Cac Phuong Phap Tinh Tich Phan

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Chuyn 1: Cc phng php tnh tch phnCc phng php tnh tch phnCc phng php tnh tch phnCc phng php tnh tch phnThng thng ta gp cc loi tch phn sau y:

+) Loi 1: Tch phn ca hm s a thc phn thc hu t.+) Loi 2: Tch phn ca hm s cha cn thc+) Loi 3: Tch phn ca hm s lng gic

+) Loi 4: Tch phn ca hm s m v logariti vi cc tch phn c th tch theo cc phng php sau:I) Phng php bin i trc tip

Dng cc cng thc bin i v cc tch phn n gin v p dng c )a(F)b(F)x(Fdx)x(fb

a

b

a

========

+) Bin i phn thc v tng hiu cc phn thc n ginV d 1. Tnh:

1.

====

2

1

3

2

dxx

x2xI ta c 12ln)21(ln)12(ln

x

2xlndx)

x

2

x

1(I

2

1

2

1

2=++=

+==

2. ++++

====

2e

1

dx

x

4x3x2J ( ) ++=+=

+=

22e

1

2e

1

2/1 7e4e3xln4x3x4dx

x

43x2

3.

====

8

13 2

3 5

dxx3

1x3x4K =

=

=

8

1

8

1

33 423/23/1

4

207xx

4

3x

3

4dxx

3

1xx

3

4

+) Bin i nh cc cng thc lng gicV d 2. Tnh:

1.

====

2/

2/

xdx5cosx3cosI

( ) 08

x8sin

2

x2sin

2

1dxx8cosx2cos

2

12/

2/

2/

2/

=

+=+=

2.

====

2/

2/

xdx7sinx2sinJ

( )45

4

9

x9sin

5

x5sin

2

1dxx9cos)x5cos(

2

12/

2/

2/

2/

=

==

3. ====

2/

2/xdx7sinx3cosK

( ) 010x10cos

4x4cos

21dxx10sinx4sin

21xdx3cosx7sin

2/

2/

2/

2/

2/

2/

=

+=+==

4. ====

0

20 xdxcosx2sinH =

=

+=

0 0

0x4cos16

1x2cos

4

1dx

2

x2cos1x2sin hoc bin i

====

0

2 xdxcosx2sinH =

=

+=

0 0

0x4cos16

1x2cos

4

1dx

2

x2cos1x2sin

5. ++++++++++++

====

2/

6/

dxxcosxsin

x2cosx2sin1G

( ) 1xsin2xdxcos2dxxcosxsin

xsinxcos)xcosx(sin2/

6/

2/

6/

2/

6/

222

===+

++=

6.

====

2/

0

4 xdxsinE

( )16

3x2sin

4

x4sinx3

8

1dxx2cos4x4cos3

8

1dx

2

x2cos12/

0

2/

0

2/

0

2

=

+=+=

=

7. ====4/

0

2 xdxtanF

( )4

4xxtandx1

xcos

1 4/0

4/

0

2

==

= . xut: ====

2/

4/

21 xdxcotF

v

====4/

0

42 xdxtanF

+) Bin i biu thc ngoi vi phn vo trong vi phnV d 3. Tnh:

1. ++++====1

0

3 dx)1x2(I 104

)1x2(

2

1)1x2(d)1x2(

2

11

0

41

0

3=

+=++=

2. ====2

1

3 dx)1x2(1J 0

)1x2(1

41

2)1x2(

21)1x2(d)1x2(

21

1

0

2

1

0

22

1

3=

=

+==


2/21



3. ====3/7

1

dx3x3K9

16)3x3(

9

2)3x3(d)3x3(

3

13/7

1

3

3/7

1

2/1===

4.

====

4

0 x325

dxH

3

13210)x325(

3

2)x325(d)x325(

3

11

0

2/1

4

0

2/1 =

==

5. ++++++++

====

2

1

dx1x1x

1G

3

123dx)1x1x(

2

1dx

)1x()1x(

1x1x2

1

2

1

=

+=

+

+=

(Nhn c t v mu vi bt lin hp ca mu s)

xut C)cax()bax()cb(a

1dx

caxbax

1G 331 ++++

++++++++++++

====

++++++++++++

==== vi cb;0a

6. ====1

0

dxx1xP =+=+=1

0

1

0

1

05

4dxx1)x1(dx1)1x(dxx1)11x(

7.

====

1

0

x1 xdxeQ2

= 1ee)x1(de2

1 1

0

x1

1

0

2x1 22

==

xut15

264dxx1xQ

1

0

231

====++++==== HD a x vo trong vi phn v thm bt (x2 + 1 - 1).

V d 4. Tnh:

1. 0dx)x2sin3x3cos2(I0

1 ====++++====

;41

xdxcosxsinI2/

0

32 ========

v 1exdxsineI2/

0

xcos3 ========

2. 2lnxdxtanJ4/

0

1 ========

; 2lnxdxcotJ2/

6/

2 ========

v 2ln3

2dx

xcos31

xsinJ

4/

0

3 ====++++

====

(a sinx, cosx vo trong vi phn)

3. 1cos1dxx

)xsin(lnK

e

11

========

; 2cos1dx

x

)xcos(lnK

2e

12

========

v 2dx

xln1x

1K

3e

13

====

++++

====

{a 1/x vo trong vi phn c d(lnx)}

4. ++++====3ln

1

x

x

1 dxe2

eH

e2

5lne2ln

3ln

1

x

+=+=

++++

====

2ln

0

x

x

2 dxe1

e1H =+=+

+=

2ln

0

2ln

0

x

x2ln

0

x

xx

3ln22ln3dxe1

e2dxdx

e1

e2e1

++++====2ln

0

x3 5e

dxH

7

12ln

5

15eln

5

1x

5

1

5e

dxe

5

1dx

5

1

5e

dx)e5e(

5

12ln

0

x

2ln

0

x

x2ln

0

2ln

0

x

xx

=

+=

+=

+

+=

++++

====

1

0xx

x

4 ee

dxeH

+=+=

+

=

1

0

21

0

x2

x2

x2

2

1eln

2

11eln

2

1

1e

dxe

+) Bin i nh vic xt du cc biu thc trong gi tr tuyt i tnh ====b

a

dx)m,x(fI

- Xt du hm s f(x,m) trong on [a; b] v chia [ ] ]b;c[...]c;c[]c;a[b;a n211 = trn mi on hm s f(x,m)gi mt du

- Tnh +++=b

c

c

c

c

a n

2

1

1

dx)m,x(f...dx)m,x(fdx)m,x(fI

V d 5. Tnh:

1. ++++====2

0

2 dx3x2xI Ta xt pt: 3x1x032xx 2 ===+ . Bng xt du f(x)


3/21



Suy ra 4dx)3x2x(dx)3x2x(dx3x2xdx3x2xI2

1

2

1

0

2

2

1

2

1

0

2=+++=+++=

2.

