calculus1 6-all

เฉลยแบบฝึกหดั Calculus 1 เจษฎา หอ่ไพศาล (พี่แบงค์) www.clipvidva.com

1

แบบฝึกหัดที ่6.1 ข้อ 1 จงอนิทเิกรต

1.1) ( sin ) cos

= (xcosx cos )

x x dx xd x

xdx

= xcosx + sinx + C

2 2

2 2

2

2

11.2) ( 2 ) 2

ln 2

1 = ( 2 2 )

ln 2

1 = ( 2 2 2 )

ln 2

1 2 = ( 2 2 )

ln 2 ln 2

x x

x x

x x

x x

x dx x d

x dx

x xdx

x xd

2

2

1

1 12

2 3

1 2 = ( 2 { 2 2 })

ln 2 ln 2

1 2 2 = ( 2 { 2 )

ln 2 ln 2 ln 2

1 2 2 = 2

ln 2 (ln 2) (ln 2)

x x x

xx x

x xx

x x dx

x x C

xx C

2 2

3

2 = { ln 2 ln 2 2}

ln 2

x

x x x C

2 2

2

2 2

3 2 2

2

2 2

2

11.3) ( ) ( 1)

2

1 = ( 1)

2

1 = {( 1) ( 1)}

2

1 = {( 1)

2

x x

x

x x

x x e dx x e dx

x de

x e e d x

x

2 2

2 2

2 2

2

2

1

2

}

1 = {( 1) }

2

1 1 = ( 1)

2 2

x x

x x

x x

e e dx

x e e C

x e e C

221 =

2

xx e C

3

2

1.4) cos ec c osec cot

(cosec cot cot cosec )

(cosec cot (cosec cot ) )

x dx x d x

x x x d x

x x x x dx

2

3

3

1

cosec cot cosec (cosec 1)

cosec cot cosec cosec

2 cos ec = cosec cot ln | cosec cot |

x x x x dx

x x x dx x dx

x dx x x x x C

3 1 1 cosec = cosec cot ln | cosec cot |

2 2x dx x x x x C


2

2 211.5) cos( ) sin( )x ax dx x d ax

a

2 2

2

2

2

2

1

1= ( sin( ) sin( ) )

1 ( sin( ) 2 sin( ) )

1 2 ( sin( ) cos( ))

1 2 ( sin( ) { cos( ) cos( ) })

1 2 1 ( sin( ) { cos( ) sin( ) })

x ax ax dxa

x ax x ax dxa

x ax x d axa a

x ax x ax ax dxa a

x ax x ax ax Ca a a

2

2 3

1 2 2 sin( ) cos( ) sin( )x ax x ax ax C

a a a

1

1 1

1

11.6) ln( ) ln( )

1

1 ( ln( ) ln( ))

1

1 ( ln( ) )

1

n n

n n

n n

x ax dx ax dxn

x ax x d axn

x ax x dxn

1

1

1

1 ( ln( ) )

1 1

nn x

x ax Cn n

1 1 {ln( ) }

1 1

nxax C

n n

11.7) sin( ) cos( )

1 ( cos( ) cos(bx) )

1 ( cos( ) cos(bx) )

ax ax

ax ax

ax ax

e bx dx e d bxb

e bx deb

e bx a e dxb

2

1 ( cos( ) sin( ))

1 ( cos( ) { sin( ) sin( ) })

1 cos( ) { sin( ) sin( )

ax ax

ax ax ax

ax ax ax

ae bx e d bx

b b

ae bx e bx bx de

b b

ae bx e bx a e bx dx

b b

2

12 2

2 2

12 2

}

1 cos( ) sin( ) sin( )

1 sin( ) = cos( ) sin( )

sin( ) =

ax ax ax

ax ax ax

ax

a ae bx e bx e bx dx C

b b b

a b ae bx dx e bx e bx C

b b b

e bx dx

2 2 2 2 cos( ) sin( )ax axb a

e bx e bx Ca b a b

2 2= { sin( ) cos( )}

axea bx b bx C

a b

11.8) cos( ) sin( )ax axe bx dx e bx

b

1 ( sin( ) sin(bx) )

1= ( sin( ) sin(bx) )

ax ax

ax ax

e bx deb

e bx a e dxb


3

2

2

12 2

1 ( sin( ) cos( ))

1 ( sin( ) { cos( ) cos( ) })

1 sin( ) { cos( ) cos( ) }

1 sin( ) cos( ) cos( )

ax ax

ax ax ax

ax ax ax

ax ax ax

ae bx e d bx

b b

ae bx e bx bx de

b b

ae bx e bx a e bx dx

b b

a ae bx e bx e bx dx C

b b b

2 2

12 2

2 2 2 2

1 cos( ) = sin( ) cos( )

sin( ) = sin( ) cos( )

ax ax ax

ax ax ax

a b ae bx dx e bx e bx C

b b b

b ae bx dx e bx e bx C

a b a b

2 2

= { cos( ) sin( )}axe

a bx b bx Ca b

1.9) arcsin( ) xarcsin(ax) arcsin( )ax dx x d ax

2

2

2

2

2

xarcsin(ax) 1 ( )

1 ( ) xarcsin(ax)

2 1 ( )

1 {1 ( ) } xarcsin(ax)

2 1 ( )

axdx

ax

d ax

a ax

d ax

a ax

21 xarcsin(ax) 1 ( )ax C

a

1.10) ln(2 3) ln(2 3) ln(2 3)x dx x x x d x

2 ln(2 3)

2 3

3 ln(2 3) (1 )

2 3

3 ln(2 3)

2 3

3 (2 3) ln(2 3)

2 2 3

xx x dx

x

x x dxx

x x x dxx

d xx x x

x

3 ln(2 3) ln | 2 3 |

2x x x x C

arccot1.11) 2 arccot

2 arccot 2 arccot

xdx x d x

x

x x x d x

2 arccot 2 arccotx x x d x 1

2 arccot 1

x x dxx

2 arccot ln | 1 | x x x C

2

2 1ln1.12) ln

xdx x dx

x

22ln 1

ln x

d xx x


4

2

2

21

2

2

2

2

ln ln 2

ln 2 ln

ln ln 1 2( ln )

ln ln 1 2 2

ln ln 2 2

x xdx

x x

xx dx

x

x xd x

x x x

x xdx

x x x

x xC

x x x

21 (ln 2ln 2)x x C

x

2 21.13) tan (sec 1) x xdx x x dx

2

2

2

sec

tan2

( tan tan )2

x dx x x dx

xx d x

xx x x dx

2

tan ln | sec | 2

xx x x C

2 3

3 3

33

2

11.14) arctan arctan

3

1 ( arctan arctan )

3

1 ( arctan )

3 1

x xdx x dx

x x x d x

xx x dx

x

3

2

2 23

2

1 ( arctan { } )

3 1

1 1 ( arctan )

3 2 2 1

xx x x dx

x

x dxx x

x

3 2 21 (2 arctan ln(1 ))

6x x x x C

1.15) cos(ln ) cos(ln ) cos(ln )

cos(ln ) sin(ln )

cos(ln ) { sin(ln ) sin(ln )}

x dx x x xd x

x x x dx

x x x x xd x

1 cos(ln ) sin(ln ) cos(ln )x x x x x dx C

1 2 cos(ln ) cos(ln ) sin(ln )x dx x x x x C

cos(ln ) {cos(ln ) sin(ln )}2

xx dx x x C

2 2 21.16) ln( 4) ln( 4) ln( 4)x dx x x xd x

22

2 ln( 4) 2

4

x

x x dxx

2

2

4 ln( 4) 2 (1 )

4

x x dxx


5

2

2

2

2

2

2

1 ln( 4) 2 8

4

1 ln( 4) 2 2

12

1 ln( 4) 2 4 { }

21

2

x x dx dxx

x x dx dxx

xx x x d

x

2 ln( 4) 2 4arctan( )2

xx x x C

1.19) sin ln(cos ) ln(cos ) cosx x dx x d x

(cos ln(cos ) cos ln(cos ))

cos ln(cos ) sin

x x xd x

x x xdx

cos {1 ln(cos )}x x C

2 21.20) ( 3 5)cos ( 3 5) sinx x xdx x x d x

2 2

2

2

2

2

2

( 3 5)sin sin ( 3 5)

( 3 5)sin (2 3)sin

( 3 5)sin (2 3) cos

( 3 5)sin {(2 3)cos cos (2 3)}

( 3 5)sin (2 3)cos 2 cos

( 3

x x x xd x x

x x x x x dx

x x x x d x

x x x x x x d x

x x x x x x dx

x x

5)sin (2 3)cos 2sinx x x x C

2 ( 3 3)sin (2 3)cosx x x x x C

ln1.21) 2 ln

2 lnx 2 ln

1 2 lnx 2

xdx xd x

x

x xd x

x dxx

2 lnx 4x x C

2

3 3

2 22 2

ln 1 ln1.22)

2( 1) ( 1)

x x xdx dx

x x

1

2 2

2 2

ln ( 1)

ln 1 ln

1 1

x d x

xd x

x x

2 2

ln 1

1 1

xdx

x x x

2

ln arcsec( )

1

xx C

x


6

2 21.23) sin sin x xe x dx x de 2 2

2

2

sin sin

sin 2 sin cos

sin sin 2

x x

x x

x x

e x e d x

e x e x x dx

e x e x dx

2

2

2

2

sin sin 2

sin ( sin 2 sin 2 )

sin sin 2 2 cos 2

sin sin 2

x x

x x x

x x x

x x

e x x de

e x e x e d x

e x e x e x dx

e x e x

2

2 2

2 2

1

2 (1 2sin )

sin sin 2 2 4 sin

5 sin = sin sin 2 2

x

x x x x

x x x x

e x dx

e x e x e dx e xdx

e x dx e x e x e C

2 21 sin = (sin sin 2 2)

5

x xe x dx e x x C

1.24) sin 2 sin

2 cos

2( cos cos )

x dx x xd x

xd x

x x xd x

2sin 2 cosx x x C

1

21.25) (1 )

(1 )

1 ( ( ))

1 1

1 ( ( ) )

1 1

1

xx

xx

xx x

x x

xedx xe d x

x

xed xe

x x

xexe e dx

x x

xe e

x

1 x( 1x ) )

1

1

(1 )1

xx

xx

x

dx

xee dx

x

xee C

x

xe C

x

1

xeC

x

22 1

2

22

1.26) ( 2)( 2)

1 ( )

2 2

xx

xx

x edx x e d x

x

x ed x e

x x

22

2

1 ( 2 )

2 2

2 2

x

x x

x x

x ex e xe dx

x x

x e xe

x x( 2x ) dx

2

2

xxx e

xe dxx


7

2

2

xxx e

xdex

2

2

( )2

2

x

x x

xx x

x exe e dx

x

x exe e C

x

2

( 1 )2

x xe x C

x

2 21.27) 2 (2 )x uxe dx u u e du

3

3 3

3 2

3 2

3 2

3 2 2

3

2 (2 )

[2{(2 ) (2 )}]

2(2 ) 2 (2 3 )

2(2 ) 4 6

2(2 ) 4 6

2(2 ) 4 6( )

2(2 ) 4

u

u u

u u

u u u

u u u

u u u u

u

u u de

u u e e d u u

u u e e u du

u u e e du e u du

u u e e u de

u u e e e u e du

u u e

2

3 2

3 2

3 2

3 2

2

6 12

2(2 ) 4 6 12

2(2 ) 4 6 12( )

2(2 ) 4 6 12 12

{4 2 4 6 12 12}

{4 2 2(2 ) 2 4 6(2 )

u u u

u u u u

u u u u u

u u u u u

u

x

e e u ue du

u u e e e u ude

u u e e e u ue e du

u u e e e u ue e C

e u u u u C

e x x x x

2

12 2 12}

{4 2 4 2 2 2 4 12 6 12 2 12}x

x C

e x x x x x x C

2 2 { 2 6 2 3 10}xe x x x x C

3 3 211.28) ln ln

2x xdx xdx

2 3 2 3

2 3 2

1 ( ln ln )

2

1 ( ln 3 ln )

2

x x x d x

x x x x dx

2 3 2 2

2 3 2 2 2 2

2 3 2 2

2 3 2 2

2 3 2 2 2

2 3 2 2 2

1 3 ( ln ln )

2 2

1 3 ( ln { ln ln })

2 2

1 3 ( ln { ln 2 ln })

2 2

1 3 3 ln ln ln

2 4 2

1 3 3 ln ln ln

2 4 4

1 3 3 ln ln ( ln

2 4 4

x x x dx

x x x x x d x

x x x x x x dx

x x x x x x dx

x x x x x dx

x x x x x

2

2 3 2 2 2

2 3 2 2 2 2

ln )

1 3 3 3 ln ln ln

2 4 4 4

1 3 3 3 ln ln ln

2 4 4 8

x x d x

x x x x x x x dx

x x x x x x x C

2

2

1

2

2

u x

du dxu

x u


8

2 3 23 4 ( ln 2ln 2ln 1)

8 3x x x x C

2 2

1 11.29) (ln(ln ) ) (ln(ln )

(ln ) (ln )x dx x dx dx

x x

2

2

1

2

2

1 ln(ln ) ln(ln )

