phan loai mot so dang tich phan dac biet

7/22/2019 Phan Loai Mot So Dang Tich Phan Dac Biet

1/23

http://ebooktoan.com 1TCH PHN

I.CC PHNG PHP TNH TCH PHN

1. Phng php i bin s

Bi ton:Tnh ( )

b

a

I f x dx ,*Phng php i bin dng I

nh l. Nu 1) Hm ( )x u t c o hm lin tc trn on ; ,

2) Hm hp ( ( ))f u t c xc nh trn ; ,

3) ( ) , ( )u a u b ,

th '( ) ( ( )) ( )b

a

I f x dx f u t u t dt

.V d 1. Hy tnh cc tch phn sau:

a)

1

2 3

0

5I x x dx b) 2

4

0

sin 1 cosJ x xdx

Gii: a) Ta c 3 25 3 t x dt x dx

Khi x=0 th t=5

Khi x=1 th t=6

1 6

2 3

0 5

53

dt

I x x dx t

16 121

2

5

6 61 1 ( ) 21 5 53 3 91

2

t

t dt t t

4 106 5

3 9 .

b) Ta c

2

4

0

(sin 1) (sin )J x d x

51 6

sin sin 25 5

0x x

V d 2. Hy tnh cc tch sau:


2/23

http://ebooktoan.com 2

a)

4

2

0

4 x dx b)1

2

01

dx

x

Gii: a) t 2sin , ;2 2

x t t

.

Khi x = 0 th t = 0. Khi 2x th2

t .

T 2sinx t 2cosdx tdt 4 2 2

2 2 2

0 0 0

4 4 4sin .2cos 4 cos x dx t tdt tdt

.

b) t tan , ;2 2

x t t .

Khi 0x th 0t , khi 1x th4

t .

Ta c:2

tancos

dt

x t dxt

.

1 4 4

2 2 2

0 0 0

1. .4

1 1 tan cos 40

dx dt dt t

x t t

Ch : Trong thc t chng ta c th gp dng tch phn trn dng tng qut hn

nh:

Nu hm s di du tch phn c cha cn dng 2 2 2 2,a x a x v

2 2x a (trong trong a l hng s dng) m khng c cch bin i no

khc th nn i sang cc hm s lng gic lm mt cn thc, c th l:

Vi 2 2a x , t sin , ;2 2

x a t t

hoc cos , 0;x a t t .

Vi 2 2a x , t tan , ;2 2

x a t t

hoc , 0; x acott t .


3/23


Vi 2 2x a , t , ; \ 0sin 2 2

ax t

t

hoc ;cos

ax

t 0; \

2t

.

*Phng php i bin dng II

nh l :Nu hm s ( )u u x n iu v c o hm lin tc trn on ;a b

sao cho '( ) ( ( )) ( ) ( )f x dx g u x u x dx g u du th( )

( )

( ) ( )

u bb

a u a

I f x dx g u du .

V d 3:Tnh

1

2 3

05I x x dx

Gii: t 3( ) 5u x x .Tac (0) 5, (1) 6u u .

T c: 6

5

61 2 2 4 106 6 5 5 6 5

53 9 9 9 9I udu u u

V d 4:Hy tnh cc tch phn sau bng phng php i bin dng II:

a) 1

5

0

2 1x dx b)2

ln

e

e

dx

x x c)1

2

0

4 2

1

xdx

x x

d)

2

2

1(2 1)

dx

x e)2

3

3

2cos(3 )

3x dx

Gii: a) t 2 1u x khi 0x th 1u . Khi 1x th 3u

Ta c 22

dudu dx dx . Do :

1 3

65 5 6

0 1

31 12 1 (3 1)

12 12 12

ux dx u du = 60 23 .

b)tlnu x

. Khix e

th1u

. Khi 2

x eth

2u.


4/23


Ta cdx

dux

2 2

1

2ln ln 2 ln1 ln 2

1ln

e

e

dx duu

x x u .

c)t 2 1u x x . Khi 0x th 1u . Khi 1x th 3u .

Ta c (2 1)du x dx . Do :1 3

2

0 1

34 2 22ln 2(ln3 ln1) 2ln3

11

x dudx u

x x u

.d)t 2 1u x . Khi 1x th 1u . Khi 2x th 3u .

