+

Li gii: Nhn xt vi mi a>0 ta c lna=-ln 1a

theo vi mi thuc [-1;1]



0 0 1

2 2 2 2 2

1 1 0

1 1

0 0

2 2 2 2 2 2 2ln ln ln ln ln ln ln

2 2 2 2 2 2 2

2 2ln

2 2

x x x x x x xx x x dx x dx x dx

x x x x x x x

t xdt dx

t x

-

- -

- + - + - + += - = - = - =

+ - + - + - -

- -= = -

+ +

K=0 1 1 1

2 2 2 2

1 0 0 0

2 2 2 2ln ln ln ln 0

2 2 2 2x x x x

x dx x dx x dx x dxx x x x-

- - - -+ = - + =

+ + + +

3/ Nu f(x) chn lin tc trn [-a;a] th ta c 0

( )( )

1

a a

xa

f xdx f x dx

a-=

+

V d 3: Tnh

1 4

11 2x

xdx

- + (HvBCVT)

Li gii: t x=-t ta c dx=-dt 0 0

1 1

x t

x t

= = = - =

Khi 0 0 0 14 4 4 4

1 1 1 0

0 1 1 1 14 4 4 44

1 0 0 0 0

2 21 2 1 2 1 2 1 2

2 11 2 1 2 1 2 1 2 5

t t

x t t t

x

x x x x

x t t tdx dt dt

x x x xL dx dx dx x dx

--

-

= - = - = + + + +

= + = + = =+ + + +

V d 4

Tnh M=2

2

s inx.sin 2 . os51 x

x c xdx

e

p

p- + (HBK)

HD: f(x)=sinx.sin2x.cos5x l hm s chn v lin tc trn ;2 2p p-

nn theo kt

qu trn M=2

2

sin x sin 2 os5 0xc xdx

p

p-

=

V d5: Tnh N=22

2

| s inx|1 2x

xdx

p

p- + (H Dc)

HD:Hm s x2|sinx| -chn v t kt qu trn ta c N=2

2

0

s inxx dx

p

tip tc

Dng tch phn tng phn hai ln ta i n kt qu N= 42

p +



4/ Nu hm s f(x) lin tc trn [ ]0; ; 0 :a a > ta c 0 0

( ) ( )a a

f x dx f a x dx= -

Vi d 6: Tnh P=4

0

ln(1 t anx)dx

p

+ (H thu li 2001) Li gii:

ta c P=4

0

ln(1 t anx)dx

p

+ =4 4 4

0 0 0

ln 22ln(1 t an( -x)) ln

4 1 t anx

dx Pdx dx

p p p

p -+ = =

+

suy ra P= ln 28p

5/Nu f(x) l hm s lin tc trn [ ];a b v tho mn f(x)=f(a+b-x) th ( ) ( )

2

b b

a a

a bxf x dx f x dx

+=

chng minh tnh cht trn ta thc hin php i bin s t t=a+b-x c bit nu ta chn: a=0,b=p , f(x) = f(sinx) th ta c

0 0

(s inx) (s inx)2

xf dx f dxp pp

= (I)

Mt khc 2

0 02

(s inx) (s inx) (s inx)f dx f dx f dx

pp p

p

= +

t x= tp - cho tch phn 2

(s inx)f dxp

p ta c

2

0 0

(s inx) 2 (s inx)f dx f dx

pp

= (II)

li t x=2

tp- ta c

2 2

0 0

(s inx) (cos )f dx f x dx

p p

= (III)

V d 7: Tnh Q = 20

sin x cosx xdxp

Li gii: Coi f(sinx)=sinx(1-sin2x) nh kt qu (I) ta thu c Q = 2 2 3 0

0 0

sin x cos os (cos ) os2 2 6 3

xdx c xd x c xp p

pp p p p= - = - =

V d 8 Tnh T=2

22

0

1tan (s inx)

os (cos )dx

c x

p

-

Li gii:



Ta c T= ( )2 2

2 2 22

0 0

1tan (s inx) 1 tan (cos ) tan (s inx)

os (cos )dx x dx

c x

p p

- = + -

Coi f(cosx) l tan2(cosx) ,f(sinx) l tan2(sinx) ta p dng tnh cht (III) ta c 2 2

2 2tan (cos ) tan (s inx)o o

x dx dx

p p

= . Nn ta c T=2

0 2dx

p

p=

6/ Nu f(x) lin tc ,tun hon vi chu k T th

2

02

( ) ( ) ( ) ,

Ta T T

Ta

f x dx f x dx f x dx a R+

-

= = " .