(ln )

1 1 ln(ln )

ln (ln )

1 ln(ln ) ( ln )

(ln ) ln

1 ln(ln )

(ln )

x x xd x dxx

x x dx dxx x

xx x dx xd x C

x x

x x dxx

2

1

ln (ln )

xdx

x x C

ln(ln )ln

xx x C

x

ข้อ 2 จงหาค่าของอนิทกิรัลต่อไปนี ้

1

1 13

0 0

3

1

113 3

0 0

113 2

0 0

2.1) = 2

= 2

= 2( )

= 2( 3 )

x u

eu

uu u

u

uu u

u

xe dx u e du

u de

u e e du

u e u e du

1

13 2

0 0

11 13 2 2

0 0 0

= 2( 3 )

= 2( 3{ })

euu u

u

u uu u u

u u

u e u de

u e u e e du

1

11 13 2

0 0 0

1 13 2

0 0 0

1 13 2

0 0

= 2 6 12

= 2 6 12

= 2 6 12(

u uu u u

u u

eu uu u u

u u

u uu u

u u

u e u e ue du

u e u e ude

u e u e ue

11

0 0)

uu u

ue du

1 1 1 13 2

0 0 0 0

1 1 1 1

1

= 2 6 12 12

= 2 6 12 12 12

= 32 12

u u u uu u u u

u u u uu e u e ue e

e e e e

e

ให้ 2

1

2

u x

du dxu

x u

2.2) cot cosec = cosec

= ( cosec cos ec )

= cosec ln | cosec cot |

x x xdx xd x

x x x dx

x x x x C

3 3 3

4 4 4

4 4 4

cot cosec = cosec ln | cosec cot |x x

x xx x xdx x x x x

3 3cos cot

3 4 4= (3cosec cos )+ln4 4 4

cos cot4 4

2 2 1= ln

2 2 1

ec

ec

ec


9

2= ln(3 2 2)

2

3 312.3) cos 2 = sin 2

2x xdx x d x

3 3

3 2

3 2

3 2 2

3 2

3 2

1= ( sin 2 sin 2 )

2

1= ( sin 2 3 sin 2 )

2

1 3= sin 2 cos 2

2 4

1 3= sin 2 ( cos 2 cos 2 )

2 4

1 3 3= sin 2 cos 2 cos 2

2 4 2

1 3 3= sin 2 cos 2 sin 2

2 4 4

1=

x x xdx

x x x xdx

x x x d x

x x x x xdx

x x x x x xdx

x x x x xd x

3 2

3 2

3 3sin 2 cos 2 ( sin 2 sin 2 )

2 4 4

1 3 3 3= sin 2 cos 2 sin 2 cos 2

2 4 4 8

x x x x x x xdx

x x x x x x x C

23 3 22

00

3 2 2

0

1 3 3 3 cos 2 = sin 2 cos 2 sin 2 cos 2

2 4 4 8

3 3 3 3 = sin cos sin cos

16 16 8 8 8

x

x

x

x

x xdx x x x x x x x

23 3 =

4 16

3 312.4) sin 4 = sin 4

3

x xe xdx x de

3 31= ( sin 4 sin 4 )

3

x xe x e d x

3 3

3 3

3 3 3

3 3 3

1= ( sin 4 4 cos 4 )

3

1 4= sin 4 cos 4

3 9

1 4= sin 4 ( cos 4 cos 4 )

3 9

1 4 16= sin 4 cos 4 sin 4

3 9 9

x x

x x

x x x

x x x

e x e xdx

e x x de

e x e x e d x

e x e x e xdx C

3 3 3

3 3 3

3 3 34 4

00

25 1 4 sin 4 = sin 4 cos 4

9 3 9

3 4 sin 4 = sin 4 cos 4

25 25

1 sin 4 = 3 sin 4 4 cos 4

25

=

x x x

x x x

xx x x

x

e xdx e x e x

e xdx e x e x

e xdx e x e x

3 3

4 41

3 sin 2 4 cos 425

e e

3

44

= ( 1)25

e


10

1

2

2

2

ln2.5) = ln

ln 1 = ( ln )

ln 1 =

ln 1 =

ln ln 1 =

e

ex

xdx xdx

x

xd x

x x

xdx

x x

xC

x x

x xdx

x x x

2 3 =

2

x e

e

e e

1 3 = ( 2)

2e

e

2

2

2 2

2

arcsin2.6) = arcsin 1

1

= ( 1 arcsin 1 arcsin )

= 1 arcsin

x xdx x d x

x

x x x d x

x x dx

2

11

222

20 0

= 1 arcsin

arcsin = 1 arcsin

1

1 1 1 = 1 arcsin

4 2 2

x

x

x x x C

x xdx x x x

x

1 3 =

2 12

2 2 22.7) ln = ln lnxdx x x x d x

2= ln 2 ln x x x dx

2

2

2

= ln 2( ln ln )

= ln 2 ln 2

= ln 2 ln 2

x x x x xd x

x x x x dx

x x x x x C

2 2

11

2

ln = ln 2 ln 2

= ln 2 ln 2( 1)

e x e

xxdx x x x x x

e e e e e

= 2e

2

2

2

2.8) arccos = arccos arccos

= arccos1

1 1 = arccos

2 1

1 1 = arccos

2 1

x dx x x xd x

xx x dx

x

x x dxx

x xx

2

2

2

(1 )

= arccos 1

d x

x x x C


11

11

222

0 0

arccos = arccos 1

1 1 3 = arccos 1

2 2 2

x

x

x dx x x x

3 = 1

6 2

2

2 2

2

2

12.9) x arcsec = arcsec

2

1 1 = arcsec arcsec

2 2

1 1 = arcsec

2 2 1

=

x dx x dx

x x x d x

xx x dx

x

2 2

2

2 2

1 1 1 arcsec

2 4 1

1 1 = arcsec 1

2 2

x x dxx

x x x C

112 2

2 2

1 xarcsecx = arcsec 1

2

1 = arcsec( 1) 4arcsec( 2) 3

2

x

x

dx x x x

3 5 =

2 6

3. จงหาพืน้ทีร่ะหว่างเส้นโค้ง ln( )y x 2 กบัแกน x บนช่วง [-3, -1]

ln( ) = 2 ln

= 2( ln | | ln )

x dx xdx

x x xd x

2

= 2( ln | | 1 )x x dx

= 2 ln | | 2x x x C

1 1 12

3 33 ln( ) = 2 ( ln | |) 2

= 2( ln | 1 | 3ln | 3 | 2( 1 3)

x dx x x x

= 6ln(3) 4


12

2 3

3

25 5

25

25 5

xx x

x x

x


2

1 11.1)

16 1 (4 1)(4 1)

1 1 1 ( )

2 4 1 4 1

1 1 1 ( ) (4 )

8 4 1 4 1

dx dxx x x

dxx x

d xx x

1 4 1 ln | |

8 4 1

xC

x

21.2)

6 ( 3)( 2)

x x

dx dxx x x x

พิจารณา ( 3)( 2)

x

x x

A B

= ( 3)( 2) 3 2

A( 2) B( 3)

x

x x x x

x x x

แทน 2 ; x จะได ้ 2B

5

แทน 3 ; x จะได ้ 3A

5

ดงันั้น 3 2 =

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

x

x x x x

2

1 3 2 ( )

6 5 3 2

x

dx dxx x x x

3 2= ln 3 ln 2

5 5 x x C

3

2 2

2

2

5 25 51.3) ( )

25 25

25 5 { }

25 ( 5)( 5)

5 25

25 ( 5)( 5)

x xdx x dx

x x

xx dx

x x x

xxdx dx dx

x x x

2

2

2

25 1 1 1 { }

2 25 2 5 5

25 25 1 1 ln | 5 | ln | 5 | ln | 5 | ln | 5 |

2 2 2 2 2

dxxdx dx

x x x

xx x x x C

2

12ln | 5 | 13ln | 5 | 2

xx x C

3 2 2

3

4 2 1 2 11.4) (1 )

4 (2 1)(2 1)

x x x xdx dx

x x x x x

พิจารณา 22 1

(2 1)(2 1)

x x

x x x

3 3 2

3

2

1 4 4 2 1

4

2 1

x x x x

x x

x x


13

2

2 2

2 1 A B C =

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

2 1 A(4 -1) B (2 1) C (2 -1)

x x

x x x x x x

x x x x x x x

แทน 0 ; x จะได ้ A 1

แทน 1 ;

2x จะได ้ B 2

แทน 1 ;

2x จะได ้ C 1

ดงันั้น 22 1 1 2 1

= (2 1)(2 1) 2 1 2 1

x x

x x x x x x

3 2

3

4 2 1 1 2 1 (1 )

4 2 1 2 1

x xdx dx

x x x x x

1 = ln | | ln | 2 1 | ln | 2 1 |

2x x x x C

3 2

2 2

2 5 2 3 4 51.5) (2 1 )

2 2 2 2

x x x xdx x dx

x x x x

พิจารณา

2

4 5

2 2

x

x x

2

4 5 A B =

( 1) 3 1 3 1 3

4 5 A( 1 3) B( 1 3)

x

x x x

x x x

แทน 1 3 ; x จะได ้ 1 4 3A

2 3

แทน 1 3 ; x จะได ้ 1 4 3B

2 3

ดงันั้น 2

4 5 1 4 3 1 4 3 =

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

x

x x x

3 2

2

2

2 5 2 3 1 4 3 1 4 3 (2 1 )

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

1 4 3 1 4 3 = ln | 1 3 | ln | 1 3 |

2 3 2 3

x x xdx x dx

x x x x

x x x x C

2 2 1 1 3 = 2 ln | 2 2 | ln | |

2 3 1 3

xx x x x C

x

3 2

1 11.6)

2 ( 2)( 1)

x xdx dx

x x x x x x

พิจารณา 1

( 2)( 1)

x

x x x

1 A B C

= ( 2)( 1) 2 1

1 A( 2)( 1) B( )( 1) ( 2)

x

x x x x x x

x x x x x Cx x


2


6

แทน 1 ; x จะได ้ 2

C3

2 3 2

3 2

2

2

2 1 2 2 2 5 2 3

2 4 4

6 3

2 2

4 5

xx x x x x

x x x

x x

x x

x


14

ดงันั้น 1 1 1 2 =

( 2)( 1) 2 6( 2) 3( 1)

x

x x x x x x

3 2

1 1 1 2 { }

2 2 6( 2) 3( 1)

xdx dx

x x x x x x

1 1 2 = ln | | ln | 2 | ln | 1 |

2 6 3x x x C

2

31.7)

( 1)

xdx

x


3( 1)

x

x

2

3 2 3

2 2

A B C =

( 1) 1 ( 1) ( 1)

A( 1) B( 1)

x

x x x x

x x x C

แทน 1 ; x จะได ้ C 1

แทน 0 ;x จะได ้ A B 1

...(1)

แทน 1 ; x จะได ้ 2A B 0

...(2)

แกส้มการ (1) และ (2)

จะได ้ A 1, B 2


3 2 3

1 2 1 =

( 1) 1 ( 1) ( 1)

x

x x x x

2

3 2 3

1 2 1 { }

( 1) 1 ( 1) ( 1)

xdx dx

x x x x

2

3 4 = ln | 1 |

2( 1)

xx C

x

4 2

3 2 2

8 21.8) { 2 4 }

2 ( 2)

x xdx x dx

x x x x


2

( 2)

( 2)

x

x x

2

2 2

2 2

2 A B C =

( 2) 2

2 A( )( 2) B( 2)

x

x x x x x

x x x x Cx

แทน 0 ; x จะได ้ B 1

แทน 2 ; x จะได ้ 1C

2


A2


2 2

2 1 1 1 =

( 2) 2 2( 2)

x

x x x x x

4

3 2 2

8 2 4 2 { 2 }

2 2

xdx x dx

x x x x x

224

= 2 2ln | 2 | 2

xx x x C

x

3 2 4

4 3

3

3 2

2

2 2 8

2

2

2 4

4 8

xx x x

x x

x

x x

x


15

2 2

20 11 20 111.9)

(3 2)( 4 5) (3 2){( 2) 1}

x xdx dx

x x x x x


20 11

(3 2){( 2) 1}

x

x x

2 2

2

20 11 A B C =

(3 2){( 2) 1} 3 2 4 5

20 11 A( 4 5) (B C)(3 2)

x x

x x x x x

x x x x x

แทน 2 ;

3x จะได ้ A 3



ดงันั้น

2 2

20 11 3 2 =

(3 2){( 2) 1} 3 2 4 5

x x

x x x x x

2 2

20 11 3 2 ( )

(3 2)( 4 5) 3 2 4 5

x xdx dx

x x x x x x

2

2

2 2

1 1 2 4= ( ) (3 2)

3 2 2 4 5

1 1 ( 4 5) 4= ( ) (3 2)

3 2 2 4 5 ( 2) 1

xd x dx

x x x

d x xd x dx dx

x x x x

21= ln | 4 5 | ln | 3 2 | 4arctan( 2)

2x x x x C

2

2

10 131.10)