Ta c 22

dudu dx dx . Do :

2 3

2 2

1 1

31 1 1 1 1( 1)

1(2 1) 2 2 2 3 3

dx du

x u u

.

e)t2

33

u x

.

Khi

3x

th

3u

,

Khi2

3x

th

4

3u

.

Ta c 33

dudu dx dx . Do :

2 4

3 3

3 3

4

2 1 1 1 43cos(3 ) cos sin sin sin3 3 3 3 3 3

3

x dx udu u

1 3 3 3

3 2 2 3

.

2.Phng php tch phn tng phn.

nh l .Nu u(x) v v(x) l cc hm s c o hm lin tc trn ;a b th:


5/23


' '( ) ( ) ( ) ( ) ( ) ( )b b

a a

bu x v x dx u x v x v x u x dx

a

hay

b b

a a

budv uv vdu

a

.

p dng cng thc trn ta c qui tc cng thc tch phn tng phn sau:

Bc 1: Vit f(x)dx di dng 'udv uv dx bng cch chn mt phn thchhp ca f(x) lm u(x) v phn cn li ' ( ) .dv v x dx

Bc 2: Tnh 'du u dx v ' ( )v dv v x dx .

Bc 3: Tnh 'b b

a a

vdu vu dx vb

uva

.

Bc 5: p dng cng thc trn.V d 5:Tnh

1

lne

x xdx

Gii: tlnu x

dv xdx

2

2

dxdux

xv

2 2 2 2

1 1

1 1ln ln

1 12 2 2 4 4

e ee ex e x e

x xdx x xdx

.

V d 6:Tnh cc tch phn sau:

a)

2

5

1

lnxdx

x b)2

0

cosx xdx

c)1

0

xxe dx d)2

0

cosxe xdx

Gii: a) t

54

ln

1

14

dxu x du

x

dv dx vx x

. Do :


6/23

http://ebooktoan.com 6222 2

5 4 5 411 1 1

ln ln 1 ln 2 1 1 15 4 ln 2

4 4 64 4 4 256

x x dxdx

x x x x

.

b) t

cos sin

u x du dx

dv xdx v x

. Do :

2 2

0 0

cos sin sin cos 12 22 2

0 0x xdx x x xdx x

.

c)tx x

u x du dx

dv e dx v e

. Do :

1 1

0 0

1 11 1

0 0x x x xxe dx xe e dx e e e e .

d) t cos sin

x xu e du e dx

dv xdx v x

2 2

0 0

cos sin sin20

x x xe xdx e x e xdx

.

t1 1

1 1sin cos

x xu e du e dx

dv xdx v x

2 2

2

0 0cos cos cos2

0

x x x

e xdx e e x e xdx

.

2 22

2

0 0

12 cos 1 cos .

2x x ee xdx e e xdx

*Cch t u v dv trong phng php tch phn tng phn.


7/23


( )b

x

a

P x e dx ( ) lnb

a

P x xdx ( )cosb

a

P x xdx cosb

x

a

e xdx u P(x) lnx P(x) xe

dv xe dx P(x)dx cosxdx cosxdx

Ch : iu quan trng khi s dng cng thc tch phn tng phn l lm th no

chn u v 'dv v dx thch hp trong biu thc di du tch phn f(x)dx. Ni

chung nn chn u l phn ca f(x) m khi ly o hm th n gin, chn

'dv v dx l phn ca f(x)dx l vi phn mt hm s bit hoc c nguyn hm

d tm.

C ba dng tch phn thng c p dng tch phn tng phn:

Nu tnh tch phn ( ) ( )P x Q x dx

m P(x)l a thc cha x v Q(x) l

mt trong nhng hm s: , cos , sinaxe ax ax th ta thng t

' ( )( )

( ) ( )

du P x dxu P x

dv Q x dx v Q x dx

Nu tnh tch phn ( ) ( )P x Q x dx

m P(x) l a thc ca x v Q(x) l

hm s ln(ax) th ta t '

( )( ) ( )

du Q x dxu Q xdv P x dx v P x dx

Nu tnh tch phn cosaxI e bxdx

hoc sinaxJ e bxdx

th

ta t 1cos sin

axax du ae dxu e

dv bxdx v bxb


8/23


hoc t 1sin cos

axax du ae dxu e

dv bxdx v bxb

Trong trng hp ny, ta phi tnh tch phn tng phn hai ln sau tr

thnh tch phn ban u. T suy ra kt qu tch phn cn tnh.