Tht vy t J1= ( )a T

a

f x dx+

, J2=0

( )T

f x dx ,J3 =2

2

( )

T

T

f x dx-

J1=0

0

( ) ( ) ( ) ( )a T T a T

a a T

f x dx f x dx f x dx f x dx+ +

= + +

t x=t+T i vi ( )a T

T

f x dx+

ta c J1=J2 .Chn a= 2T- ta c J1=J3

V D 9:

Tnh H=2010

2009sino

xdxp

Li gii: Nhn thy f(x)=sin2009x l hm s tun hon vi chu k T= 2p v H=

2 4 20102009 2009 2009 2009

0 2 2008

sin sin .... sin 1005 sin 0xdx xdx x xdxp p p p

p p p-

+ + + = =

Do hm s f(x) =sin2009x l hm s l.

V d 10: Tnh G=2

2ln(s inx 1 sin )o

x dxp

+ + (0lympic sv-2007)

HD:Nhn xt f(x)=ln(sinx+ 21 sin )+ lin tc ,tun hon vi chu k T= 2p Do G= ( ) 0f x dx

p

p-

= (Do f(x) l hm s l) Bi Tp Tng t



1/I =1 4

21

s inx1

xdx

x-

++ ; 2/ J =

2

2

2

osx4 sinx c

dxx

p

p-

+- 3/K =

1 2010

1 2009 1x

xdx

- + ;

4/L = 6 64

4

sin os2009 1x

x c xdx

p

p-

++ ; 5/M=

4

20090

log (1 t anx)dx

p

+ ; 6/ N=4

20

sin x2 osx dx

c x

p

+

Sau y l cc bi tp v tch phn c la chn t cc thi i Hc trong nhiu nm cho ti k thi nm hc 2009-2010.Gip hc sinh vn dng thnh tho cc phng php v k thut in hnh nh trn gii

1. / 4 2

0

1 2sin1 sin 2

xdx

x

p -+ 2. -

2

0

2 dxxx

3. -+2

1 11dx

x

x 4. +e

dxx

xx

1

ln.ln31

5. -3

2

2 )ln( dxxx 6. / 2

0

sin 2 sin

1 3cos

x xdx

x

p ++

7. / 2

0

sin 2 cos1 cos

x xdx

x

p

+ 8. / 2

sin

0

( cos ) cosxe x x dxp

+

9. / 2

2 20

sin 2

cos 4sin

xdx

x x

p

+ 10. -

1

0

2)2( dxex x

11. -+ -5ln

3ln 32xx ee

dx 12. / 2

0

sin sin 2 sin 3x x x dxp

13. / 2

4 4

0

cos 2 (sin cos )x x x dxp

+ 14. / 2

5

0

cos xdxp

15. -+3

248

7

21dx

xxx 16.

e

dxxx1

22 ln

17. +3

0

23 1 dxxx 18. +e

xdxx

x

1

3

ln1

19. -9

1

3 1 dxxx 20. +-3

1

2 12 dxxx

21. +1

0

2 )1( dxex x 22. +

+3

02

35

1

2dx

x

xx

23. /3

2/ 4 cos 1 cos

tgxdx

x x

p

p + 24.