(2 1)( 2)

x xdx

x x


2

10 13

(2 1)( 2)

x x

x x

2

2 2

2 2

10 13 A B C =

(2 1)( 2) 2 1 2

10 13 A( 2) (B C)(2 1)

x x x

x x x x

x x x x x

แทน 1 ;





2 2

10 13 4 3 8 =

(2 1)( 2) 2 1 2

x x x

x x x x

2

2 2

2 2

10 13 4 3 8 ( )

(2 1)( 2) 2 1 2

4 3 8 =

2 1 2 2

x x xdx dx

x x x x

xdx dx dx

x x x

23 = 2ln|2x 1| ln | 2 | 4 2 arctan( )

2 2

xx C

2

2

11 131.11)

( 3)( 2)( 3)

x x

dxx x x


2

11 13

( 3)( 2)( 3)

x x

x x x


16

2

2 2

2 2 2

11 13 A B C D =

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

11 13 A( 2)( 3) B( 3)( 3) (C D)( 3)( 2)

x x x

x x x x x x

x x x x x x x x x



แทน 0 ; x จะได ้ D 4



2 2

11 13 1 2 4 =

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

x x x

x x x x x x

2

2 2

2 2

11 13 1 2 4 ( )

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

1 2 4 =

3 2 3 3

x x xdx dx

x x x x x x

xdx dx dx dx

x x x x

21 4 = ln|x+3| 2 ln | 2 | ln | 3 | arctan

2 3 3

xx x C

2 2

35 471.12)

(3 5) ( 3 6)

xdx

x x x


35 47

(3 5) ( 3 6)

x

x x x

2 2 2 2

2 2 2

35 47 A B C D =

(3 5) ( 3 6) 3 5 (3 5) 3 6

35 47 A(3 5)( 3 6) B( 3 6) (C D)(3 5)

x x

x x x x x x x

x x x x x x x x

แทน 5 ;

3x จะได ้ B 3



แทน x = –1; จะได ้ A = 3

ดงันั้น 2 2 2 2

35 47 3 3 1 =

(3 5) ( 3 6) 3 5 (3 5) 3 6

x x

x x x x x x x

2 2 2 2

35 47 3 3 1 ( )

(3 5) ( 3 6) 3 5 (3 5) 3 6

x x

dx dxx x x x x x x

2 2 2

3

3 3 1 12= 3 5 (3 5) 3 6 2 3 6

x

dx dx dx dxx x x x x x

21 1

= ln|3x+5| ln | 3 6 |3 5 2

x xx

1 2 3

arctan{ ( )}215 15

x C

2 2

21.13)

( 1)( 1)

xdx

x x


2

( 1)( 1)

x

x x


17

2 2 2 2

2 2 2

2 A B C D =

( 1)( 1) 1 ( 1) 1

2 A( 1)( 1) B( 1) (C D)( 1)

x x

x x x x x

x x x x x x


หา A จาก 2

2 2 2

1 1

2 2( 1)A 0

1 ( 1)x x

d x x

dx x x




2 1 1 =

( 1)( 1) ( 1) 1

x

x x x x

2 2 2 2

2 1 1 { }

( 1)( 1) ( 1) 1

1 = arctan

1

xdx dx

x x x x

x Cx

2 2

2 11.14)

(4 9)( 4)

xdx

x x


2 1

(4 9)( 4)

x

x x

2 2 2 2

2 2

2 1 A B C D =

(4 9)( 4) 4 9 4

2 1 (A B)( 4) (C D)(4 9)

x x x

x x x x

x x x x x

แทน 0 ; x จะได ้ 4B 9D 1

...(3)

แทน 1 ; x จะได ้ 5A 5B 13C 13D 3

...(4)

แทน 1 ; x จะได ้ 5A 5B 13C 13D 1

...(5)

แทน 2 ; x จะได ้ 16A 8B 50C 25D 5

...(6)

แกส้มการ (3), (4), (5) และ (6)

จะได ้ 8 4 2 1A , B ,C ,D

7 7 7 7


2 1 8 4 2 1 =

(4 9)( 4) 7(4 9) 7( 4)

x x x

x x x x

2 2 2 2

2 2 2 2

2 1 8 4 2 1 { }

(4 9)( 4) 7(4 9) 7( 4)

8 4 1 2 1 1 =

7 4 9 7 4 9 7 4 7 4

x x xdx dx

x x x x

x xdx dx dx dx

x x x x

2 2

2 22 2

(4 9) ( 4)1 4 1 1 1 1 =

27 4 9 63 7 4 28( ) 1 ( ) 1

3 2

d x d xdx dx

x xx x

2 21 2 2 1 1 = ln(4 9) arctan( ) ln( 4) arctan( )

7 21 3 7 14 2

x xx x C

2 2

3 3 3

2 1 2 11.15)

27 1 27 1 27 1

x x x xdx dx dx

x x x


2 1

27 1

x

x


18

2 2

2

2 1 A B C =

(3 1)(9 3 1) 3 1 9 3 1

2 1 A(9 3 1) (B C)(3 1)

x x

x x x x x x

x x x x x

แทน 1 ;

3x จะได ้ 1

A9

แทน 0 ; x จะได ้ 8C

9


B3

ดงันั้น 2 2

2 1 1 3 8 =

(3 1)(9 3 1) 9(3 1) 9(9 3 1)

x x

x x x x x x

2 2

3 3 2

3

3 22

2 1 1 3 8 { }

27 1 27 1 9(3 1) 9(9 3 1)

13

1 1 (3 1) 1 5 12 = 1 33 27 1 27 3 1 9 9 3 1 6

(3 )2 4

x x x xdx dx dx

x x x x x

xdx d x

dx dxx x x x

x

3 2

12

3 2

1 1 1 10 1 = ln|27 1| ln | 3 1 | ln(9 3 1)

6 181 27 54 9( ) 1

3

1 1 1 5 3 6 1 = ln|27 1| ln | 3 1 | ln(9 3 1) arctan( )

81 27 54 27 3

x x x x dx Cx

xx x x x C

25 2 5 3 6 1 = ln(9 3 1) ln | 3 1 | arctan( )

162 81 27 3

xx x x C

3 2

3 2 2

2 8 6 18 461.16) = {2 }

3 9 27 ( 3)( 9)

x x xdx dx

x x x x x


2

6 18 46

( 3)( 9)

x x

x x

2

2 2

2 2

6 18 46 A B C =

( 3)( 9) 3 9

6 18 46 A( 9) (B C)( 3)

x x x

x x x x

x x x x x


9


C3


9


2 2

6 18 46 23 31 69 =

( 3)( 9) 9( 3) 9( 9)

x x x

x x x x

3

3 2 2

2 8 23 31 69 {2 }

3 9 27 9( 3) 9( 9)

x x

dx dxx x x x x

2 2

2

22

23 1 31 23 1= 2

9 3 9 9 3 9

23 1 31 ( 9) 23 1= 2

9 3 18 9 27( ) 13

xdx dx dx dx

x x x

d xdx dx dx

xx x

223 31 23= 2 ln | 3 | ln( 9) arctan( )

9 18 9 3

xx x x C

3 2 3

3 2

2

2 3 9 27 2 8

2 6 18 54

6 18 46

x x x x

x x x

x x


19

3

2 21.17)

( 2 10)

xdx

x x


2 2( 2 10)

x

x x

3

2 2 2 2 2

3 2

Ax B C D =

( 2 10) ( 2 10) 2 10

Ax B (C D)( 2 10)

x x

x x x x x x

x x x x

แทน 0 ; x จะได ้ B 10D 0

...(7)

แทน 1 ; x จะได ้ A B 9C 9D 1

...(8)

แทน 1 ; x จะได ้ A B 13C 13D 1

...(9)

แทน 2 ; x จะได ้ 2A B 20C 10D 8

...(10)

แกส้มการ (7), (8), (9) และ (10)

จะได ้ A 6, B 20,C 1, D 2


2 2 2 2 2

6x 20 2 =

( 2 10) ( 2 10) 2 10

x x

x x x x x x

3

2 2 2 2 2

6 20 2 { }

( 2 10) ( 2 10) 2 10

x x x

dx dxx x x x x x

2 2 2 2 2

1 1 1= 6 26

( 2 10) ( 2 10) 2 10

x x

dx dx dxx x x x x x

2

13

2 10

dxx x

2 2

2 2 2 2 2

( 2 10) 1 1 ( 2 10)= 3 26

( 2 10) (( 1) 9) 2 2 10

d x x d x x

dxx x x x x

2

1 ( 1)

13( ) 1

3

d x

x


1

(( 1) 9)dx

x

ให้ 2

2

1 3tan θ

3sec θ θ

( 1) 9 3secθ

x

dx d

x


2 2 4

1 1 1 = sec θ θ

(( 1) 9) 27 sec θdx d

x

2

1

1 = cos θ θ

27

1 = (1 cos 2θ) θ

54

1 1 = (θ sin 2θ)

54 2

1 = (arcta

54

d

d

C

12

1 3( 1)n )

3 2 10

x xC

x x

เพราะฉะนั้น 3

2

2 2 2 2

3 13 1 3( 1) 1 (arctan ) ln( 2 10)

( 2 10) 2 10 27 3 2 10 2

x x xdx x x

x x x x x x

θ

1x

3

2

(1)

9

x


20

1

arctan( )3

xC

2

2

14 1 40 13 1 = arctan ln( 2 10)

27 3 9( 2 10) 2

x xx x C

x x

2

2 2

4 2 81.18)

( 2)

x xdx

x x


2 2

4 2 8

( 2)

x x

x x

2

2 2 2 2 2

2 2 2 2

4 2 8 A B C D E =

( 2) 2 ( 2)

4 2 8 A( 2) (B C)( )( 2) (D E)

x x x x

x x x x x

x x x x x x x x


แทน 1 ; x จะได ้ 3B 3C D E 4 ...(11)

แทน 1 ; x จะได ้ 3B 3C D E 8 ...(12) แทน 2 ; x จะได ้ 12B 6C 2D E 22 ...(13)

แทน 3 ; x จะได ้ 33B 11C 3D E 64 ...(14) แกส้มการ (11), (12) ,(13) และ (14)

จะได ้ B 2, C 0,D 0,E 2


2 2 2 2 2

4 2 8 2 2 2 =

( 2) 2 ( 2)

x x x

x x x x x

2

2 2 2 2 2

2 2 2

2

2

4 2 8 2 2 2 { }

( 2) 2 ( 2)

2 = 2 2

2 ( 2)

2 ( 2) = 2

2

x x xdx dx

x x x x x

x xdx dx dx

x x x

d xdx

x x

2 2

1

( 2)dx

x


1

( 2)dx

x

ให้ 2

2

2 tan θ

2 sec θ θ

2 2 secθ

x

dx d

x


2 2 4

1 2 1 = sec θ θ

( 2) 4 sec θdx d

x

2

1

12

2= cos θ θ

4

2= (1 cos 2θ) θ

8

2 1= (θ sin 2θ)

8 2

2 2= (arctan )

8 22

d

d

C

x xC

x

เพราะฉะนั้น 2

2

2 2 2

4 2 8 2 = 2ln|x| ln( 2) arctan

( 2) 4 2( 2)2

x x x xdx x C

x x x

θ

x2

2x

2


21

5 2

2 2 2 2

8 ( 2)1.19) = { }

( 4) ( 4)

x x xdx x dx

x x


2 2

8 ( 2)

( 4)

x x

x

2

2 2 2 2 2

2 2

8 ( 2) A B C D =

( 4) 4 ( 4)

8 ( 2) (A B)( 4) C D

x x x x

x x x

x x x x x

แทน 0 ; x จะได ้ 4B D 0 ...(15)

แทน 1 ; x จะได ้ 5A 5B C D 24 ...(16)

แทน 1 ; x จะได ้ 5A 5B C D 24 ...(17) แทน 2 ; x จะได ้ 16A 8B 2C D 96 ...(18)

แกส้มการ (11), (12) ,(13) และ (14)

จะได ้ A 8, B 0,C 16,D 0


2 2 2 2 2

8 ( 2) 8 16 =

( 4) 4 ( 4)

x x x x

x x x

5

2 2 2 2 2

2 2

2 2 2

8 16 { }

( 4) 4 ( 4)

( 4) ( 2) = 4 8

4 ( 2)

x x xdx x dx

x x x

d x d xxdx dx

x x

22

2

8 = 4ln( +4)

2 2

xx C

x

2 2

181.20)

(4 9)dx

x


9 1

98( )

4

dx

x

ให้ 2

2

3tan θ

2

3sec θ θ

2

9 3secθ

4 2

x

dx d

x


42 2

9 1 1 1 = sec θ θ

98 3 sec θ( )

4

dx d

x

2

1

2

1= cos θ θ

3

1= (1 cos 2θ) θ

6

1 1= (θ sin 2θ)

6 2

1 2 3= (arctan

96 32( )

4

d

d

C

x xC

x

2

1 2= arctan

6 3 4 9

x xC

x

4 2 5

5 3

3

8 16

8 16

8 16

xx x x

x x x

x x

θ

x

2

9

4x

3

2


22

2 2

2 2 2 2

13 69 65 13 69 651.21) =

(2 11 15) (2 5) ( 3)

x x x xdx dx

x x x x


2 2

13 69 65

(2 5) ( 3)

x x

x x

2

2 2 2 2

2 2 2 2 2

13 69 65 A B C D =

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

13 69 65 A( 3) B(2 5)(x+3) +C(2 5) D(2x+5) (x+3)

x x

x x x x x x

x x x x x

แทน 5 ;