II.TCH PHN MT S HM S THNG GP

1. Tch phn hm s phn thc

a)Tnh tch phn dng tng qut sau:

2 0dx

I aax bx c

.

(trong 2 0ax bx c vi mi ;x )

Xt 2 4b ac .

+)Nu 0 th 2

2

dxI

ba x

a

tnh c.

+)Nu 0 th 1 2

1 dxI

a x x x x

,

(trong 1 2;2 2

b bx x

a a

)

1

1 2 2

1ln

x xI

a x x x x

.

+) Nu 0 th22 2

22 4

dx dx

Iax bx c b

a xa a

t 22 21

tan 1 tan

2 4 2

bx t dx t dt

a a a

, ta tnh c I.


9/23


b) Tnh tch phn: 2 , 0mx n

I dx aax bx c

.

(trong 2

( ) mx n

f x

ax bx c

lin tc trn on ; )

+) Bng phng php ng nht h s, ta tm A v B sao cho:

cbxax

B

cbxax

baxA

cbxax

nmx

222

)2(

+)Ta c I=

dcbxax

Bdx

cbxax

baxAdx

cbxax

nmx

222

)2(

. Tch phn dcbxax

baxA

2)2(

=

cbxaxA 2ln

Tch phn2

dx

ax bx c

tnh c.

c) Tnh tch phn( )

( )

b

a

P xI dx

Q x vi P(x) v Q(x) l a thc ca x.

Nu bc ca P(x) ln hn hoc bng bc ca Q(x) th dng php chia athc.

Nu bc ca P(x) nh hn bc ca Q(x) th c th xt cc trng hp:+ Khi Q(x) ch c nghim n

1 2, ,..., n th t

1 2

1 2

( )...

( )n

n

AA AP x

Q x x x x

.

+ Khi 2 2( ) , 4 0Q x x x px q p q th t

2

( ).

( )

P x A Bx C

Q x x x px q

+ Khi 2

( )Q x x x vi th t


10/23


2

( )

( )

AP x B C

Q x x x x

.

V d 7. Tnh tch phn:

1

2

0

4 115 6

x dxx x

.

Gii:

Cch 1.Bng phng php ng nht h s ta c th tm A, B sao cho:

2 2 2

2 54 11, \ 3; 2

5 6 5 6 5 6

A xx Bx

x x x x x x

2 22 54 11 , \ 3; 2

5 6 5 6Ax A Bx x

x x x x

2 4 2

5 11 1

A A

A B B

Vy

2 2 22 2 54 11 1

, \ 3; 25 6 5 6 5 6

xxx

x x x x x x

.

Do

1 1 1

2 2 2

0 0 0

4 11 2 52

5 6 5 6 5 6

x x dxdx dx

x x x x x x

21 12 9

2ln 5 6 ln ln0 03 2

xx x

x

.

Cch 2. V 2 5 6 2 3x x x x nn ta c th tnh tch phn trn bng

cch:

Tm A, B sao cho:

24 11

, \ 3; 25 6 2 3

x A Bx

x x x x

2 2

34 11, \ 3; 2

5 6 5 6

A B x A Bxx

x x x x

4 33 2 11 1A B AA B B


11/23


Vy 24 11 3 1

, \ 3; 25 6 2 3

xx

x x x x

.

Do

1 1 1

2

0 0 0

4 113

5 6 2 3

x dx dxdx

x x x x

1 1 93ln 2 ln 3 ln

0 0 2x x .

V d 8:Tnh tch phn:

1

2

01

dx

x x .Gii:

Do

1 1

22

0 01 1 3

2 4

dx dx

x xx

t 21 3 3

tan , ; 1 tan2 2 6 3 2

x t t dx t dt

Vy 21 3 3

22

06 6

3 1 tan 2 3 2 3 33231 3 3 9(1 tan )4 6

t dtdxdt t

x x t.

V d 9. Tnh tch phn:

12 3

2

01

xdx

x .Gii:

1 1 1 1

2 2 2 23

2 2 2

0 0 1 01 1 1

x x xdxdx x dx xdx

x x x

22

1 11 1 1 3

ln 1 ln2 2

2 2 8 2 40 0

xx .