-

+-2

1

2

21

dxxx



25. 2

0

sin1 cos

x xdx

x

p

+ 26. +1

0 1xe

dx

27. / 4

2

0

xtg x dxp

28. +

dxx

x1

29. -

--+5

3

)22( dxxx 30. +2

02

2

)2(dx

xex x

31. - ++

4

1 45

2

x

dx 32. --1

0

22 )124( dxexx x

33. +

2

05

4

1dx

x

x 34. -2

0

22 4 dxxx

35. -++2

1 22 xx

xdx 36. -

+0

1

1 dxxx

37. ++1

02 252 xx

dx 38. +2

12

)1ln(dx

xx

39. / 2

0

sin 2cos 1

xdx

x

p

+ 40. / 2

0

sin1 3cos

xdx

x

p

+

41. +3

0

32 .1 dxxx 42. +1

02)1(x

xdx

43. / 2 2004

2004 20040

sinsin cos

xdx

x x

p

+ 44. / 2 3

0

4sin1 cos

xdx

x

p

+

45. +1

0

23 3 dxxx 46. - +++

-3

1 313

3dx

xx

x

47. -1

0

25 1 dxxx 48. / 2

3

0

sin 5xe x dxp

49. +3

0

33 .1 dxxx 50. / 4 2

0

1 2sin1 sin 2

xdx

x

p -+

51. - ++

0

12 42xx

dx 52. e

dxx

x

12

ln

53. ++3/7

03 13

1dx

x

x 54. / 2

0

cos3sin 1

xdx

x

p

+

55. / 2

2 20

sin

sin 2cos cos2

xdxx

x x

p

+ 56.

/3 2

20

sinsin 2 cos

x xdxx x

p



57. e

dxxx1

ln 58. 2 / 4

0

.cosx x dxp

59. ++++2

02

23

4942

dxx

xxx 60. +1

03)1(x

xdx

61. -e

xx

dx

12ln1

62. / 2 3

0

4sin1 cos

xdx

x

p

+

63. / 4

0 (sin cos )cosdx

x x x

p

+ 64. ( )1

2 3

0

1xx e x dx+ -

65. 2ln

0

5 2 dxex x 66. dxx

xx

+

+1

03 2

2

)1(

67. / 4

0

(1 )sin2x

tgx tg xdxp

+ 68. +1

0

2 )1ln( dxxx

69. +2

12

)1ln(dx

xx 70. +

1

0

2 1 dxxx

71. +1

021 x

xdx 72. / 2

/ 4

sin cos

1 sin 2

x xdx

x

p

p

-+

73. +3

0

2 )5ln( dxxx 74. / 2

30

cos 2(sin cos 3)

xdx

x x

p

- +

75. / 4

0

( 1) cosx x dxp

- 76. / 4

0

cos 21 2sin 2

xdx

x

p

+

77. +

2ln

0

2

2dx

e

ex

x

78. / 2 4

0

4sin1 cos

xdx

x

p

+

79. / 2

20

cos7 5sin cos

x dxx x

p

- - 80. / 4

20 cos

xdx

x

p

81. - +++

-3

13 31

3dx

xx

x 82. -9

1

3 1 dxxx

83. +e

dxxx

x

1

3

ln).1

( 84. + dxxx 32 2

85. +1

02

3

1dx

xx 86.

+

3ln

03)1( x

x

e

dxe

87. -

++0

1

32 )1( dxxex x 88. +2/

0

56 3 cossincos1n

dxxxx



89. / 4

0 1 cos 2x

dxx

p

+ 90. -1

0

23 1 dxxx

91. -

5ln

ln

2

2 1dx

e

ex

x

92. 1

0

3 2 dxex x

93. +e

dxxx

x

1

2

.ln.1 94. +

3

13xx

dx

95. +8ln

3ln

2.1 dxee xx 96. 2

0

sinx x dxp

97. +3

1

2

1ln

lnedx

xx

x 98. / 2

2

0

(2 1)cosx x dxp

-

99. +++6

2 1412 xx

dx 100. / 2

0

( 1)sin 2x x dxp

+

101. --10

5 12 xx

dx 102. +-e

dxxx

x

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ln23

103. 3 21

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x xdx 104. 46

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dxcos x

p

105. ( )

4

0

sin4

sin 2 2 1 sin cos

x dx

x x x

p p -

+ + + 106. 2

31

ln xdx

x

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