แทน 0 ; x จะได ้ 3B 5D 109 ...(19) แทน 1 ; x จะได ้ 4B 7D 109 ...(20)


จะได ้ B 218,D 109


2 2 2 2

13 69 65 105 218 25 109 =

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

x x

x x x x x x

2

2 2 2 2

13 69 65 105 218 25 109 { }

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

x x

dx dxx x x x x x

2 2

1 1 1= 105 218 25

(2 5) 2 5 ( 3)

dx dx dxx x x

1109

3

dxx

105 2 5 25= 109ln | |

2(2 5) 3 3

xC

x x x

2 2 2 2 2 2

64 128 21.22) = 64

( 8) ( 8)

x xdx dx

x x x x

พิจารณา 2 2 2

2

( 8)

x

x x

2 2 2 2 2 2 2

2 2 2 2 2 2 2

2 A B Cx D E F =

( 8) ( 8) 8

2 A( 8) B( )( 8) +(C D)( ) (E F)( ) ( 8)

x x

x x x x x x

x x x x x x x x x


32

หา B จาก 2

2 2 2 3

00

2 3 8 8 1B

( 8) ( 8) 64xx

d x x x

dx x x

แทน 1 ; x จะได ้ 51C D 9E 9F

64 ...(21)

แทน 1 ; x จะได ้ 17C D 9E 9F

64 ...(22)

แทน 2 ; x จะได ้ 8C 4D 96E 48F 5 ...(23) แทน 2 ; x จะได ้ 8C 4D 96E 48F 0 ...(24)

แกส้มการ (21), (22) ,(23) และ (24)

จะได ้ 1 1 1 1C ,D ,E ,F

8 4 64 32

4 2 5

5 3

3

8 16

8 16

8 16

xx x x

x x x

x x


23

ดงันั้น 2 2 2 2 2 2 2

2 1 1 x 2 2 =

( 8) 32 64 8( 8) 64( 8)

x x

x x x x x x

2 2 2 2 2 2 2

2 1 1 x 2 2 { }

( 8) 32 64 8( 8) 64( 8)

x x

dx dxx x x x x x

2 2 2 2

1 1 1 1 1 1= ln | |

32 64 64 8 32 8 8 ( 8)

x x

x dx dx dxx x x x

12 2

1 1

4 ( 8)

dx Cx

2

2

1 1 1 2 1= ln | | ln( 8) arctan

32 64 128 128 16( 8)2 2

xx x

x x

12 2

1 1

4 ( 8)

dx Cx


2 2

1

( 8)dx

x

ให้ 2

2

2 2 tan θ

2 2 sec θ θ

2 2 2 secθ

x

dx d

x


2 2 4

1 2 1 = sec θ θ

( 8) 32 sec θdx d

x

22

= cos θ θ32

2= (1 cos 2θ) θ

64

d

d

1

12

2 1= (θ sin 2θ)

64 2

2 2 2= (arctan )

64 82 2

C

x xC

x

2

2 2 2 2

2 2 1 3 2 4 64 ln | | ln( 8) arctan

( 8) 2 4 82 2

x x xdx x x C

x x x x

5 2

2 3

1001.23)

( 11)

x x xdx

x


5 2

2 3 2 3 2 2 2

5 2 2 2 2

100 A B C D E F =

( 11) ( 11) ( 11) 11

100 A B (C D)( 11) (E F)( 11)

x x x x x x

x x x x

x x x x x x x x

แทน 0 ; x จะได ้ B 11D 121F 0 ...(25)

แทน 1 ; x จะได ้ A B 12C 12D 144E 144F 98 ...(26) แทน 1 ; x จะได ้ A B 12C 12D 144E 144F 100 ...(27)

แทน 2 ; x จะได ้ 2A B 30C 15D 450E 225F 164 ...(28) แทน 2 ; x จะได ้ 2A B 30C 15D 450E 225F 172 ...(29)

แทน 3 ; x จะได ้ 3A B 60C 20D 1200E 400F 48 ...(30)

แกส้มการ (25), (26) ,(27), (28), (29) และ (30)

จะได ้ A 21, B 11,C 22, D 1, E 1, F 0

θ

x2

8x

2 2


24


2 3 2 3 2 2 2

100 21 11 22 1 =

( 11) ( 11) ( 11) 11

x x x x x x

x x x x

5 2

2 3 2 3 2 2 2

100 21 11 22 1 { }

( 11) ( 11) ( 11) 11

x x x x x x

dx dxx x x x

2 3 2 2 2 3 2 2

11 1= { } 21 22

( 11) ( 11) ( 11) ( 11)

x x

dx dx dxx x x x

2 11

x

dxx

2 2

2 3 2 2 2 3 2 2

11 1 21= { } 11

( 11) ( 11) 2 ( 11) ( 11)

dx dx

dxx x x x

2

2

1

2 11

dx

x พิจารณา

2 3 2 2

11 1{ }

( 11) ( 11)dx

x x

ให้ 2

2

11 tan θ

11sec θ θ

11 11secθ

x

dx d

x

ดงันั้น 2 3 2 2 2 3 2 2

11 1 1 1{ } = 11

( 11) ( 11) ( 11) ( 11)dx dx dx

x x x x

2 2

2 6 2 4

11 sec θ 11 sec θ= θ θ

11 sec θ 11 sec θd d

2 4 2 2

11 θ 11 θ=

11 sec θ 11 sec θ

d d

4 2

2 2

11 11= cos θ θ cos θ θ

11 11d d

2

2 2

2

2 2

11 11= (1 cos 2θ) θ (1 cos 2θ) θ

4 11 2 11

11 11= (1 2cos 2θ cos 2θ) θ (1 cos 2θ) θ

4 11 2 11

d d

d d

2 2

2 2

2 2

2 2

11 11= θ (1 cos 4θ) θ

4 11 8 11

11 11= θ (cos 4θ) θ

8 11 8 11

11 11= θ sin 4θ

8 11 32 11

11 11= θ sin 2θcos2θ

8 11 16 11

d d

d d

5 22

2 3 2 3 2 2 2 2 2

100 11 11 11 21 1 11 1 ln( 11)

( 11) 6( 11) 4( 11) 4 ( 11) 11 2

x x xdx x C

x x x x x

3 2

cosθ sinθ1.24) θ =

sinθ sin θ sinθ(1 sin θ)

dd


1

sin θ(1 sin θ)

2 2

2

1 A sin θ C =

sin θ(1 sin θ) sin θ 1 sin θ

1 A(1 sin θ) (Bsinθ C)(sinθ)

B

แทน sinθ 0 ; จะได ้ A 1

แทน sinθ 1 ; จะได ้ B C 1 ...(31)

θ

x2

11

x

11


25

แทน sinθ 1 ; จะได ้ B C 1 ...(32)


จะได ้ B 1,C 0


1 1 sin θ =

sinθ(1 sin θ) sinθ 1 sin θ

3 2

cosθ 1 sinθ = { } sinθ

sin θ sin θ sinθ 1 sin θdx d

21 = ln|sin θ| ln | 1 sin θ |

2C

tanθ sinθ

1.25) θ = θ2 sin θ cosθ(2 sinθ)

d d

ให้ 2

2 2 2

2 1 2sinθ , cosθ , θ

1 1 1

z zd dz

z z z

2

2 2

2 2

2

tanθ 21θ = 1 22 sin θ 1

(2 )1 1

z

zd dzz z z

z z

2 2

2

4=

(1 )(2 2 2)

= 2(1 )(1 )( 1)

zdz

z z z

zdz

z z z z

พิจารณา 2(1 )(1 )( 1)

z

z z z z

2 2

2 2 2

2 A B C D =

(1 )(1 )( 1) 1 1 1

2 A(1 )( 1) B(1 )( 1) (C D)(1 )

z z

z z z z z z z z

z z z z z z z z z

แทน 1 ; z จะได ้ B 1

แทน 1 ; z จะได ้ 1A

3

แทน 0 ; z จะได ้ 2D

3

แทน 2 ; z จะได ้ 4C

3


2 1 1 4 2 =

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

z z

z z z z z z z z

2 2

2 1 1 4 2 { }

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

z z

dz dzz z z z z z z z

2

1 1 1 2 2 1=

3 1 1 3 1

z

dz dz dzz z z z

21 2= | 1 | | 1 | | 1 |

3 3 n z n z n z z C

21 2= | 1 tan | | 1 tan | | tan

3 2 2 3 2

x x xn n n

tan 1 | 2

x

C


26

3

3 3

1 cosθ 11.26) θ = θ

sin θ(cosθ 1) sin θ

cosθ 1 = θ θ

sin θ sin θ

d d

d d


3

1θ = cosec θ θ

sin θd d

2

2

3

= cosecθ cot θ

(cosecθ cot θ cot θ cosecθ)

cosecθ cot θ cosecθ cot θ θ

cosecθ cot θ cosecθ(cosec θ 1) θ

cosecθ cot θ cosec θ θ cos θ θ

d

d

d

d

d ec d


12 θ = cosecθcot θ cos θ θ

sin θd ec d

3

1 cosθ 1 1 θ = θ cosecθcot θ cos θ θ

sinθ(cosθ 1) sin θ 2 2d d ec d

3

sinθ 1 1= cosecθcot θ cosecθ θ

sin θ 2 2

dd

2

1 1 1= cosecθcot θ ln | cosecθ cot θ |

2sin θ 2 2 C

2 2

1 1 cosθ 1 1 cosθ= ln | |

2sin θ 2 sin θ 2 sinθ

C

2

1 cosθ 1 1 cosθ= ln | |

2sin θ 2 sinθ

C

1 1 1 cosθ= ln | |

2(1 cosθ) 2 sinθC

2 2

2

1 11.27) θ = θ

cosθ(cos θ 4sinθ 5) cosθ(1 sin θ 4sinθ 5)

1 = θ

cosθ(sin θ 4sinθ+4)

=

d d

d

2

1 θ

cosθ(sin θ 2)d

ก าหนดให้ 2sin θ ==> cosθ 1

θcosθ

x x

dxd

ดงันั้น 2 2 2

1 1θ =

cosθ(cos θ 4sinθ 5) ( 1)(x 2)d dx

x

2

1 =

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

x


1

( 1)(x 1)(x 2)x

2 2

2 2 2 2

1 A B C D =

( 1)(x 1)(x 2) 1 1 2 ( 2)

1 A( 1)( 2) B( 1)( 2) C( 2)(x 1) D(x 1)

x x x x x

x x x x x


18


A2


27

แทน 2 ; x จะได ้ 1D

3

หา C จาก 2 2 2

2 2

1 2 4C

1 ( 1) 9x x

d x

dx x x

ดงันั้น

2 2

1 1 1 4 1 =

( 1)(x 1)(x 2) 2( 1) 18( 1) 9( 2) 3( 2)x x x x x

2 2

1 1 1 4 1 { }

( 1)(x 1)(x 2) 2( 1) 18( 1) 9( 2) 3( 2)dx dx

x x x x x

2

1 1 1 1 4 1 1 1=

2 1 18 1 9 2 3 ( 2)

dx dz dxx x x x

1 1 4 1

= | 1 | | 1 | | 2 | 2 18 9 3( 2)

n x n x n x Cx

1 1 4= | sin θ 1 | | sinθ 1 | | sinθ 2 |

2 18 9 n n n

1

3(sin θ 2)

C

2 2 2 2

3 2

4 2

3

(1 sec θ)sec θ (tan θ 2)(tan θ 1)1.28) θ = θ

1+tan θ (tan θ 1)(tan θ tan θ+1)

tan θ 3tan θ 2 = θ

tan θ 1

d d

d

2

3

3tan θ tan θ 2= (tan θ+ ) θ

tan θ 1d


3

3tan θ tan θ 2θ

tan θ 1d

2 2

3 2 3

3tan θ tanθ 2 3 2θ =

tan θ 1 ( 1)( 1)

u ud du

u u


2

2 3

3 2

( 1)( 1)

u u

u u

2

2 3 2 2

2 3 2 2 2

3 2 A B C D E =

( 1)( 1) 1 1 1

3 2 (A B)( 1) C( 1)( 1)+(D E)( 1)( 1)

u u u u

u u u u u u

u u u u u u u u u u

แทน 1 ; u จะได ้ C 1

แทน 0 ; u จะได ้ B E 1 ...(33) แทน 1 ; u จะได ้ A B 2D 2E 1 ...(34)

แทน 2 ; u จะได ้ 6A 3B 10D 5E 1 ...(35) แทน 2 ; u จะได ้ 14A 7B 10D 5E 19 ...(36)

แกส้มการ (33), (34) ,(35) และ (36)

จะได ้ A 1,B 0,D 0,E 1


2 3 2 2

2 2 1 1 =

( 1)( 1) 1 1 1

u u u

u u u u u u

2

3 2 2

3tan θ tanθ 2 1 1 (tan θ ) θ = tanθ θ ( )

tan θ 1 1 1 1

ud d du

u u u u

22

1 1= tan θ θ ( )