====

1

3

3 dxxx4J tnh tng t ta c 16dxxx4dxxx4dxxx4J1

0

3

0

2

3

2

3

3=++=

3.2ln

14dx42K

3

0

x++++========

4. ====2

0

1 dxx2cos1H 22dxxsin2dxxsin2dxxsin22

0

2

0

=+==

====

0

2 dxx2sin1H

{Vit (1 sin2x) v bnh phng ca mt biu thc ri khai cn}

22dxxcosxsindxxcosxsindxxcosxsindxxcosxsin

2/3

4/3

4/3

4/

4/

00

=+++++=+=

II) Phng php i bin s

A - Phng php i bin s dng 1:

Gi s cn tnh tch phn =b

a

dx)x(fI ta thc hin cc bc sau:

- Bc 1. t x = u(t)- Bc 2. Ly vi phn dx = u(t)dt v biu th f(x)dx theo t v dt. Chng hn f(x)dx = g(t)dt- Bc 3. i cn khi x = a th u(t) = a ng vi t = ; khi x = b th u(t) = b ng vi t =

- Bc 4. Bin i =

dt)t(gI (tch phn ny d tnh hn th php i bin mi c ngha)

Cch t i bin dng 1.

Cch t 1. Nu hm s cha 2x1 th t ]2/;2/[t;tsinx ==== hoc t ];0[t;tcosx ====

V d 1. Tnh:

1.

====

1

2/2

2

2

dxx

x1A ta t ]2/;2/[t;tsinx = dx = cost.dt; i cn khi x = 2 /2 th t = 4/ ; khi x

= 1 th t = 2/ . Khi 4

4dt.

tsin

tsin1dt.

tsin

tcosdt.tcos

tsin

tsin1A

2/

4/

2

22/

4/

2

22/

4/

2

2

=

==

=

2.

====

1

02

2

dxx4

xB ta vit

=

1

02

2

dx)2/x(12

xB .

t ];0[t;tcos)2/x( = tdtsin2dxtcos2x ==

i cn suy ra ( )2

3

3

dtt2cos12tdtcos4)tdtsin2(

tcos12

)tcos2(B

2/

3/

2/

3/

2

3/

2/2

2

=+==

=

3. ====1

0

22 dxx34xC Trc ht ta vit

=

1

0

2

2 dx2

x.31x2C .

t ]2/;2/[t;tsinx2

3= a tch phn v dng:

12

1

27

32dt

2

t4cos1

33

4tdtcostsin

33

16C

3/

0

3/

0

22+=

==

Ch :


4/21



- Nu hm s cha 0a,xa 2 >>>> th ta vit2

2

a

x1axa

= v t

=

=

];0[t;tcosa

x

]2/;2/[t;tsina

x

- Nu hm s cha 0b,a,bxa 2 >>>> th ta vit

2

2 xa

b1abxa

= v t

=

=

];0[t;tcosxa

b

]2/;2/[t;tsinxab

V d 2. Tnh:

1.

====

2

3/22

dx1xx

1E {Vit tch phn v dng 2X1 }

ta vit( )

=

2

3/2

22dx

x/11x

1E v t [ ]2/;2/t;tsin

x

1= suy ra

12dtE

3/

4/

==

2.

====

3/22

3/2

3

2

dxx

4x3G {Vit tch phn v dng 2X1 }

ta vit( )

=

3/22

3/2

3

2

dxx

x3/21.x.3G v t [ ]2/;2/t;tsin

x3

2= suy ra tch phn c dng

16

)336(3tdtcos

2

33G

3/

4/

2 +==

{Nu tch phn c dng bax 2 th vit v dng 2X1 }

Cch t 2. Nu tch phn c cha 2x1 ++++ hoc (((( ))))2x1 ++++ th ta t (((( ))))2/;2/t;ttanx ==== hoc( );0t;tcotx =

V d 3. Tnh:

1. ++++====3

3/1

2dx

x1

1M ta t ( )2/;2/t;ttanx = suy ra

6dtM

3/

6/

==

2. ++++

====

3

122

dxx1.x

1N ta t ( )2/;2/t;ttanx = suy ra

3

3218dt

.tsin

tcosN

3/

4/

2

==

3. ++++====a

0

2220a;dx

)xa(

1P

ta vit

+

=

a

0

2

24

dx

)a

x

(1a

1P v t ;ttan

a

x=

3

4/

0

2

3a4

2tdtcos

a

1P

+==

4. ++++++++====1

0

2dx

1xx

1Q

ta vit

++

=

1

0

2dx

)2

1x(

3

21

1

3

4Q v t ( )2/;2/t;ttan

2

1x

3

2=

+

9

3dt

2

3

3

4Q

1

0

==

Ch : Nu gp tch phn cha 2bxa ++++ hoc 2bxa ++++ th ta vit:

+=+

2

2x

a

b1axba hoc

2

2 x

a

b1abxa

+=+ v ta t ( )2/;2/t;ttanx

a

b=


5/21



Cch t 3. Nu tch phn c chaxaxa

++++

hoc

xaxa

++++th ta t ta t ]2/;0[t;t2cosax ==== v lu vn dng

=+

=

tcos2t2cos1

tsin2t2cos1

2

2

V d 4. Tnh:1.