2. Tch phn cc hm lng gic


12/23

http://ebooktoan.com 122.1.Dng 1: Bin i v tch phn c bn

V d 10:Hy tnh cc tch phn sau:

a)

2

2

sin2 sin7J x xdx

;

b)

2

4 4

0

cos (sin cos )K x x x dx

;

c)

2 3

0

4sin

1 cos

xM dx

x

.

Gii

a)

2 2

2 2

1 1cos5 cos9

2 2J xdx xdx

1 1 42 2sin5 sin910 18 45

2 2

x x

.

b) Ta c 24 4 2 2 2 2cos (sin cos ) cos sin cos 2sin cosx x x x x x x x

21 1 3 1

cos 1 sin 2 cos 1 1 cos4 cos cos cos42 4 4 4

x x x x x x x

3 1

cos cos5 cos34 8

x x x .

2 2 2 2

4 4

0 0 0 0

3 1 1cos (sin cos ) cos cos5 co3

4 8 8K x x x dx xdx xdx xdx


13/23


3 1 1 3 1 1 11sin sin5 sin32 2 2

4 40 24 4 40 24 150 0 0

x x x

.

c)

3 2 24sin 4sin sin 4(1 cos )sin4(1 cos )sin1 cos 1 cos 1 cos

x x x x xx xx x x

2M .

2.2.Dng 2: i bin s hu t ha tch phn hm lng gic

2.2.1.Tnhcos

dxI

asinx b x c

Phng php:

t2

2tan2 1

x dtt dxt

Ta c:2

2sin

1

tx

t

v

2

2

1cos

1

tx

t

22

cos 2

dx dt I

asinx b x c c b t at b c

bit cch tnh.

V d 11. Tnh4cos 3sin 5

dxx x

Gii: t 22

1 2tan 1 tan

2 2 2 1

x x dtt dt dx dx

t

2

2 2

2 2

21

1 2cos 3sin 3 3 23 31 1

dtdx dt t

t tx x t tt t

tan 11 2ln ln2 tan 2

2

xt

C Cxt

.

2.2.2. Tnh 2 2sin sin cos cos

dx

I a x b x x c x d


14/23


Phng php: 2 2sin sin cos cos

dxI

a d x b x x c d x

2

2

costan tan

dxx

a d x b x c d

t2cos

dxt tgx dt

x

2dt

Ia d t bt c d

tnh c.

V d 12. Tnh:2 2sin 2sin cos 3cos

dxI

x x x x

.

Gii:Ta c2

2 2 2

cossin 2sin cos 3cos tan 2 tan 3

dxdx xI

x x x x x x

t2

tancos

dx

t x dt x

21 1 1 tan 1

ln ln2 3 1 3 4 3 4 tan 3

dt dt t x

I C Ct t t t t x

2.2.3. Tnhsin cossin cos

m x n x pI dxa x b x c

.

Phng php:

+)Tm A, B, C sao cho:

sin cos sin cos cos sin ,m x n x p A a x b x c B a x b x C x

+) Vysin cos

sin cos

m x n x pI dx

a x b x c

=

=

cxbxa

dxCdx

cxbxa

xbxaBdxA

cossincossin

sincos

Tch phn dx tnh c

Tch phn Ccxbxadx

cxbxa

xbxa

cossinln

cossin

sincos


15/23


Tch phn cxbxadx

cossin tnh c.

V d 13. Tnh:cos 2sin

4cos 3sin

x xI dx

x x

.

Gii:

Bng cch cn bng h s bt nh, tm A v B sao cho:

cos 2sin 4cos 3sin 4sin 3cos , x x A x x B x x x

cos 2sin 4 3 cos 3 4 sin ,x x A B x A B x x

2

4 3 1 53 4 2 1

5

A

A BA B

B

2 1 4sin 3cos 2 1. ln 4cos 3sin

5 5 4cos 3sin 5 5

x xI dx x x x C

x x

.2.3.Dng 3: i bin s a vtch phn hm lng gic n gin hn

(Xem v d 17, 20, 21)

2.4.Ch : Nguyn hm dng sin , cosR x x dx , vi sin , cosR x x l mthm hu t theo sinx, cosx

tnh nguyn hm trn ta i bin s v a v dng tch phn hm hu t

m ta bit cch tnh tch phn.