311 1( )

44

ud du

u uu

2

tan θ

(1 ) θ

u

du u d

3 4 2

4

2

tan θ tan θ 1 tan θ 3tan θ +2

tan θ tan θ

3 tan θ tan θ 2


28

21

= | secθ tanθ | ( 1) | 1 |2

n n u n u

2 2 1

arctan{ }( )3 3 4

Cu

2

2

1= | sec θ tan θ | (tan θ 1) | tan θ 1 |

2

2 2 1 arctan{ tan θ }( )

43 3

1= | sec θ tan θ | (tan θ 1) | tan θ 1 |

2

n n n

C

n n n

2 2 1 arctan{ tan θ }( )

43 3C

2

2

1 21.29) =

( 1)1

1 = 2

1

1 = 2

( 1)( 1)

1 1 = ( )

1 1

udx du

u ux x

duu

duu u

duu u

-1 = ln| |

1

uC

u

1 1 = ln| |

1 1

xC

x

1 11.30) =

( 1) ( 1)

1 1 =

1

dx d nxx nx nx nx nx

d nxnx nx

1

= | | nx

nx Cnx

3 2 2

2 2

1 11.31) =

( 1)

1 =

( 1)

x x x x x xdx dx

e e e e e e

duu u u


1

( 1)u u u

2 2 2 2

2 2 2

1 A B C D =

( 1) 1

1 A ( 1) B( 1) (C D)

u

u u u u u u u

u u u u u u u

แทน 0 ; u จะได ้ B 1

หา A จาก 2 2 2

0 0

1 2 1A ( ) 1

1 ( 1)u u

d u

du u u u u

แทน 1 ; u จะได ้ C D 1 ...(37)

แทน 1 ; u จะได ้ C D 1 ...(38)


2

1

1

2

x u

x u

dx udu

xu e

du udx


29

จะได ้ C 1,D 0


1 1 1 =

( 1) 1

u

u u u u u u u

2 2 2 2

1 1 1 = { }

( 1) 1

udu du

u u u u u u u

2 22

1 1 0.5 1 1 = { }

1 31 2( )

2 4

udu du

u u u uu

21 1 1 2 1 = | | | 1 | arctan{ ( )}

2 23 3n u n u u u C

u

21 1 1 2 1 = | 1 | arctan{ ( )}

2 23 3

x x x

xx n e e e C

e

3 3

1 1 11.32) =

(1 2 ) 2 ( 1)xdx du

n u u


1

(1 )u u

3 3 2

3 2

1 A B C D =

( 1) ( 1) ( 1) 1

1 A( 1) B C ( 1) D ( 1)

u u u u u u

u u u u u u

แทน 0 ; u จะได ้ A 1 แทน 1 ; u จะได ้ B 1

หา C จาก จะได ้ 2

1 1

1 1C ( ) 1

u u

d

du u u

หา D จาก จะได ้ 2

2 2 31 1 1

1 1 1 1 1D ( ) 1

2 2u u u

d d d

du u du u du u


1 1 1 1 1 =

( 1) ( 1) ( 1) 1u u u u u u

3 3 2

1 1 1 1 1 1 1 = { }

2 ( 1) 2 ( 1) ( 1) 1du du

n u u n u u u u

3 2

3 2

1 1 1 1 1 = { }

2 ( 1) ( 1) 1

1 1 1 1 1 = { }

2 ( 1) ( 1) 1

dun u u u u

dun u u u u

2

1 1 1 = { | | | 1 |}

2 2( 1) 1n u n u C

n u u

2

1 1 1 = { (2) | 2 1 |}

2 2(2 1) 1 2

x

x xx n n C

n

2 2

2

2

1 11.33) = 2

(1 ) (1 )

1 = 2

(1 )

1 = 2

(1 )

x x

u

dx d xx e e

due

dzz z


1

(1 )z z

2

2

xu

du u n dx

u

u x

z e

dz zdu


30

2 2

2

1 A B C =

(1 ) ( 1) 1

1 A( 1) Bz Cz( 1)

z z z z z

z z

แทน 0 ; z จะได ้ A 1 แทน 1 ; z จะได ้ B 1

หา C จาก จะได ้ 2

1 1

1 1C ( ) 1

z z

d

dz z z


1 1 1 1 =

(1 ) ( 1) 1z z z z z

2 2

1 1 1 1 = { }

(1 ) ( 1) 1du du

z z z z z

1= | |

1 1

zn C

z z

1= | 1 |

1

x

xx n e C

e

ข้อ 2 จงหาค่าของอนิทกิรัลต่อไปนี้

2

5 7 5 72.1) =

2 3 ( 3)( 1)

x xdx dx

x x x x


( 3)( 1)

x

x x

5 7 A B

= ( 3)( 1) 3 1

5 7 A( 1) B( 3)

x

x x x x

x x x

แทน 1 ; x จะได ้ B 3 แทน 3 ; x จะได ้ A 2

ดงันั้น 5 7 2 3 =

( 3)( 1) 3 1

x

x x x x

5 7 2 3 = { }

( 3)( 1) 3 1

xdx dx

x x x x

= 2 | 3 | 3 | 1 |n x n x C

0 0

22

5 7 = 2 | 3 | 3 | 1 |

( 3)( 1)

= 2 (3) 3 (3)

x

xdx n x n x

x x

n n

= 3n 3 2 2

3 2

11 10 9 62.2) = 1

2 4 ( 2)( 2 2)

x x x x x

dx dxx x x x x


2

9 6

( 2)( 2 2)

x x

x x x

2

2 2

2 2

9 6 A B C =

( 2)( 2 2) 2 2 2

9 6 A( 2 2) (B C)( 2)

x x x

x x x x x x

x x x x x x


3 3 2

3

2

1 2 4 11 10

2 4

9 6

x x x x x

x x

x x


31

แทน 0 ; x จะได ้ C 1 แทน 1 ; x จะได ้ B 3


2 2

9 6 3 1 2 =

( 2)( 2 2) 2 2 2

x x x

x x x x x x

2

2 2

9 6 3 1 2 = { }

( 2)( 2 2) 2 2 2

x x x

dx dxx x x x x x

2

2 2

2

3 1 2=

2 2 2

3 2 2 1 2= 2

2 2 2 ( 1) 1 2

1= | 2 2 | 2arctan( 1) 2 | 2 |

2

xdx dx

x x x

xdx dx dx

x x x x

n x x x n x C

121

2

200

9 6 1 = | 2 2 | 2arctan( 1) 2 | 2 |

( 2)( 2 2) 2

1 1 3 = ( ) 2 ( )

2 2 2 2

x

x xdx n x x x n x

x x x

n n

3 3

4 2

5 4 5 42.3) =

16 ( 2)( 2)( 4)

x x x xdx dx

x x x x


2

5 4

( 2)( 2)( 4)

x x

x x x

3

2 2

3 2 2 2

5 4 A B C D =

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

5 4 A( 2)( 4) B( 2)( 4) (C D)( 4)

x x x

x x x x x x

x x x x x x x x

แทน 2 ; x จะได ้ A 1 แทน 2 ; x จะได ้ B 1 แทน 0 ; x จะได ้ D 0



2 2

5 4 1 1 3 =

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

x x x

x x x x x x

3

2 2

5 4 1 1 3 = { }

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

x x xdx dx

x x x x x x

2 23= | 4 | | 4 |

2n x n x C

434

2 2

433

5 4 3 = | 4 | | 4 |

16 2 x

x xdx n x n x

x

12 3 20 = ( ) ( )

5 2 13n n

2

2

12.4) =

1 ( 1)( 1)

1 1 1 { }

2 1 1

1 1

2

tt

t t t

t

t t

t

edt de

e e e

dee e

n e C

33

2

222

1 = 1

1 2

ntn

t

tnx n

edt n e

e

3 3 2

3

2

1 2 4 11 10

2 4

9 6

x x x x x

x x

x x


32

1

= { (8) (3)}2

n n

1 8

= ( )2 3

n

3 2

2sec θtanθ 12.5) θ = 2 secθ

sec θ secθ secθ(sec θ 1)d d


1

secθ(sec θ 1)

2 2

2

1 A Bsecθ C =

secθ(sec θ 1) secθ sec θ 1

1 A(sec θ 1) (Bsecθ C)(secθ)

แทน secθ 0 ; จะได ้ A 1 แทน secθ 1 ; จะได ้ B C 1 ...(39) แทน secθ 1 ; จะได ้ B C 1 ...(40)


จะได ้ B 1,C 0


1 1 secθ =

secθ(sec θ 1) secθ sec θ 1

2 2

1 1 secθ 2 secθ = 2 { } secθ

secθ(sec θ 1) secθ sec θ 1d d

2

2

= 2 | secθ | (sec θ 1)

1= | |

cos θ 1

n n C

n C

33

3 2θ4

4

2secθtanθ 1 θ = | |

sec θ secθ cos θ 1d n

2

2

1 cos ( )4 | |

1 cos ( )3

n

6

= ( )5

n


33

แบบฝึกหัด 6.3 ข้อ 1 จงอนิทเิกรต

2

4

2

1 cos 21.1) sin =

2

1 = {1 2 cos 2 cos 2 }

4

1 1 = {1 2 cos 2 (1 cos 4 )}

4 2

1 3 1 = { 2 cos 2

4 2 2

xxdx dx

x x dx

x x dx

x

cos 4 )}x dx 3 1 1

= sin 2 sin 48 4 32

x x x C

5 4

2 2

2 4

1.2) cos = (cos ) sin

= (1 sin ) sin

= {1 2sin sin } sin

xdx x d x

x d x

x x d x

3 52 1 = sin sin sin

3 5x x x C

7 6

2 3

2 4 6

11.3) sin = (sin ) cos

1 = (1 cos ) cos

1 = {1 3cos 3cos cos } cos

xdx x d x

x d x

x x x d x

3 5 71 3 1 = (cos cos cos cos )

5 7x x x x C

8 2 4

4 2 2

1.4) sin 3 = (1 cos 3 )

= (cos 3 2cos 3 1)

xdx x dx

x x dx

2

2 2

1 = ( (1 cos 6 ) (1 cos 6 ) 1)

4

1 = ( cos 6 (1 2 cos 6 cos 6 ))

4

x x dx

x x x dx

2

2

2 2

1 1 1 = ( cos 6 (1 cos12 ))

4 2 8

3 1 1 = ( cos 6 cos12 ))

8 2 8

3 1 3 1 1 1 = {( cos 6 ) 2( cos 6 )( cos12 ) cos

8 2 8 2 8 64

x x dx

x x dx

x x x

2 2

12 }

9 3 1 3 1 1 = { cos 6 cos 6 cos12 cos 6 cos12 cos 12 }

64 8 4 32 8 64

9 3 1 3 1 1 = { cos 6 (1 cos12 ) cos12 (cos18 cos 6 ) (1 cos 24

64 8 8 32 16 128

x dx

x x x x x x dx

x x x x x

)}

35 7 7 1 1 = { cos 6 cos12 cos18 cos 24 )}

128 16 32 16 128

x dx

x x x x dx


34

35 7 7 1 1 = sin 6 sin12 sin18 sin 24

128 96 384 288 3072x x x x x C

8 2 41.5) cos 2 = (1 sin 2 )xdx x dx

2 4 2

2 2

2 2

2

2

2

= (1 2sin 2 sin 2 )

1= (1 (1 cos 4 ) (1 cos 4 ) )

4

1= (cos 4 (1 2cos 4 cos 4 ))

4

1 1 1= ( cos 4 (1 cos8 ))

4 2 8

3 1 1= ( cos 4 cos8 ))

8 2 8

3 1 3 1= {( cos 4 ) 2( cos 4 )

8 2 8 2

x x dx

x x dx

x x x dx

x x dx

x x dx

x x 2

2 2

1 1( cos8 ) cos 8 }8 64

9 3 1 3 1 1= { cos 4 cos 4 cos8 cos 4 cos8 cos 8 }

64 8 4 32 8 64

9 3 1 3 1= { cos 4 (1 cos8 ) cos8 (cos12 cos 4 )}

64 8 8 32 16

1 (1 cos16 )

128

35 5 7= { cos 4 cos8

128 16 32

x x dx

x x x x x x dx

x x x x x dx

x dx

x1 1

cos12 cos16 )}16 128

x x x dx

35 5 7 1 1= sin 4 sin 8 sin12 sin16

128 64 256 192 2048x x x x x C

2 2 21.6) sin 3 cos 3 = (sin 3 cos3 )x xdx x x dx

2

2

1= (sin 6 )

4

1= (1 cos 6 )

4

1 1= (1 (1 cos12 ))

4 2

x dx

x dx

x dx

1

= (1 cos12 ))8

x dx

1 1

= sin128 96

x x C

3 3

cos 11.7) = sin

sin sin

xdx d x

x x

2

1 =

2sinC

x

3

2

3

2

1.8) cos sin 2 = 2 sin cos

= 2 cos cos

x xdx x xdx

xd x

5

24

= cos5

x C


35

2 24 (1 sin ) coscos1.9) =

1 sin 1 sin

x xxdx dx

x x

2

2 2

2

= (1 sin ) cos

= (cos sin cos )