>>>>

++++====

0

a

0a;dxxaxa

I ta t ]2/;0[t;t2cosax = suy ra +

=

4/

2/

dt)t2sina2(t2cos1

t2cos1I

4

4a

=

2. ++++

====

2/2

0

dxx1

x1J ta t ]2/;0[t;t2cosx = suy ra

+=

8/

4/

dt)t2sin2(t2cos1

t2cos1J

=

4

224tdtcos4J

4/

8/

2 +==

{c th t tx1x1====

++++suy rra tch phn J v dng tch phn ca hm s hu t}

B - Phng php i bin s dng 2:

Gi s cn tnh tch phn ====b

a

dx)x(fI ta thc hin cc bc sau:

- Bc 1. t t = v(x)- Bc 2. Ly vi phn dx = u(t)dt v biu th f(x)dx theo t v dt. Chng hn f(x)dx = g(t)dt- Bc 3. i cn khi x = a th u(t) = a ng vi t = ; khi x = b th u(t) = b ng vi t =

- Bc 4. Bin i =

dt)t(gI (tch phn ny d tnh hn th php i bin mi c ngha)

Cch t i bin dng 2.Cch t 1. Nu hm s cha n mu th t t = mu s.V d 1. Tnh:

1. ====2/

0

2dx

xcos4

x2sinI

ta c th t t = 4 - cos2x suy ra3

4ln

t

dtI

4

3

==

2. ++++====4/

0

22dx

xcos2xsinx2sinJ

t xcos1xcos2xsint 222 +=+= suy ra ==2

2/34

3ln

t

dtJ

{c th h bc bin i tip mu s v cos2x sau a sin2x vo trong vi phn}

xut: ++++====2/

0

22221dx

xcosbxsina

xcosxsinJ

vi 0ba 22 >+

3. ++++====2ln

0

xdx

5e

1K ta t 5et x += 5tex = dtdxex = sau lm xut hin trong tch phn biu

thc dxex 7

12ln

5

1

t

5tln

5

1

)5t(t

dt

)5e(e

dxeK

7

6

7

6

2ln

0

xx

x

=

=

=+

=

{C th bin i trc tip7

12ln

5

1dx

5e

e

5

1dx

5e

5e

5

1dx

5e

e5e

5

1K

2ln

0

x

x2ln

0

x

x2ln

0

x

xx

=+

+

+=

+

+= }

4. ++++++++

====

2/

0

2dx

)4x2cosxsin2(

xcosx2sinH

ta t 4x2cosxsin2t += 21

2dt

t

1

2

1H

7

3

2==

{I khi khng t c MS}

5. ++++====2/

0

2

3

dxxcos1

xcosxsinG

ch rng tch m 3 = 2 +1 t

xcos1t 2+= 1txcos2 = dtxdxcosxsin2 = khi :2

2ln1)tlnt(

2

1dt

t

)1t(

2

1G

2

1

2

1

=

=

=


6/21



6. ++=4/

0

dx2xcosxsin

x2cosM

ta t 2xcosxsint ++= dx)xsinx(cosdt = lu cos2x = (cosx+sinx)(cosx-sinx)

( ) 322

ln12tln2tt

dt)2t(

dx2xcosxsin

)xsinx)(cosxsinx(cos

M

22

3

22

3

4/

0

++==

=

++

+=

+

+

7. ++

=

4/

0

3dx

)2xcosx(sin

x2cosN

t 2xcosxsint ++= suy ra

)21(2

1

9

2

3

1

9

1

22

1

)22(

1

t

1

t

1

t

dt)2t(N

2

22

3

22

3

23+

=+

+

+

=

=

=

+ +

xut: +=4/

0

1 dx2xcosxsinx2cos

M

v +

=

4/

0

31dx

)2xcosx(sin

x2cosN

8.

Cch t 2. Nu hm s cha cn thc n )x( th t n )x(t ==== sau lu tha 2 v v ly vi phn 2 v.

V d 1. Tnh:

1. ++++++++

====

1

0

dx1x32

3x4I ta t 1x3t += ( )1t

3

1x

2= tdt

3

2dx = khi a tch phn v dng:

( )3

4ln

3

4

27

2dt

t2

6

9

2dt3t8t4

9

2dt

t2

t13t4

9

2I

2

1

2

1

2

2

1

3

=+

+=+

=

2. +

=

7

03 2

3

dxx1

xJ ta t 3 2x1t += 1tx 32 = dtt3xdx2 2=

20

141dt)tt(

2

3J

2

1

4==

3. +

=

2

1

2dx

x1x

1K ta t 2x1t += 1tx 22 = tdtxdx =

5

2

5

2

2 1t

1tln

2

1

t)1t(

tdtJ

+

=

=

4. +

=

2

13

dxx1x

1H ta t 3x1t += 1tx 23 = tdt2dxx3 2 = nhn c t v mu s vi x2 ta c:

2

12ln

3

2

1t

1tln

3

1

1t

dt

3

2

x1x

xdxH

3

2

3

2

2

2

132

+=

+

=

=

+

=

5. +

+=

3

02

35

dx1x

x2xG ta t 2x1t += 1tx 22 = tdtxdx = nhm x2.x.(x2 +2) ta c:

5

26t

5

t

t

tdt)1t)(1t(dx

1x

x.x)2x(G

2

1

52

1

223

02

22

=

=

+=

+

+=

6.