Trng hp chung: t 22

tan 2 1 x dt

t dx t

Ta c2

2 2

2 1sin ;cos

1 1

t tx x

t t

Nhng trng hp c bit:+) Nu sin , cosR x x l mt hm s chn vi sinx v cosx ngha l

sin , cos sin ,cosR x x R x x th t tant x hoc cott x ,sau a tch phn v dng hu t theo bin t.


16/23

http://ebooktoan.com 16+) Nu sin , cosR x x l hm s l i vi sinx ngha l:

sin ,cos sin ,cosR x x R x x th t cost x .

+) Nu sin , cosR x x l hm s l i vi cosx ngha l:

sin , cos sin ,cosR x x R x x th t sint x .

3.Tch phn hm v t

3.1 .Dng 1: Bin i v tch phn v t c bn

V d 14. Tnh tch phn:

1

01

dxI

x x

.

Gii

1 1

3322

0 0

121 1

031

dx

I x x dx x xx x

2 2 2 23

V d 15:Tnh tch phn

1 3

20 1

x dx

x x .

Gii:

1 13 3 2 4

20 0

2 2 1( 1 )151

x dx x x x dxx x

.

3.2.Dng 2: Bin i v tch phn hm lng gic

(xem v d 2)

3.3Dng 3: Bin i lm mt cnGm: i bin s t l ton b cn thc

Vit biu thc trong cn di dng bnh phng ngV d 15:Tnh

1

0

23 1 dxxxI

Gii:

1

0

22

1

0

23 .11 xdxxxdxxxI


17/23


t t= 22222 111 txxtx

Ta c: xdx=-tdt, Khi x= 0 th t =1,khi x = 1 th t =0

Vy

152

53)1(

1

0

530

1

22

ttdtttI

4.Tch phn cha du gi tr tuyt i

Phng php:Chng ta phi ph du gi tr tuyt i

V d 16:Tnh

2

2

2

1J x dx

Gii: Lp bng xt du ca 2 1x trn on 2;2

x -2 -1 1 2

2 1x + 0 - 0 +

Do 2 1 1 2

2 2 2 2

2 2 1 1

1 1 1 1I x dx x dx x dx x dx

3 3 31 1 2

42 1 13 3 3

x x xx x x

.

III.TCH PHN MT S HM C BIT

1.Cho hm s ( )y f x lin tc v l trn on ;a a . Khi

( ) 0

a

a

I f x dx

.

V d 17:Chng minh

2

2

2

04 sin

xdxI

x

.

Gii: t x t dx dt . Khi x= 2

th t = - 2

, khi2

x

th2

t


18/23


Do : I= It

tdt

2

2

2sin4

Suy ra : 2I = 0. Ta c2

2

2

04 sin

xdxIx

.

2.Cho hm s ( )y f x lin tc v chn trn on ;a a . Khi

0

( ) 2 ( )a a

a

I f x dx f x dx

.

Chng minh : Ta c

0

0

( ) ( ) ( )a a

a a

I f x dx f x dx f x dx

(1)

Ta tnh

0

( )a

J f x dx

bng cch t 0x t t a dx dt 0 0

0 0

( ) ( ) ( ) ( )

a a

a a

J f x dx f t dt f t dt f x dx

(2)

Thay (2) vo (1) ta c

0

( ) 2 ( )a a

a

I f x dx f x dx

V d 18:Tnh tch phn:

2

2

2

cos

4 sin

x x

I dxx

Gii: Ta c

2 2 2

2 2 2

2 2 2

cos cos

4 sin 4 sin 4 sin

x x x xI dx dx dx

x x x

Do 1 2( ) 4 sinxf x

x

l hm s l trn ;

2 2

nn

2

2

2

04 sin

x dxx


19/23


v 2 2cos

( )4 sin

xf x

x

l hm s chn trn ;

2 2

nn ta c:

2 2 2

2 20

2 2

cos cos (sin )2 2

4 sin 4 sin (sin 2) sin 2

x x d xdx dx

x x x x

Vy1 sin 2 1

ln ln322 sin 2 2

0

xI

x

.