1= (1 cos 2 ) cos cos

2

x xdx

x x x dx

x dx x d x

31 1 1= sin 2 cos

2 4 3x x x C

3 3

2 2 2 2

2 2

sin cos sin cos1.10) = { }

sin cos cos sin

1 1 = cos sin

cos sin

x x x xdx dx

x x x x

d x d xx x

1 1 =

cos sinC

x x

5 3 5 3

2

5 3

2

2 3

cos sin 1 cos sin1.11) = { }

1 cos 2 2 cos

1 cos sin = { }

2 cos

1 = {cos sin } sin

2

x x x xdx dx

x x

x xdx

x

x x d x

3 51 = {sin sin } sin

2x x d x

4 61 1 = sin sin

8 12x x C

2

2 6 4

4 2

4 2

1 sin1.12) = { }

cot sec sec

= {cos sin }

= {cos (1 cos )}

xdx dx

x x x

x x dx

x x dx

4 6

2 3

= {cos cos )}

1 1= { (1 cos 2 ) (1 cos 2 ) }

4 8

x x dx

x x dx

2 2 31 1= { (1 2cos2 cos 2 ) (1 3cos2 3cos 2 cos 2 )}

4 8x x x x x dx

2 3

3

2

3

1 1 1 1= { cos 2 cos 2 cos 2 }

8 8 8 8

1 1 1 1= { cos 2 (1 cos 4 ) cos 2 }

8 8 16 8

1 1 1 1= { cos 2 cos 4 } cos 2 sin 2

16 8 16 16

1 1 1 1 1= sin 2 sin 4 (sin 2 sin 2 )

16 16 64 16 3

x x x dx

x x x dx

x x dx xd x

x x x x x C

31 1 1= sin 4 sin 2

16 64 48x x x C


36


2

2

1 cos 62.1) sin 3 = ( )

2

1 1 = sin 6

2 12

1 1 sin 3 = sin 6

2 12

1 1 = ( ( ))

2 1

x

x

xxdx dx

x x C

xdx x x

{sin 6 sin( 6 )} 2

=

2 2 2

2 22

2

12.2) sin cos = (sin 2 )

4

1 = (1 cos 4 )

8

1 1 = sin 4

8 32

1 1 sin cos = s

8 32

x xdx x dx

x dx

x x C

x xdx x

2

2

in 4

1 1 = ( ( )) {sin 2 sin( 2 )}

8 2 2 32

x

x

x

=

8

4 2 4 2

4 6

2.3) sin cos = (sin )(1 sin )

= (sin sin )

x xdx x x dx

x x dx

2 3

2 2 3

1 1 = { (1 cos 2 ) (1 cos 2 ) }

4 8

1 1 = { (1 2cos 2 cos 2 ) (1 3cos 2 3cos 2 cos 2 )}

4 8

x x dx

x x x x x dx

2 31 1 1 1 = { cos 2 cos 2 cos 2 }

8 8 8 8x x x dx

2

2

1 1 1 1 = { cos 2 (1 cos 4 )} cos 2 sin 2

8 8 16 16

1 1 1 1 = { cos 2 cos 4 } (1 sin 2 ) sin 2

16 8 16 16

x x dx xd x

x x dx x d x

3

24 2 32

44

1 1 1 1 = sin 2 sin 4 sin 2

16 8 64 48

1 1 1 1 sin cos = sin 2 sin 4 sin 2

16 8 64 48

1 1 = ( ) (sin s

16 2 4 8

x

x

x x x x C

x xdx x x x x

3 31 1in ) (sin 2 sin ) (sin sin )

2 64 48 2

1 1 =

64 8 48

5 =

64 48


37

4 2

2

12.4) sin = (1 cos 2 )

4

1 = (1 2 cos 2 cos 2 )

4

1 1 = {1 2 cos 2 (1 cos 4 )}

4 2

1 3 1 = { 2 cos 2

4 2 2

xdx x dx

x x dx

x x dx

x

242

00

cos 4 )}

3 1 1 = sin 2 sin 4

8 4 32

3 1 1 sin = sin 2 sin 4

8 4 32

3 1 1 = (sin sin 0) (sin 2 sin 0

16 4 32

x

x

x dx

x x x C

xdx x x x

)

3

= 16

2 22.5) (1 sin ) = (1 2sin sin )

1 = (1 2sin (1 cos 2 ))

2

x dx x x dx

x x dx

3 1 = 2cos sin 2

2 4x x x C

2

00

3 1 (1 sin ) = 2cos sin 2

2 4

3 1 = 2(cos cos0) (sin 2 sin 0)

2 4

x

x

x dx x x x

3 = 4

2

3 2 3

2 2

2.6) sin 3 (cos 4 sin 3 cos 3 ) = (sin 3 cos 4 sin 3 cos 3 )

1 1 = (sin 7 sin ) sin 3 cos 3 sin 3

2 3

x x x x dx x x x x dx

x x dx x xd x

2 2

3 5

1 1 = (sin 7 sin ) sin 3 (1 sin 3 ) sin 3

2 3

1 1 1 1 = cos cos 7 sin 3 sin 3

2 14 9 5

sin 3

x x dx x x d x

x x x x C

x

3 3 5

00

3 3

1 1 1 1(cos 4 sin 3 cos 3 ) = cos cos 7 sin 3 sin 3

2 14 9 5

1 1 1 = (cos cos 0) (cos 7 cos 0) (sin 3 sin 0

2 14 9

x

x

x x x dx x x x x

5 5

)

1 (sin 3 sin 0)

5

6 =

7


38


2

4 21.1) tan = sec 1xdx x dx

4 2

2

2

= sec 2sec 1

= sec tan 2 tan

= (tan 1) tan 2 tan

x x dx

xd x d x dx

x d x d x dx

31= tan tan

3x x x C

7 2 3

6 4 2

6 4 2

1.2) tan = (sec 1) tan

sec = (sec 3sec 3sec 1) tan

sec

sec 3sec 3sec 1 = ( ) sec

sec

xdx x xdx

xx x x x dx

x

x x xd x

x

6 4 21 3 3 = sec sec sec ln | sec |

6 4 2x x x x C

7 2 3

6 4 2

cosec1.3) cot = (cosec 1) cot cos

cosec

cosec 3cosec 3cosec 1 = ( ) cosec

cosec

xxdx x x d x

x

x x xd x

x

6 4 21 3 3

= cosec cosec cosec ln | cosec | 6 4 2

x x x x C

8 2 6

2 6 6

6 2 4

1.4) cot = (cosec 1) cot

= (cosec ) cot cot

= cot cot (cosec 1) cot

=

xdx x x dx

x x dx x dx

x d x x x dx

6 2 4 4

6 4 2 2

6 4 2 2

cot cot (cosec ) cot cot

= cot cot cot cot (cosec 1) cot

= cot cot cot cot (cosec ) cot

x d x x x dx x dx

x d x x d x x x dx

x d x x d x x x d 2

6 4 2 2

7 5 3

cot

= cot cot cot cot cot cot cosec

1 1 1 = cot cot cot cot

7 5 3

x x dx

x d x x d x x d x x dx dx

x x x x x C

5 3

3 3

3 2 3

3 2

1.5) sec = sec tan

= tan sec tan sec

= tan sec 3 tan sec

= tan sec 3 (sec 1

xdx xd x

x x xd x

x x x xdx

x x x 3

3 5 3

)sec

= tan sec 3 sec 3 sec

xdx

x x xdx xdx

3

3

1 3 = tan sec sec tan

4 4

1 3 = tan sec (sec tan tan sec )

4 4

x x xd x

x x x x xd x


39

3 21 3 = tan sec (sec tan sec tan )

4 4 x x x x x xdx

3 2

3 3

1 3 3 = tan sec sec tan sec (sec 1)

4 4 4

1 3 3 = tan sec sec tan (sec sec )

4 4 4

x x x x x x dx

x x x x x x dx

3 3

3

1 3 3 3 = tan sec sec tan sec sec

4 4 4 4

1 3 3 = tan sec sec tan sec

4 8 8

x x x x xdx xdx

x x x x xdx

31 3 3 = tan sec sec tan ln | sec tan |

4 8 8x x x x x x C

6 4

2 2

4 2

1.6) sec = sec tan

= (tan 1) tan

= (tan 2 tan 1) tan

xdx xd x

x d x

x x d x

5 31 2 = tan tan tan

5 3x x x C

8 2 3

6 4 2

1.7) cosec = (1+ cot ) cot

= ( cot 3 cot 3cot 1) cot

xdx x d x

x x x d x

7 5 31 3 = cot cot cot cot

7 5x x x x C

11 11

6 2 25 5

11

2 2 5

11

2 4 5

1.8) sec tan = (1 tan ) tan tan

= (1 tan ) tan tan

= (1 2 tan tan ) tan tan

x xdx x xd x

x xd x

x x xd x

11 21 31

5 5 5 = (tan 2 tan tan ) tanx x x d x 16 26 36

5 5 55 5 5

= tan tan tan16 13 36

x x x C

7 7 6

6 2 3

6 6 4 2

1.9) sec tan = (sec tan ) sec

= sec (sec 1) sec

= sec (sec 3sec 3sec 1) sec

x xdx x x d x

x x d x

x x x x d x

12 10 8 6 = (sec 3sec 3sec sec ) secx x x x d x

13 11 9 71 3 1 1 = sec sec sec sec

13 11 3 7x x x x C

45

2 2

4 2

cot1.10) cosec cot = { } cosec

cosec

(cosec 1) = cosec

cosec

cosec 2cosec 1 =

cos

xx xdx d x

x

xd x

x

x x

cosecec

d xx


40

9 5 1

2 2 22 4

= cosec cosec 2cosec9 5

x x x C

3 24

2

3

tan sec1.11) = {tan sec }

sin

= sec sec

x xdx x x dx

x

xd x

41 = sec

4C

3 2 3 2

3 3 2 3

1.12) tan (2 sec ) = tan (4 4 sec sec )3 3 3 3 3

= {4 tan 4 sec tan sec tan }3 3 3 3 3

=

x x x x xdx dx

x x x x xdx

3 3 2 3

2 2 3

2

4 tan 4 sec tan sec tan3 3 3 3 3

= 4 (sec 1) tan 12 tan sec 3 tan tan3 3 3 3 3 3

= 4 (sec

x x x x xdx dx dx

x x x x x xdx d d

x

2 3tan 4 tan 12 (sec 1) sec 3 tan tan3 3 3 3 3 3 3

x x x x x xdx dx d d

2 3 43

= 6tan 12 ln | sec tan | 4sec 12sec tan3 3 3 3 3 4 3

x x x x x xC


5 5

4

2 2

4 2

sec2.1) tan = (tan )

sec

tan = ( ) sec

sec

(sec 1) = { } sec

sec

sec 2sec 1 = { }

sec

xxdx x dx

x

xd x

x

xd x

x

x xd

x

4 2

sec

1 = sec sec ln | sec |

4

x

x x x C

35 4 23

66

4 4 2 2

1 tan = sec sec ln | sec |

4

sec1 3 = (sec sec ) (sec sec )+ln| | 4 3 6 3 6

sec6

x

x

xdx x x x

8 1 = ln 3

9 2

4 2

2

5

2

2.2) tan sec = tan sec tan

= tan (1 tan ) tan

= ( tan tan ) tan

=

x xdx x xd x

x x d x

x x d x

3 7

2 22 2

tan tan3 7

x x C

3 7 444 2 2

0

0

2 2 tan sec = tan tan

3 7

x

x

x xdx x x


41

3 3 7 7

2 2 2 22 2

= (tan tan 0) (tan tan 0) 3 4 7 4

20 =

21


38


2

2

1 1 21.1) =

22 sin 12

1

dx dzzx z

z

2

2

1=

1

1=

1 3( )

2 4

dzz z

dz

z

2

2

4 1=

2 13( ( )) 1

23

2 1 2 1= { ( )}

2 1 23 3( ( )) 1

23

2 2 1= arctan ( )

23 3

dz

z

d z

z

z C

2 2 1 = arctan{ (tan )}

2 23 3

xC

2 2

2

1 1 21.2) =

13 2 cos 13 2( )

1

dx dzzx z

z

2

2

2

2=

5

2 1=

5( ) 1

5

2 1=

5 5( ) 1

5

2= arctan

5 5

dzz

dzz

zd

z

zC

2 1= arctan{ tan }

25 5

xC

2 2 cos1.3) =

sin tan sin (1 cos )

xdx dx

x x x x

2

2

2 2

2 2

2

3

1

21= 22 1 1

(1 )1 1

= (1 )

= z3

z

z dzz z z

z z

z dz

zC

31= tan tan

2 3 2

x xC


39

2 2

2 2

1 1 21.4) =

1 2cos sin 1 11

1 1

1 =

1

= ln | 1 |

dx dzz zx x z

z z

dzz

z C

= ln | 1 tan | 2

xC

2 2

2 2

2

2

1 1 21.5) =

2 1sin cos 3 13

1 1

1 =

2 1

1 1 =

12

2 2

dx dzz zx x z

z z

dzz z

dzz

z

2

1 1 =

1 72( )