+++

=

6

13

dx1x91x9

1M ta t 6 1x9t += ( )1t9

1x

6= dtt

3

2dx

5= lu tha bc hai v bc ba

ta c:

+=

++=

+=

+= 3

2ln

6

11

3

2dt)

1t

11tt(

3

2

1t

dtt

3

2

tt

dtt

3

2M

2

1

2

2

1

32

1

23

5

V d 2. Tnh:

1. [H.2005.A] +

+=

2/

0

dxxcos31

xsinx2sinP

ta t xcos31t += )1t(3

1xcos 2 = tdt

3

2xdxsin = nhm

nhn t sinx ta c: +

+=

2/

0xcos31

xdxsin)1xcos2(P

( )27

34t

3

t2

9

2dx1t2

9

22

1

32

1

2=

+=+=


7/21



2. dx.xsin31

x2sinx3cosQ

2

0

+

+=

ta t xsin31t += )1t(3

1xsin

2= tdt

3

2xdxcos = p dng cng thc

nhn i v nhn 3 ta vit: dx.xsin31

xcosxsin2xcos3xcos4Q

2

0

3

+

+=

xdxcos.xsin31

xsin23xsin442

0

2

+

+=

Vy +=2

1

24 dt)1t14t4(27

2Q

405

206tt

3

14t

5

4

27

22

1

35=

+=

3. [H.2006.A] +

=

2

022

dxxsin4xcos

x2sinR

ta t xsin31t 2+= )1t(3

1xsin 22 =

tdt3

2xdx2sin = . khi :

3

2t

3

2

t

tdt

3

2R

2

1

2

1

===

4.V d 3. Tnh:

1. +=

e

1

dxx

xln31xlnP

Ta t xln31t += )1t(3

1xln

2= tdt

3

2

x

dx= khi : ( )

135

116dxtt

9

2P

2

1

4==

2. +

=

e

1

dxxln21x

xln23Q

Ta t xln21t += )1t(2

1xln

2= tdt

x

dx= . Khi :

3

1139

3

tt4dt)t4(

t

tdt)1t(3Q

3

1

32

1

2

2

1

2

=

==

=

3. +

=

2ln2

2lnx 1e

dxR . Ta t 1et x += suy ra tdt2dxex =

+

+

=

=

5

3

213

13.

15

15ln

1t

dt2R

4. +

=

3

03 xe1

dxS . Ta t 3 xet = suy ra

1e

e2ln3

)1t(t

dx3S

e

1+

=+

=

5. +

=

5ln

0

x

xx

3e

dx1eeX

Cch t 3. Nu hm s cha cc i lng xsin , xcos v2

xtan th ta t

2

xtant = khi

2t1t2xsin+

= , 2

2

t1t1xcos

+

=

V d 4. Tnh:

1. dx.5xcos3xsin5

1Q

2/

0

++=

Ta t2

xtant =

2t1

dt2dx

+= v

5

8ln

3

1

4t

1tln

3

1dt

4t5t

1Q

1

0

1

0

2=

+

+=

++=

2. dx.2xcos

2

xtan

L

3/

0

+=

ta t2

xtant =

2t1

dt2dx

+= v

9

10ln3tln

3t

tdt2L

3/1

0

2

3/1

0

2=+=

+=


8/21



3. ++=4

0

dx1x2sinx2cos

x2cosV

ta t xtant = 2t1

dtdx

+= v

++

+=

+

+=

1

0

2

1

0

2

1

0

2 )t1(2

tdt

)t1(2

dt

)t1(2

dt)t1(V

1

0

2

1 1tln4

1V ++= ta tnh

8)t1(2

dtV

ytant1

0

21

==

+= suy ra 8

2ln2dx

1x2sinx2cos

x2cosV

4

0

+=

++=

4. ++

+=

4

0

22

2

dx1xsinx2sinxcos

xtan1N

ta vit +++

=

4

0

2

dx1x2sinx2cos

xtan1N

v t

xtant = 2t1

dtdx

+= suy ra

4

2ln231tlnt

2

t

2

1dt

1t

t1

2

1N

1

0

21

0

2+

=

+++=

+

+=

5. [H.2008.B] +++

=

4

0

dx)xcosxsin1(2x2sin

4xsin

F

ta vit( )

+++

=

4

0

dx)xcosxsin1(2xcosxsin2

xcosxsin

2

1F

da

vo mi quan h gia xcosxsin + v xcosxsin ta t xcosxsint += dx)xsinx(cosdt = v

2

1txcosxsin

2

= khi

+=

+=

++=

++

=

2

1

2

1

2

2

1

222

1

22

1

1t

1

2

1

1t2t

dt

2

1

)t1(21t

dt

2

1F

Cch t 4. Da vo c im hai cn ca tch phn.

Nu tch phn c dng

=

a

a

dx)x(fI th ta c th vit +=

a

0

0

a

dx)x(fdx)x(fI t t = - x bin i

=

0

a

1 dx)x(fI

Nu tch phn c dng =

0

dx)x(fI th ta c th t t = - x

Nu tch phn c dng =2

0

dx)x(fI th ta c th t t = 2 - x

Nu tch phn c dng =

2/

0

dx)x(fI

th ta c th t t = 2

- x

Nu tch phn c dng =b

a

dx)x(fI th ta c th t t = (a + b) - x

V d 4. Tnh:

1.

=

1

1

2008 xdxsinxI ta vit +=

0

1

2008 xdxsinxI BAxdxsinx

1

0

2008+= . Ta t t = -x th A = - B. vy I = 0.

2. +=

0

2dx

xcos1

xsinxJ ta t xt = khi ++=

0

2

0

2dt

tcos1

tsintdt

tcos1

tsinJ ta i bin tip:

2dttcos1

tsin

J

2utantcos

0

21

=

====+= vJdttcos1

tsint

J

xt

0

22

=

===+=

.Vy 4JJ2J

22 ==

Cch t 4. Nu tch phn c cha 0a;cbxax 2 >++ th ta c th t cbxaxxat 2 ++= sau tnh x theo tv tnh dx theo t v dt.{Php th le}V d 5. Tnh:

1. +

=

1

02 1xx

dxI ta t 1xxxt 2 +=

1t2

t1x

2

+

= 3ln

1t2

dt2I

2

1

=

=

2. +

=

1

02 1x2x9

dxJ ta t 1x2x9x3t 2 +=

)1t3(2

1tx

2

=

2

126ln

3

1

1t3

dtJ

22

1

=

=

III)Phng php tch phn tng phn


9/21



-Gi s cn tnh tch phn =b

a

dx)x(fI . Khi ta thc hin cc bc tnh:

Bc 1. Vit tch phn di dng: ==b

a

b

a

dx)x(h).x(gdx)x(fI

Bc 2. t

=

=

dx).x(hdv

)x(gu

=

=

dx).x(hvdx)x('gdu

Bc 3. p dng cng thc: hay =b

a

b

a

b

a

du.vv.udv.u

Cc cch t tch phn tng phn:

+Cch t 1. Nu tch phn c dng =b

a

dx.axsin).x(PI th ta s t

=

=

dx.axsindv

)x(Pu

=

=

a

axcosv

dx)x('Pdu

Nu tch phn c dng

b

a

dx.axcos).x(P th ta t

=

=

dx.axcosdv

)x(Pu

=

=

aaxsinv

dx)x('Pdu

Nu tch phn c dng b

a

ax dx.e).x(P th ta t

=

=

dx.edv

)x(Pu

ax

=

=

a

ev

dx)x('Pdu

ax

V d 5. Tnh:

1. =

0

dx.x2sin).1x3(I ta t

=

=

dx.x2sindv

1x3u

=

=

2

x2cosv

dx3du

2

3dx.x2cos

2

3

2

x2cos)1x3(I

00

=+=

2. +=2/

0

2 dx.xcos).1x(J

ta t

=

+=

dx.xcosdv

1xu 2

=

=

xsinv

xdx2du

1

2

0

2/

0

2 J24

4dx.xsin..x2xsin)1x(J

+=+=

ta tnh =2/

0

1 dx.xsin.xJ

bng cch t

=

=

dx.xsindv

xusau suy ra 1xdxcosxcosxJ

2/

0

2/

01=+=

.Vy4

42

4

4J

22

=+

=

3. +=1

0

x32 dx.e).1xx(L ta t

=

+=

dx.edv

1xxu

x3

2

1

31

0

x3

1

0

x32 L31

31edx.e).1x2(

31e)1xx(

31L

=+=

Tnh tip =1

0

x3

1 dx.e).1x2(L t

=

=

dx.edv

1x2u

x3

9

4e4L

3

1

= suy ra

27

5e5L

3

=

4. =

0

2 dx.)xsinx(M ta vit =

==

00

2

00

2 xdx2cosx2

1

4

xdx.

2

x2cos1xdx.xsinxM

xt 0dx.x2cosxMxu

xdx2cosdv0

1 ====

=

=

. vy ta c4

M2

=

5.

=

4/

0

2

dx.xsinM

ta i bin xt = a

=

2/

0

tdtsint2M

bng cch t

=

=

dt.tsindv

t2u 2M =


10/21



+Cch t 2. Nu tch phn c dng =b

a

ax dx.bxsineI th ta t

=

=

dx.edv

bxsinu

ax

=

=

a

ev

bxdxcosbdu

ax

Nu tch phn c dng

=

b

a

ax dx.bxcoseI th ta t

=

=

dx.edv

bxcosu

ax

=

=

aev

bxdxsinbdu

ax

V d 6. Tnh:

1. =2/

0

x2dx.x3sin.eI

ta t

=

=

dxedv

x3sinu

x2

=

=

2

ev

xdx3cos3du

x2

10

x2

2/

0

x2

I2

3

2

edx.x3cose

2

3

2

ex3sinI ==

(*). Ta xt =

0

x2

1 dx.x3coseI v t

=

=

dxedv

x3cosu

x2

I2

3

2

1dx.x3sine

2

3

2

ex3cosI

0

x2

2/

0

x2

1 +=+=

thay vo (*) ta c:

+= I

2

3

2

1

2

3

2

eI

13

3e2I

+=

2. =

0

2x dx.)xsin.e(F ta vit =

=

0

x2

0

x2

0

x2 dx.x2cose2

1dx.e

2

1dx.

2

x2cos1eF

Ta xt2

1edx.e

2

1F

2

0

x2

1

==

. Sau hai ln tch phn tng phn ta tnh c4

1edx.x2cose

2

1F

2

0

x2

2

==

.

Vy ta c:8

1edx.)xsin.e(F

2

0

2x ==

+Cch t 3. Nu tch phn c dng [ ]=b

a

dx)x(Q.)x(PlnI th ta t[ ]

=

=

dx).x(Qdv

)x(Plnu

=

=

dx)x(Qv

dx)x(P

)x('Pdu

V d 7. Tnh:

1. =5

2

dx)1xln(.xI ta t[ ]

=

=

dx.xdv

1xlnu

=

=

2

xv

dx1x

1du

2 =

5

2

25

2

2

dx2x2

x)1xln(

2

xI

4

272ln48 +=

2. ++=3

0

2 dx)x1xln(J ta t

=

++=

dxdv

x1xlnu 2

=

+

=

xv

dxx1

1du

2 1)23ln(3J +=

3. =e

1

2xdxln.xK ta t

=

=

xdxdvxlnu 2 suy ra =

e

1

e

1

22xdxln.xxln

2xK . Xt =

e

1

1 xdxln.xK v t

=

=

xdxdv

xlnuth

4

1eK

4

1eK

22

1

=

+= .

4. =2

1

5dx

x

xlnH ta t

=

=

dxxdv

xlnu

5suy ra

256

2ln415dxx

4

1xln

x4

1H

e

1

5

2

14

=+=

.

5. =3/

6/

2dx

xcos

)xln(sinG

t

=

=

dxxcos

1dv

)xln(sinu

2

=

=

xtanv

xdxcotdu

=

3/

6/

3/

6/ dx)xln(sinxtanI

6

2ln343ln33 =


11/21



6. dx)x(lnoscFe

1

=

t

=

=

dxdv

)xcos(lnu

=

=

xv

dxx

)xsin(lndu

+=

e

1

e

1dx)xsin(ln)xcos(lnxI (*). Ta xt

=

e

1

1 dx)xsin(lnF t

=

=

dxdv

)xsin(lnu

=

=

xv

dxx

)xcos(lndu

Fdx)xcos(ln)xsin(lnxFe

1

e

11==

thay

vo (*) ta c:2

1eFF1eF

+==

.

III)Phng php tm h s bt nh

A- Khi gp tch phn: ==== dx)x(Q)x(P

I vi P(x), Q(x) l cc a thc cax.

Bc 1: Nu bc ca P(x) bc ca Q(x) th ta ly P(x)chia cho Q(x) c thngA(x) v dR(x),tc l P(x) = Q(x).A(x) + R(x), vi bc R(x) < bc Q(x).