3.Cho hm s ( )y f x lin tc v chn trn on : . Khi

dxxfdxa

xfIx

)(21

1)(

Chng minh: t t= -x dt= - dx

Ta c f(x) = f(-t)= f(t); ax+1= a-t+1=1t

t

a

a

Khi x= - th t = ; x = th t =-

Vy

dttfa

adt

a

tfadx

a

xfI

t

t

t

t

x)(

1

11

1

)(

1

)(

Idxxfdta

tfdttf

t)(

1

)()(

Suy ra

dxxfdxa

xf

I x )(2

1

1

)(

V d 19:Tnh tch phn:1

4

12 1xx

I dx

.

Gii:t t= -x dt= - dx

Khi x= - 1 th t = 1 ; x =1 th t =-1

Vy

1

1

1

`1

441

1

4

122

1212 dttdt

tdx

xI

t

t

tx


20/23


1

1

1

1

1

1

44

4

12 Idxxdt

tdtt

t

Suy ra 5

1

52

1

2

11

1

51

1

4

x

dxxI

4.Cho f(x) lin tc trn on 0;2

.Khi

2 2

0 0

(sin ) (cos )f x dx f x dx

.Chng minh:

t2

t x dx dt

Khi x = 0 th2

t , khi

2x

th t = 0

Do

02 2 2

0 0 02

(sin ) (sin( ) (cos ) (cos )2

f x dx f t dt f t dt f x dx

.

Nhn xt: Bng cch lm tng t ta c cc cng thc*Nu f(x) lin tc trn 0;1 th (sin ) (sin )

2xf x dx f x dx

*Nu f(x) lin tc trn 0;1 th2 2

(cos ) (cos )

xf x dx f x dx

V d 20:Chng minh: I=

2

0

sin

sin cos 4

n

n n

xdx

x x

.

.

Gii :


21/23

http://ebooktoan.com 21Tng t nh trn ta c:

I=

2 2

0 0

sin cos

sin cos sin cos

n n

n n n n

x xdx dx

x x x x

=J

+) Vy I+J=

2 2

0 0

sin cos

sin cos sin cos 2

n n

n n n n

x xdx dx

x x x x

Vy I=

2

0

sin

sin cos 4

n

n n

xdx

x x

.

V d 21:Tnh tch phn:2

0

sin1 cosx x

dxx

.

Gii: t 0x t t dx dt .

Khi

0

2 2

0

sinsin

1 cos 1 cos

t tx xdx dt

x t

2 2

0 0

2 2

0 0

sin sin

1 cos 1 cos

sin sin

1 cos 1 cos

t t tdt dt

t t

x x xdx dx

x x

2 2

0 0

sin sin

2 1 cos 1 cos

x x x

dx dxx x

Vy2

2 2

0 0

sin sin

1 cos 2 1 cos 4

x x xdx dx

x x

.

BI TP NGHBi 1.Tnh cc tch phn sau


22/23


2

022 sin4cos

2sin)

dxxx

xIa

( H-KA-2006)

2

0 cos31sin2sin)

dxxxxIc

(H-KA-2005)

2

0 cos1

cos.2sin)

dx

xxIe

(H-KB-2005)

2

4

2sin1 cossin)

dxxxxIg

2

03)3cos(sin

2cos)

dxxx

xIi

2

0

sin)

dxxxIb

2

0

2

.cos)12()

dxxxId

4

0 2cos1)

dxx

xIf

3

4

2cos1cos

tan)

dxxx

xIh

4

0

2 .tan)

dxxxIk

Bi 2.Tnh cc tch phn sau

3

0

2

35

1

2) dx

x

xxIa

4

0 121

12) dx

x

xIc

dxxxIe 3

1

23 1.)

32

52 4

)xx

dxIg

3

1

22

)1(

)

xx

dxIb

1

2

12

11

1) dx

xxId

3

13

)xx

dxIf

5

3

22) dxxxIh

Bi 3. Tnh cc tch phn sau

1

0

2 )1() dxexIa x

1

0 1)

xe

dxIc

2

02

2

)2(

.

) dxx

ex

Ie

x

2

12

)1ln() d

x

xIb

e

dxxx

xId

1

3

.ln1

)

3

2

2

).ln() dxxxIf


23/23


0

1

32 )1() dxxexIg x 2

0

sin .cos)cos()

dxxxeIh x

phan loai mot so dang tich phan dac biet

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