4 16

dz

z

2

2

8 1 =

4 17{ ( )} 1

47

2 1 4 1 = { ( )}

4 1 47 7{ ( )} 1

47

2 4 1 = arctan( ( ))

47 7

dz

z

d z

z

z

C

2 4 1 = arctan( (tan ))

2 47 7

xC

2

2

2

2 2

2

sin 211.6) = 22 sin 1

21

2 =

(1 )( 1)

z

x zdx dzzx z

z

zdz

z z z


2

(1 )( 1)

z

z z z

2 2 2 2

2 2

2 A B C D =

(1 )( 1) 1 1

2 (A B)(z z+1) (C D)(1 )

z z z

z z z z z z

z z z z

แทน 0 ; z จะได ้ B D 0

...(39)

แทน 1 ; z จะได ้ A B 2C 2D 2

...(40)

แทน 1 ; z จะได ้ 3A 3B 2C 2D 2

...(41)

แทน 2 ; z จะได ้ 6A 3B+10C 5D 4

...(42)

แกส้มการ (39), (40), (41) และ (42)

จะได ้ A 0, B 2,C 0,D 2


40


2 2 2 =

(1 )( 1) 1 1

z

z z z z z z

2 2 2 2

22

2 2 2 ( )

(1 )( 1) 1 1

2 2 = ( )

1 31( )

2 4

zdz dz

z z z z z z

dzz

z

22

2 4 1 2 1 = { ( )}

2 11 23 3{ ( )} 1

23

4 2 1 = 2 arctan arctan{ ( )}

23 3

dz d zz

z

z z C

4 2 1 = arctan{ (tan )}

2 23 3

xx C

2

2

2 2

2

2

2 2

1 cos1.7) =

3sec 1 3 cos

1

21 = 1 1

31

1 =

(2 1)( 1)

xdx dx

x x

z

z dzz z

z

zdz

z z


2 2

1

(2 1)( 1)

z

z z

2

2 2 2 2

2 2 2

1 A B C D =

(2 1)( 1) 1 2 1

1 (A B)(2z +1) (C D)( 1)

z z z

z z z z

z z z z

แทน 0 ; z จะได ้ B D 1

...(43)


...(44)


...(45)

แทน 2 ; z จะได ้ 18A 9B+10C 5D 3

...(46)

แกส้มการ (43), (44), (45) และ (46)

จะได ้ A 0, B 2,C 0,D 3


2 2 2 2

1 2 3 =

(2 1)( 1) 1 2 1

z

z z z z

2

2 2 2 2

2 2

1 2 3 ( )

(2 1)( 1) 1 2 1

2 3 1 = ( 2 )

1 2 ( 2 ) 1

zdz dz

z z z z

dz d zz z

3

= 2 arctan arctan 22

z z C

3 = arctan( 2 tan )

22

xx C


41

2

2

2

2 2

2

2

cot cos1.8) =

1 sin sin (1 sin )

1

21 = 2 2 1

(1 )1 1

1 =

( 1)

x xdx dx

x x x

z

z dzz z z

z z

zdz

z z


2

1

( 1)

z

z z

2

2 2

2 2

1 A B C =

( 1) 1 ( 1)

1 A(z+1) B( )( 1) C

z

z z z z z

z z z z

แทน 0 ; z จะได ้ A 1

แทน 1 ; z จะได ้ C 0

หา B จาก 2

2

11

1 1B 1 2

zz

d z

dz z z


2

1 1 2 =

( 1) 1

z

z z z z

2

2

1 1 2 ( )

( 1) 1

= ln | | 2 ln | 1 |

zdz dz

z z z z

z z C

= ln | tan | 2 ln | tan 1 | 2 2

x xC

2 2

2 2

2

2 2

sec 11.9) =

1 sin cos (1 sin )

1 2 =

1 2 1(1 )

1 1

1 = 2

(1 )( 1)

xdx dx

x x x

dzz z z

z z

zdz

z z


2 2

1

(1 )( 1)

z

z z

2

3 2 3

2 3 2

1 A B C D =

(1 )( 1) 1 1 ( 1) ( 1)

1 A(z+1) B(1 )( 1) C(1 )( 1) (1 )

z

z z z z z z

z z z z z D z

แทน 1 ; z จะได ้ 1A

4

แทน 1 ; z จะได ้ D 1

หา C จาก 2 2

2

1 1

1 2 1 1C

1 (1 ) 2z z

d z z z

dz z z

หา B จาก 2 2 2

2 2 3

1 1 1

1 1 1 2 1 1 4 1B

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

d z d z z

dz z dz z z


3 2 3

1 1 1 1 1 =

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

z

z z z z z z


42

2

3 2 3

2

1 1 1 1 1 2 2 ( )

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

1 1 1 1 = ln | 1 | ln | 1 |

2 2 ( 1) ( 1)

zdz dz

z z z z z z

z z Cz z

2

1 tan1 1 12 = ln | |2

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

x

Cx x x

2

2 2

2 2

2

tan sin1.10) =

1 tan sec cos sin 1

2

21 = 1 2 1

11 1

2 =

(1 )(1 )

x xdx dx

x x x x

z

z dzz z z

z z

zdz

z z


2

(1 )(1 )

z

z z

2 2

2

2 A B C =

(1 )(1 ) 1 1

2 A( +1) (B C)( 1)

z z

z z z z

z z z z

แทน 1 ; z จะได ้ A 1

แทน 0 ; z จะได ้ C 1 แทน 1 ; z จะได ้ B 1


2 1 1 =

(1 )(1 ) 1 1

z z

z z z z

2 2

2 2

2 1 1 ( )

(1 )(1 ) 1 1

1 1 =

1 1 1

= l

z zdz dz

z z z z

zdz dz dz

z z z

21n | 1 | ln | 1 | arctan

2z z z C

21 = ln | tan 1 | ln | tan 1 |

2 2 2 2

x x xC

ขอ้ 2 จงหาค่าของ 6

2 2

0

1

4 3cos 5sindx

x x

2 2 2 2 2

2

1 1 =

4 3cos 5sin 4 2sin 3(cos sin )

1 =

1 cos 24 2 3(cos 2 )

2

1 =

4 1 cos 2 3(cos 2 )

dx dxx x x x x

dxx

x

dxx x

1 =

5 4cos 2 dxx


43

ก าหนดให้ 2x = u

22

2

2 2

2

1 1 1 =

5 4cos 2 2 5 4cos

1 1 2 =

2 115 4

1

1 =

5 5 4 4

1 =

1 9

dx dux u

dzzz

z

dzz z

dzz

2

1 1 = (3 )

3 1 (3 )

1 = arctan(3 )

3

1 = arctan(3tan )

3 2

1 = arctan(3tan )

3

d zz

z C

uC

x C

6

6

02 2

0

1 1 = arctan(3tan )

4 3cos 5sin 3

1 = arctan(3tan )

3 6

1 = arctan( 3)

3

dx x

x x

= 9


44


3 2 2 3

4

2 6 4 2

4

8 6 4 2

4

21.1) 7 2 = ( 2)

7

2 = ( 6 12 8)

7

2 = ( 6 12 8 )

7

2 =

7

x x dx u u du

u u u u du

u u u u du

9 7 5 3

4

1 6 12 8{ }9 7 5 3

u u u u C

9 7 5 3

2 2 2 24

2 1 6 12 8 = { (7 2) (7 2) (7 2) (7 2) }

7 9 7 5 3x x x x C

43

4

2 4

6 2

7 3

( 5)11.2) = 4

93 5

1 = 4 ( 5)

9

1 = (4 20 )

9

1 4 20 = { }

9 7 3

uxdx u du

ux

u u du

u u du

u u C

7 3

4 44 1 5

= { (3 5) (3 5) }9 7 3

x x C

2

2

2

2

2

51.3) = 5 2

5 5

= 2 55

5 = 2 5 {1 }

5

1 = 2 5{ 1 }

( ) 15

x udx udu

x u

udu

u

duu

du duu

= 2 5{ 5 arctan }5

uu C

= 2 5 10 arctan

5

xx C

1 11.4) = 2

2 32 1 3

3 = (1 )

2 3

3 = ln | 2 3 |

2

dx uduux

duu

u u C

3 = 1 ln | 2 1 3 |

2x x C

2

7 2

1( 2)

7

2

7

u x

x u

dx udu

4

4

3

3 5

1( 5)

3

4

3

u x

x u

dx u du

2

2

u x

x u

dx udu

2

1

1

2

u x

x u

dx udu


45

2

2

2

1 11.5) = 2

( 9)( 7) 2

1 = 2

9

2 1 =

3 3( ) 13

2 = arctan

3

dx uduu ux x

duu

ud

u

3

uC

2 1 = arctan 2

3 3x C

43

4 2

2

2

2

1.6) = 4

= 41

1 = 4 (1 )

1

= 4( arctan )

x udx u du

u ux x

udu

u

duu

u u C

4 4 = 4( arctan )x x C

4

8 635

3 9

2

4 2

5 3

2 21.7) = 6

= 6 ( 2 )

2 = 6( )

5 3

x x u udx u du

ux

u u du

u uC

5 1

6 22 = 6( )

5 3

x xC

11

11 3 2

64

6 4 2

2

7 5 3

1 11.8) = 12

(1 )(1 )

1 = 12 ( 1 )

1

= 12( + arctan )7 5 3

dx u duu u

x x

u u u duu

u u uu u C

7 5 1

1 112 12 412 12 = 12( + arctan )

7 5 3

x x xx x C

1

325

4 6 8

3

2

2

2

1.9) = 6

= 61

1 = 6 (1 )

1

= 6( arctan )

x udx u du

u ux x

udu

u

duu

u u C

6 6 = 6( arctan )x x C

2

2

2

2

u x

x u

dx udu

4

4

34

u x

x u

dx u du

6

6

56

u x

x u

dx u du

12

12

1112

u x

x u

dx u du

6

6

56

u x

x u

dx u du


46

1

43

1 4 2

2

2 2

2

1 11.10) = 4

1 = 4 (1 )

1 1

1 = 4 ln | 1 | arctan

2

x udx u du

u ux x

udu

u u

u u u C

4 41 = 4 ln | 1 | arctan

2x x x C

65

3 23

8 2

2

6 4 2

2

3 1 3 11.11) = 6

(1 )(1 )

3 = 6

1

4 = 6 (3 3 3 4 )

1

x udx u du

u ux x

u udu

u

u u u duu

7 5 33 3 = 6( 4 4 arctan )

7 5u u u u u C

7 5 1 11

6 6 6 623 3

= 6( 4 4 arctan )7 5

x x x x x C

4

4

34

u x

x u

dx u du

6

6

56

u x

x u

dx u du


47


2 2

1.1) a x

dxx

พิจารณา 2 2a x

dxx

ให้ 2 2

sin θ

cosθ θ

cosθ

x a

dx a d

a x a

ดงันั้น 2 2 cosθ

= cosθ θsinθ

a x adx a d

x a

2

2

cos θ= θ

sin θ

1 sin θ= θ

sin θ

= (cosecθ sin θ) θ

= a{ln|cosecθ cotθ|+cosθ}

a d

a d

a d

C

2 22 2= a{ln| | }

a a xa x C

x

3

2 2

11.2)

(25 )

dx

x


2 2

1

(25 )

dx

x

ให้ 2 2

5sin θ

5cosθ θ

5 5cosθ

x

dx d

x


2 2

1 1 cosθ = θ

25 cos θ(25 )

dx d

x

2

2

1 1= θ

25 cos θ

1= sec θ θ

25

1= tanθ

25

d

d

C

2

1=

25 25C

x

3

3

2 2

1.3)

( 4)

xdx

x


3

2 2( 4)

xdx

x

ให้ 2

2

2 tan θ

2sec θ θ

4 2secθ

x

dx d

x

θ

xa

2 2a x

θ

x5

2 25 x

θ

x2

4x

2


48


2

3 22 2

8tan θ = 2sec θ θ

8sec θ( 4)

xdx d

x

3

2

2

2

= 2 tan θ cos θ θ

= 2 tan θsinθ θ

= 2 (sec θ 1) cos θ

1= 2 ( 1) cos θ

cos θ

2= 2 cos θ

cos θ

d

d

d

d

C

2

2

4= 4

4x C

x

2

3

161.4)

xdx

x


3

16xdx

x

ให้ 2

4secθ

4secθtanθ θ

16 4 tan θ

x

dx d

x


3 3

16 4tanθ = (4secθtanθ) θ

64sec θ

xdx d

x

2 2

2

1= tan θcos θ θ

4

1= sin θ θ

4

d

d

1= (1 cos 2θ) θ

8d

1 1= (θ sin 2θ)

8 2C

1= (θ sinθcosθ)

8C

2

2

1 1 16= arcsec

8 4 2

x xC

x

2

5

2 2

1.5)

(9 )

xdx

x


5

2 2(9 )

xdx

x

ให้ 2

3sin θ

3cosθ θ

9 3cosθ

x

dx d

x


5 5 52 2

9sin θ = (3cosθ) θ

3 cos θ(9 )

xdx d

x

2

4

2 2

1 sin θ= θ

9 cos θ

1= tan θsec θ θ

9

d

d

θ

2 16x x

4

θ

x3

29 x


49

2

3

1= tan θ tan θ

9

1= tan θ

27

d

C

3

3

2 2

1=

27(9 )

xC

x

2

2

5 2 11.6)