Suy ra :)x(Q)x(R

)x(A)x(Q)x(P

++++==== ++++==== dx)x(Q)x(R

dx)x(Adx)x(Q)x(P

Bc 2:

Ta i tnh : ====

dx)x(Q

)x(R

I , vi bc R(x) < bc Q(x).C th xy ra cc kh nngsau :

+Kh nng 1: Vi cbxax)x(Q 2 ++++++++==== ,( 0a ) th bc R(x) < 2 R(x) = M.x+N vcbxax

Nx.M)x(Q)x(R

2++++++++

++++====

TH1 : Q(x) c 2 nghimx1, x2, tc l: Q(x) = a(x x1)(x x2).

Chn hng sA, B sao cho:2121 xx

Bxx

A)xx)(xx(a

Nx.M)x(Q)x(R

++++

====

++++====

TH2 : Q(x) c nghim kpx0, tc l:2

0 )xx(a)x(Q ==== .

Chn hng sA, B sao cho:2

002

0 )xx(

Bxx

A

)xx(a

Nx.M)x(Q)x(R

++++

====

++++====

TH3 : Q(x) v nghim. Chn hng sA, B sao cho: B)x('Q.A)x(R ++++==== v)x(Q

B)x(Q

)x('Q.A)x(Q)x(R

++++====

+Kh nng 2: Vi dcxbxax)x(Q 23 ++++++++++++==== ,( 0a ) th bc R(x) < 3TH1: Q(x) c 3 nghim .x,x,x 321 tc l: )xx)(xx)(xx(a)x(Q 321 ====

Chn hng sA, B, Csao cho:321321 xx

Cxx

Bxx

A)xx)(xx)(xx(a

)x(R)x(Q)x(R

++++

++++

====

====

TH2: Q(x) c 1 n0 n 1x , 1 n0 kp 0x , tc l:2

01 )xx)(xx(a)x(Q ====

Chn hng sA, B, Csao cho:2

0012

01 )xx(

Cxx

Bxx

A

)xx)(xx(a

)x(R)x(Q)x(R

++++

++++

====

====

TH3: Q(x) c mt nghim0

x (bi 3), tc l: 30

)xx(a)x(Q ====

Chn hng sA, B, Csao cho:3

02

003

0 )xx(

C

)xx(

Bxx

A

)xx(a

)x(R)x(Q)x(R

++++

++++

====

====

TH4:Q(x) c ng mt nghim n 1x , tc l: )xax)(xx()x(Q2

1 ++++++++==== (trong 0a42


12/21



1.

++++

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

0

1

2

2

dx2x3x

1xxI ta vit

+

+=

0

1

2dx

2x3x

1x41I v vit

2x

B

1x

A

2x3x

1x42

+

=

+

Sau chn c

A = -3; B = 7. Khi : ( ) 3ln72ln102xln71xln3xI0

1=+=

.

2. ++=

1

0

2 dx1xx

xJ ta vit x = A(x2 + x + 1) + B suy ra A = 1/2; B = - 1/2. Vy 21 JJJ += vi

3ln2

1

1xx

)1xx(d

2

1J

1

0

2

2

1 =++

++= v

+

+

=++

=

1

0

2

1

0

22

12

1x

3

2

dx

3

4.

2

1

1xx

dx

2

1J

Ta t utan2

1x

3

2=

+ suy ra

9

3du

3

32J

3/

6/

2

== .

3. +

=

3

1

3dx

x3x

1K ta vit

3x

cBx

x

A

x3x

123+

++=

+sau chn c A = 1/3, B = - 1/3, C = 0. V th vit c

3ln61dx

)3x(3xdx

x31K

3

1

2

3

1

=+

= {V a c x vo trong vi phn}.4.

B Khi gp tch phn ++

=

dxxcosdxsinc

xcosbxsinaI (c, d 0) th ta vit TS = A.(MS) + B.(MS) tc l chn A, B sao cho:

dcosx)'B(csinxdcosx)A(csinxbcosxasinx +++=+ hoc t2

xtant =

2t1

t2xsin

+=

2

2

t1

t1xcos

+

=

V d 1. Tnh:

1. ++

=

2/

0

dxxcosxsin

xcos5xsin3I

ta vit sinx)-B(cosxcosx)A(sinxcosx53sinx ++=+ suy ra A = 4; B = 1.

Khi : ( )

2xcosxsinlnx4xcosxsin

)xcosx(sinddx4I

2/

0

2/

0

2/

0

=++=+

++=

2. ++

=

2/

0

3dx

)xcosx(sin

xcosxsin3J

ta vit sinx)-B(cosxcosx)A(sinxcosx3sinx ++=+ suy ra A = 2; B = -1.

Khi : 2)xcosx(sin2

1)

4xcot(

)xcosx(sin

)xcosx(sinddx

)xcosx(sin

2I

2/

0

2

2/

0

3

2/

0

2=

+++=

+

+

+=

C Khi gp tch phn ++++

=

dxnxcosdxsinc

mxcosbxsinaI (c, d 0) th ta vit TS = A.(MS) + B.(MS) + C. Chn A, B,C sao cho:

Cn)'dcosxB(csinxn)dcosxA(csinxmbcosxasinx ++++++=++ hoc c th t

2

xtant =

2t1

t2xsin

+=

2

2

t1

t1xcos

+

=

V d 1. Tnh:

1. +++

=

2/

0

dx5xcos3xsin4

7xcosxsin7I

ta vit C3sinx)-B(4cosx)5cosx3A(4sinx7cosx7sinx ++++=+

Khi A = 1; B = -1; C = 2 v +++++++

=

2/

0

2/

0

2/

0

dx5xcos3xsin4

2dx

5xcos3xsin4

)5xcos3xsin4(ddxI


13/21



Xt ++=2/

0

1 dx5xcos3xsin4

2I

t2

xtant =

2t1

t2xsin

+=

2

2

t1

t1xcos

+

= suy ra

3

1dt

)2t(

12I

1

0

21=

+= . Vy ( ) 8

9ln

3

1

2I5xcos3xsin4lnxI 1

2/

0+=+++=

V)Phng php dng tch phn lin ktV d 1. Tnh:

1. +=2

0xcosxsin

xdxsinI

ta xt thm tch phn th hai: +=2

0xcosxsin

xdxcosJ

Khi :2

JI

=+ (*).