25

x xdx

x


2

5 2 1

25

x xdx

x

ให้ 2

5sin θ

5cosθ θ

25 5cosθ

x

dx d

x


2

5 2 1 5(25sin θ) 10sinθ 1 = 5cos θ

5cosθ25

x xdx d

x

2

2 2

= (125sin θ 10 sin θ 1) θ

= 125 (cos 2θ) θ 10 sin θ θ 126 θ

125= 126θ sin 2θ 10 cos θ

2

= 126arcsin 5 25 2 255

d

d d d

C

xx x x C

2= 126arcsin (5 2) 255

xx x C

2 2

2

2

1 11.7) =

12 4 ( 4 12)

1 =

( 4 4 16)

1 =

16 ( 2)

dx dyy y y y

dyy y

dyy


1

16 ( 2)dx

y

ให้ 2

2

2 4 tan θ

4sec θ θ

16 ( 2) 4secθ

y

dy d

y


2

1 4sec θ = θ

4secθ16 ( 2)dx d

y

= sec θ θ

= ln|secθ tan θ|

d

C

22 12 4

= ln| |4

y y yC

2 2

2 2

2

2

2

2

1.8) = 9 8 ( 8 9)

= ( 8 16 25)

= 25 ( 4)

y ydx dy

y y y y

ydy

y y

ydy

y

θ

x5

225 x

θ

2y

2

16(

2)y

4


50


225 ( 4)

ydx

y

ให้ 2

4 5sin θ

5cosθ θ

25 ( 4) 5cosθ

y

dy d

y


2

(5sinθ 4) = 5cosθ θ

5cosθ25 ( 4)

ydx d

y

2

2

= (5sinθ 4) θ

= (25sin θ 40 sin θ 16) θ

25= (1 cos 2θ) θ 40 sin θ θ 16 θ

2

57 25= θ sin 2θ 40 cos θ

2 2

d

d

d d d

C

2457= arcsin( ) ( 12) 9 8

2 5

yy y y C

3 3

2 22 2

3

2 2

1.9) ( 6 13) = ( 6 9 4)

= (( 3) 4)

y y dy y y dy

y dy


2 2(( 3) 4)y dy

ให้ 2

2

3 4 tan θ

4sec θ θ

( 3) 4 4secθ

y

dy d

y


2 3 22(( 3) 4) = 256 sec θ sec θ θy dy d

5

3

3 3

3 2 3

3 2 3

= 256 sec θ θ

= 256 sec θ tanθ

= 256{sec θ tanθ tanθ sec θ}

= 256{sec θ tanθ 3 tan θ sec θ θ}

= 256{sec θ tanθ 3 (sec θ 1)sec θ θ}

d

d

d

d

d

3 5 3

3 5 3

3 3

= 256sec θ tan θ 768 (sec θ sec θ) θ

= 256sec θ tan θ 768 sec θ θ 768 sec θ θ

256 768= sec θ tan θ sec θ θ

769 769

d

d d

d

3 1 sec {tan sec | cos( ) sin( ) | | cos( ) sin( ) |}

2 2 2 2 2

x x x xxdx x x n n C


2 32

cos( ) sin( )256 384 2 2( 6 13) = sec θ tanθ {tan sec | |}769 769

cos( ) sin( )2 2

x x

y y dy x x n Cx x

2

11.10)

( 1) 1dx

y y


1

( 1) 1dx

y y

θ

4y 5

225 ( 4)y

θ

3y

2

25(

3)y

4

θ

2 1y y

1


51

ให้ 2

secθ

secθtanθ θ

1 tan θ

y

dy d

y


1 secθ tanθ = θ

(secθ 1) tanθ( 1) 1dx d

y y

2

2

2

2

sec θ= θ

sec θ 1

1= ( ) θ

1 cos θ

1 cos θ= ( ) θ

1 cos θ

1 cos θ= ( ) θ

sin θ

sin θ= cosec θ θ

sin θ

1= cot θ

sin θ

d

d

d

d

dd

C

2

2

11=

1

yC

yy

2 32 3

2 3

2 3

1 1 11.11) =

8 27( 4 24 27)( 6 )

4

1 1 =

8 9( 6 9 )

4

1 1 =

8 9( ( 3) )

4

dx dxy y

y y

dx

y y

dx

y


1

9( ( 3) )

4

dx

y

ให้

2

33 secθ

2

3secθtanθ θ

2

34 24 27 tan θ

2

y

dy d

y y


3secθ tan θ

1 1 1 2 = θ38 89

( ) tan θ( ( 3) )24

dy d

y

2

2

2

1 sec θ= θ

18 tan θ

1 cosθ= θ

18 sin θ

1 sin θ=

18 sin θ

1=

18sin θ

d

d

d

C

2

3=

918 ( 3)

4

yC

y

θ

2 9( 3)

4y

3y

3

2


52

2 2 2

11.12) , 0

( )du a

a u


1

( )du

a u

ให้ 2

2 2

tan θ

sec θ θ

secθ

u a

du a d

a u a


2 2 2 4 4

1 sec θ = θ

( ) sec θ

adu d

a u a

3 2

2

3

1 1= θ

sec θ

1= cos θ θ

da

da

3

1= (1 cos 2θ) θ

2d

a

3 3

1 1= θ sin 2θ

2 4C

a a

3 2 2 2

1= arctan

2 2 ( )

u uC

a a a a u

2 2 2

11.13) , 0

( )du a

a u


1

( )du

a u

ให้ 2 2

sin θ

cosθ θ

cosθ

u a

du a d

a u a

ดงันั้น 2 2 2 4 4

1 cosθ = θ

( ) cos θ

adu d

a u a

3 3

1 1= θ

cos θd

a

3

3

3

3

2

3

2

3

3

3 3

1= sec θ θ

1= secθ tanθ

1= {secθ tanθ tanθ secθ}

1= {secθ tanθ secθ tan θ θ}

1= {secθ tanθ secθ(sec θ 1) θ}

1 1= secθ tanθ (sec θ secθ) θ

da

da

da

da

da

da a

3 3

3 3 3

3

3 3 3

3

1 1 1 sec θ θ = secθ tanθ (sec θ secθ) θ}

2 1 1 sec θ θ = secθ tanθ+ secθ θ

1 1 sec θ θ = secθ tan θ ln | secθ tanθ |

2 2

d da a a

d da a a

d C

2 2 2 2 2 2 3 2 2

1 1 = ln | |

( ) 2 ( ) 2

u a udu C

a u a a u a a u

θ

u2

2

au

a

θ

ua

2 2a u


53

2

4 4 4 4 4 4

2 2 4

11.14) =

2

1 =

2

u dudu

u u a u u a

dz

z z a


dz

z z a

ให้

2

2

2

2 4 2

secθ

secθtanθ θ

tan θ

z u

z a

dz a d

z a a


4 2 22 2 4

1 1 secθtanθ = θ

2 2 sec θ( tanθ)

dz ad

a az z a

4

4

4

2 4

4

1 θ=

2 secθ

1= cosθ θ

2

1 sin θ

2

2

d

a

da

Ca

z aC

za

4 4

2 4

2

u aC

u a

3 3

2 4 4 22 2

2 2

3

2 2 2

2

(3 2 ) ( ( 2 1 4)1.15) =

1 1

{4 ( 1) } =

1

u u u u u udu du

u u

u udu

u


2 2 2

2

{4 ( 1) }

1

u udu

u

ให้ 2

2 2

1 2sin θ

2 2cosθ θ

4 ( 1) 2cosθ

u

udu d

u


2 2 2 22 22

2 2

{4 ( 1) } 1 {4 ( 1) } =

1 2 1

u u udu du

u u

3

4

2

2 2

2

3

1 8cos θ= 2cosθ θ

2 2sinθ 2

cos θ= 4 θ

1 sinθ

4 cos θ(1 sinθ) θ

4 (cos θ cos θsinθ) θ

2 (1 cos 2θ) θ 4 cos θ cosθ

4 2θ sin 2θ cos θ

3

d

d

d

d

d d

C

32

2 2 4 2 2 21 1 1

2arcsin( ) ( 1) 3 2 {4 ( 1) }2 2 6

uu u u u C

θ

2 4z a

z

2a

θ

2 1u 2

2 24 ( 1)u


54

3 3

2 22 2

3

2 2

1.16) =

(4 25) (4 25)

=

(4 25)

t t

t t

e dedt

e e

du

u


2 2(4 25)

du

u

ให้ 2

2

2 5 tan θ

5sec θ θ

2

(2 ) 25 5secθ

te u

u

du d

u


3 32 2

5 sec θ = θ

2 125sec θ(4 25)

dud

u

2

1= cosθ θ

50

1= sinθ

50

1

25 4 25

d

C

uC

u

2

25 4 25

t

t

eC

e

2 2

11.17)

16t tdt

e e


1

16t tdt

e e

ให้ 2

4secθ

4secθtanθ θ

16 4tanθ

t

t

t

e

e dt d

e


1 tanθ = θ

16sec θ(4tanθ)16t tdt d

e e

2

2

2

1= cos θ θ

64

1= (1+cos2θ) θ

128

1 1 (θ sin 2θ)

128 2

1 1 4 16 arcsec( )

128 4 128

t t

t

d

d

C

e eC

e

2

2

1 1 16 arcsec( )

128 4 32

t t

t

e eC

e

3

2

11.18)

( 1)t

dt

e


2

1

( 1)t

dt

e

θ

2u

2

(2)

25

u

5

θ

2 16te

te

4

θ

2

t

e

1te

1


55

ให้

2

22

tan θ

2sec θ θ

1 secθ

t

t

t

e

e dt d

e


3 3

2

1 2sec θ = θ

tanθ(sec θ)( 1)t

dt d

e

2

2

1= 2 θ

secθ tan θ

cos θ= 2 θ

sin θ

1 sin θ 2 θ

sin θ

1 2 θ 2 sinθ θ

sin θ

2ln|cosecθ cotθ| 2cosθ

d

d

d

d d

C

2

1 1 2 2ln| |

1

t

t t

eC

ee

ข้อ 2 จงหาค่าอนิทกิรัลต่อไปนี้ 2 2

3 2 3 2

0 02.1) 2 = 1 ( 1)x x x dx x x dx

พิจารณา 3 21 ( 1)x x dx

ให้ 2

1 sin θ

cosθ θ

1 ( 1) cosθ

x

dx d

x

ดงันั้น 3 2 3 21 ( 1) = (sinθ 1) cos θ θx x dx d

3 2

5 4 3 2

= (sin θ 1) (1 sin θ) θ

= ( sin θ 3sin θ 2sin θ 2sin θ+3sin θ 1) θ

d

d

พิจารณา 5 4sin θ θ = sin θsinθ θd d

2 2

2 4

3 5

= (1 cos θ) cosθ

= (1 2cos θ cos θ) cosθ

2 1 = cosθ cos θ cos θ

3 5

d

d

C

พิจารณา 4 3 1 1sin θ θ = θ sin 2θ sin 4θ

8 4 32d C (จากแบบฝึกหัดท่ี 6.3)

พิจารณา 3 31sin θ θ = cosθ cos θ

3d C

3 2 5

5

2 22

2 2

7 1 1 3 1 ( 1) = θ cos θ sin 2θ sin 4θ

8 5 4 32

7 1 1 = arcsin( 1) (2 ) ( 1) 2

8 5 2

3 ( 1) 2 { 2

16

x x dx C

x x x x x x

x x x x 4 1} x C

52

3 2 2 22

0

2

2 2

0

7 1 1 2 = arcsin( 1) (2 ) ( 1) 2

8 5 2

3 ( 1) 2 { 2 4 1}

16

x x x dx x x x x x x

x x x x x

θ

1x 1

21 ( 1)x


56

7 = {arcsin(1) arcsin( 1)}

8

7 = { ( )}

8 2 2

7 =

8

3 33 3

3 35 54 2 4 22 2

2.2) =

( 2 3) {( 1) 4}

x xdx dx

x x x


3

4 2 2{( 1) 4}

xdx

x

ให้ 2

2

2 2

1 2secθ

2secθtanθ θ

( 1) 4 2 tan θ

x

dx d

x


2

3 3

4 2 4 22 2

1 =

2{( 1) 4} {( 1) 4}

x xdx dx

x x

3

2

2

1 (2secθ 1)(secθtanθ)= θ

8 tan θ

1 2sec θ secθ= θ

8 tan θ

d

d

2

2 2

2

4 2

24 2

1 sec θ 1 secθ θ θ

4 tan θ 8 tan θ

1 1 sinθ cotθ

4 8 sin θ

1 1 cotθ

4 8sinθ

1 2 3

8( 1)2 2 3

d d

dd

C

x xC

xx x

3

3 4 23

3 24 254 2 2 5

1 2 3 =

8( 1)2 2 3{( 1) 4}

1 60 1 12 =

64 322 60 2 12

=

x x xdx

xx xx

1 15 1 3

32 164 15 4 3

1 15 15 3 3 = { }

4 15 8 3 4

θ

2 2( 1) 4x

21

x

2

calculus1 6-all

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