Mt khc 0xcosxsin

)xcosx(sind

xcosxsin

dx)xcosx(sinJI

2

0

2

0

=+

+=

+

=

(**). GiI h (*) v (**) suy ra I = J =4

.

2. dxxcosxsin

xsinI

2

0

nn

n

n +

=

ta xt dxxcosxsin

xcosJ

2

0

nn

n

n +

=

. Khi :2

JI nn

=+ (*)

Mt khc nu t x =2

- t th n

2

0nn

n2

0nn

n

n Jdxxcosxsin

xcosdt

tcostsin

tcosI =

+

=

+

=

(**). T (*), (**) ta c4

In

=

3. dxxcosxsin

xsinI

2

0

nn

n

n +

=

tng t xt dxxcosxsin

xcosJ

2

0

nn

n

n +

=

v suy ra4

JI nn

==

4. dxxcos3xsin

xsinE

6

0

2

+

=

v dxxcos3xsin

xcosF

6

0

2

+

=

ta c 3ln4

1dx

xcos3xsin

1FE

6

0

=

+

=+

(*)

Li c 31dx)xcos3x(sinF3E6

0

==

(**). GiI h (*), (**) ta c:4

313ln

16

1E

= v

4

313ln

16

3F

+= . M rng tnh ==

+

= EFdxxcos3xsin

x2cosE

6

0

2

313ln

8

1 +

xut dxxcos3xsin

x2cosL

6

0

=

Cc bi ton tng t.Cc bi ton tng t.Cc bi ton tng t.Cc bi ton tng t.A Phng php bin i trc tip

1. [HNNI.98.A] ++++++++

====

1

0

x2

2x

e1

dx)e1(M

+ Bnh phng v phn tch thnh 2 phn s n gin.+ Bit i bin.

Gii: ++++++++++++++++

====

1

0

x2

x1

0

x2

x2

e1

dxe2

e1

dxe1M ta tnh ++++====

1

0

x2

x

1e1

dxe2M t (((( ))))2/;2/t,ttane x ==== khi vi tan =e v

+

=

4/

221tcos)ttan1(

tdttan2M =

2

e1ln

ttan1

1ln2tcosln2tdttan2

2

4/

24/

4/

+=

+==

2. [HTCKT.97] +2

0

3

xcos1xdxsin3

+

Gii:

2. [HTCKT.97] +2

0

3

xcos1xdxsin3

+

Gii:


14/21



2. [HTCKT.97] +2

0

3

xcos1xdxsin3

+

Gii:

2. [HTCKT.97] +2

0

3

xcos1xdxsin3

+

Gii:

1. [HNNI.98.A] +

+1

0

x2

2x

e1

dx)e1(

2. [HTCKT.97] +2

0

3

xcos1xdxsin3

3. [HBK.98]

+

2

0

44 dx)xcosx(sinx2cos

4. [HDL.98] ++

2

11x1x

dx

5. +

6

0

dx)

6xcos(.xcos

1

6. +

2e

e

dxx

)xln(lnxln

7. [HM.00] +

3

6dx

)6

xsin(xsin

1

8. 3

0

4 xdx2sinxcos

9. [HNN.01] +4

0

66dx

xcosxsin

x4sin

10. [HNNI.01] 2

4

4

6

dxxsinxcos

11.

3

4

4xdxtg

12. [CGTVT.01]

+

3

2

2 dx.x3x

13. [CSPBN.00] +3

0

2 dx4x4x

14.

0

dxxsinxcos

15. +3

6

22 dx2xgcotxtg


15/21



16. +3

0

23 dxxx2x

17.

1

1

dxx4 p: )35(2

18.

1

1

dxxx 3

22

19. ( )

+

5

3

dx2x2x 8

20. +

+3

0

2

2

dx.2xx

1xx

21. +

0

dxx2cos22 4

22.

0dxx2sin1 22

23. +2/

0

dxxsin1

24

24. 1

0

Ra;dxax


16/21



B Phng php i bin

1. [CBN.01] +1

0

32

3

dx)x1(

x HD t fsf

2. [PVB.01] 1

0

23 dx.x1x

3. +

3ln

0

x 2e

dx

4. [CXD.01] +2

0

2dx

xcos1x2sin

5. [HKTQD.97] 1

0

635 dx)x1(x xut: 1

0

72 dx)x1(x

6. [HQG.97.B] +

1

0 x1

dx

7. [H.2004.A] +

2

1 1x1

xdx

``8. [H.2003.A] +

32

52 4xx

dx

9. [HSPHN.00.B]

0

222 dxxax

10. [HBK.00] +

2ln

0x

x2

1e

dxe

11. ++23

14 2x58xdx

12.( ) +

2

0

2dx

xsin2

x2sin

13. 4

0

3 xcos

dx

14. ++

6

2

dx1x4x2

1

15. +3

0

25 dxx1x

16. +

2e

e

dxx

)xln(lnxln

17. +

4

2

dxx

1x

18. ++

+++1

022

23

dx1x)x1(

x101x3x10


17/21



19.

e

12

dxxln1x

1

20.

+

e

1

dxxln1x

xln

21. +++

4

03

dx1x21x2

1

23. +

4

72 9x.x

dx

24. ++

7

2

dx.2xx

1

25.( ) +

1

0

22x31

dx

26. [GTVT.00]

++++

2

22 dxxsin4xcosx

27. [HAN.97] ++++

0

2 xcos1

xdxsinx

28. [HLN.00] ++++++++2

0

dxxcosxsin2

1

29. [HH.00] ++++4

0

dxtgx1

1

30. [HVH.01] ++++

4

0 dxx2cosx2sin

xcosxsin

31. [HVBCVT.98] ++++2

0

2

3

xcos1

xdxcosxsin

32.

++++====

1

122

dx)1x(

1I

33. [HTN.01] ++++

++++

++++2)51(

124

2

dx1xx

1x

34. [HTCKT.00] ++++++++1

0

24dx

1xx

x

35. [HVKTQS.98]

++++++++++++

1

12 )x1x1(

dx

36. [PVBo.01] 1

0

23 dx.x1x

37. [H.2004.B] ++++

e

1

dxx

xlnxln31

38. [H.2005.A] ++++

++++2

0

dxxcos31

xsinx2sin


18/21



39. [H.2006.A] ++++

2

022

dxxsin4xcos

x2sin

40. [H.2005.B] ++++2

0

dxxcos1

xcosx2sin

41. . [H.2005.B] ++++

5ln

3lnxx 3e2e

dx

42. [H.2003.A] ++++

32

52 4xx

dx

43. [H.2004.A] ++++

2

1

dx1x1

x

44. [H.2008.A] 6/

0

4

dxx2cos

xtan

45. [ thi th H] ++++4/

0

66dx

xcosxsinx4sin

46. [ thi th H]

++++

++++

e

1

2 dx.xln.xxln1.x

1

47. [ thi th H] ++++

4

0

1x2 dxe

48. [ thi th H] ++++2

0

3dx

)xsin1(2

x2sin

HD: t xsin1t ++++==== 81

t2

dt)1t(22

1

3====

49.

====

8

4

2

dxx 16xI

50. ++++++++

====

4

2

dxx

1x1xJ

51. ++++++++

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

1

022

23

dx1x)x1(

x101x3x10K

52.

====

2ln

2lnx2

x

dxe1

eH

53.

++++====

3ln

0x2 1e

dxG

54. ++++++++====2

0xcos3xsin53

dxF

55. ++++====2

0 24

3

dx3xcos3xcos

xcosD

56. ++++

====

5ln

0

x

xx

3e

dx1eeS

57. ++++

====

2 xcosxsin2

dxT


19/21



58. ++++

====

4

72 9x.x

dxR

59. ++++++++

====

7

2

dx.2xx

1E

60. ++++++++

++++====

4

0

dx.1x21

1x2W

61. ++++====2/

0xcos2

dxQ

t2

xtant ==== th

9

3

t)3(

dtQ

utan3t1

022

====================

++++

====

62. ++++++++

====

2

02 1x6x3

dxM

C Phng php tch phn tng phn

1. [HC.97] ++++

1

0

x22 dxe)x1(

2. [HTCKT.98] 4

0

2 dx)1xcos2(x

3. ++++2

0

23 dx)1xln(x v 10

0

2 xdxlgx

4. [PVBo.98] e

1

2 dx)xlnx(

5. [HVNH.98]

0

2 xdxcosxsinx

6. [HC.00] ++++

2

12x

dx)1xln(

7. [HTL.01] ++++4

0

dx)tgx1ln(

8. 2

0

2xdxxtg

9. [HYHN.01] 3

2

2 dx.1x

10. [ thi th] 2

1

2 dx)xx3ln(x

11. [H.2007.D] e

1

22 xdxlnx

12. [H.2006.D] 1

0

x2 dxe)2x(

13.

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

0

1

3x2 dx)1xe(xI

14. ++++====

2e

e

dxx

)xln(lnxlnJ


20/21



15. ====

0

2 dx)xsinx(K

16. ++++====1

0

2dx

)1x2(sin

xH

17. ====4

0

3dx

xcos

xsin.xG

18. ====2ln

0

x5 dxe.xF2

19. ++++

++++====

4

1

dxxx

)1xln(D

20.

++++

++++

====

e

1

2 dxxlnxln1x

xlnS

21. ====2

2 4/

2 dx.xcosA

22. (((( ))))====

0

2x dxxcos.eP

23. dx.xsin.xU

2

0====

24. dx.xcos.xsin.xY 2

0====

25. ++++++++

====

3/

0

xdxexcosxsin

xsin1T

t ====++++

++++==== dv;xcosxsin xsin1u ..Xt ++++====

3/

0

x1 dxexcos1 xsinT

v t tptp suy ra

3

e

xcos1

xsineT

33/

0

x

====++++

====

26. ++++====1

0

2 dxx1R s:2

)21ln(2 ++++++++

27. ++++====1

0

2 dx)1xln(xE S:2

12ln

28.

++++====

2/

0

dx)xcos1ln(xcosW

s: 12

29. ++++====e

e/12

dx)1x(

xlnQ s:

1e

e2++++

30. ++++++++

====

2/

3/

dxxcos1xsinx

M

Vit M = M1+ M2. Vi

23

lndxxcos1

xsinM

2/

3/

1 ====++++

====

& ++++====3/

6/

2 dxxcos1

xM

.

t

++++====

====

dxxcos1

1dv

xu

====

====

2

xcot2v

dxdu 4ln

3

)323(M 2

====

Vy

8

3ln

3

)323(M ++++

====


21/21



31. ++++++++

====

2/

6/

dxx2cos1x2sinx

M

B Phng php h s bt nh

1. [HYHN.00] ++++2

12

2

12x7xdxx

2. [HNNI.00] ++++2

1

3dx

)1x(x

1v ++++

1

0

3dx

1x

3

3. [HXD.98] ++++++++

4

0

dxxsin3xcos4

xsin2xcos

4. [HTM.00] ++++2

0

3dx

)xcosx(sin

xsin4

5. [HTN.98]

++++++++

1

0 n nn x1)x1(

dx

6. ++++++++

====

2

0

2

4

dx4x

1xxI

7. ++++++++++++

====

1

03

2

dx1x

7x3x2J

8. ++++++++++++

====

1

0

2dx

2x3x

5x4K

9.(((( ))))

====

1

0

22 4x3x

dxL

10. ++++++++

====

2

0

2

4

dx.4x

1xxZ

D Phng php tch phn lin kt

1. ++++

2

0

dxxcosxsin

xcos

2. [ thi th] ++++

++++

====

32

32

x

1x

3

4

dxex

1xI

2

HD: )()1

( xFx

F = . Suy ra

0)32

1()32( =+

+= FFI

[HTN.00] CMR: Zn , ta c

0dx)nxxsin(sin2

0

====++++

[HVKTQS.01] ++++

b

0

22

2

dx)xa(

xa, 0b,a >>>>

[HLN.01] ++++++++

1

0

2

x2

dx)1x(

e)1x(

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