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Nội dung

các tích phân sau đây (sử dụng tích phân từng phần): a. ;b. ;c. ;d. ;e. ;f. ;g. ;h. ;i. ;j. ;k. ;l. ;m. ;n. ;o. ;p. ;q. ;r. ;s. t. (). Hướng dẫn:Câu a: Tính . Đặt Tính .Vậy .Câu b: . Đặt .Câu c: Tính: . Suy ra: .Vậy suy ra: .Câu d: . Đặt . Câu e: …Đặt Câu f: …Đặt .Đặt

Trang 1

Bài 1 Tính các tích phân sau đây (sử dụng tích phân từng phần):

e

xdx x

x I

1

2ln

1

;

2 1

)1ln( x dx

12

e e

dx x x

2 1 2

ln

dx x

x

2 1 2)1ln(

1

2)

5 2

)1ln(

3 6

2cos

)ln(sin

dx x

x

2 1

)cos(lnx dx

2 0

(*))

ln(sin

dx x

 0

2tan

xdx x

1 0

3 4

2sin

dx x

x

3 0 2cossin

dx x

x x

2 0cos

xdx e

1 0

3 2

dx e x

;

 0

2 2 2)1(

2

dx x

e

dx x

x xdx

x xdx x

x I

1 1

1

lnln

1

e

xdx x

I

1

ln 2 2 ln 2 1

ln 2 2 2

1 2 2

ln)(lnlnln

1 1

2 1

e e

e

x x

xd dx

14

2 2

3 3

ln

1ln

1ln

1ln

1ln

1

I dx x

dx x

dx x

dx x x I

e e

e e

e e

dx x x du dx

dv

dx x u dx x

1 ln

1

3

3 3 3

2 1

ln

1ln

ln

e

e e

e e

dx x x

x dx x

3 3 3

3

2 2

1

1 ln

ln

1 ln

e e

e

e e e

e

dx x x

x dx x I

dx x

3

3

.

Trang 2

x dx du

2

1ln

1 2 2 1

2 1

x

x

) 1 ln(

1 ) 1

1 2 2 2

ln 2 ln 2 ln 2 ln

Đặt e x xdx x e x dx x e x e

1 2 1 2

1

2

ln 2 2

ln2

lnln

2

1 2 1 2 1

2 2 1

1

2 2

e e

.

Câu g: 

e

dx x I

1

2)

1 1 2 2

ln 2 ln 2 ln ln 2 ln

1 1

Câu h:  

5

2

)1ln(

) 1 ( ) 1 ln(

) 1 ( 1 2 ) 1 ln(

x dx x x x I x dx du x dv x u

5 2

2 1)ln( 1) ( 1)

2

27 4 ln 24 2

) 1 ln(

)ln(sin

dx x

cos ) ln(sin tan tan

3ln3632

1ln3

32

3ln3)

ln(sin

3 6

3 6

1

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

sin(ln

)

cos(ln

dx x x x dx x I x

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

sin(ln

dx x x x dx x x x du dx

1

2 1

2

)cos(ln

2

1)2cos(ln)

2sin(lnsin

)cos(ln

ln(sin

dx x I

0 sin ) ln(sin sin

)

ln(sin

dx x x x x

)ln(sin

xdx x

x x

Trang 3

) ln(sin ) ln(sin cot ) ln(sin cot

dx x x xdx x v dx xdx dv x u

2 0

2

)ln(sin

xdx x

x x

2 0 2

2 0

x x x

xdx x

2

3

cos 3 cos cos

33

3sin12

13

coscos

33cos12

13

cos

0 0

3 0

4 0 2 4

0 2 4

x x dx x

x xdx x

4

)(costan

x

x d x

x

4 0

2ln42cos

lntan

)1(tantan

2 4

0

2 4 0

4 0 4

0

4 0 2 4

t t x x

1 0

sin

dx x

x I

cot cot

3

4 sin

coscot

x

2

3ln2

136

)349(sin

ln36

)349(sin

)(sin

4 3

dx x

x x I

Trang 4

3 cos 3

23ln21

3cos31sin

1sinln21

3cos3sin1

)(sin

3cos3sin1

3 0

2 3

x d x

0

2 2

2

0

sin sin sin

1cos

2cos

1sin

2 0 2

2 0

2 0 2

2 0 2 2

e xdx e

xdx e

e xdx e

e xdx

Câu r:  

1 0 2 1

0

dx xe x dx e x

Đặt txdtxdxdtxdx

22

0

t t x

x

1 0 1

0 1

0

2

12

2

dt e t dt

e t dx xe

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

cos cos )1 sin )1 cos

sin

1

xdx x x xdx x x

2)1(

2

dx x

1 2

1 ) 1 ( 2

x dx x I x v dx x xdx dv x

t t x

2tan

1

)tan1(1

4 4

4 4

2

2 1

1 2

dt t x

4 1)

(x x

dx

1 0 3

x

I   1

Trang 5

k.    

4 3

35

dx x x

x

1 0

114

dx x x

2 0 2

2

0

14

1)

4(16

dx x

x

dx I

31

2ln313

32

12

1 0 1

0

2 2

x

dx I

2 2

1 0

2 1

0

2

2

3 2

1 4

3 2 1 1

x

dx x

dx x

x

dx I

.

2

3 tan

2

3 2

4.2

3)1(tan43

)1(tan2

3)

1(tan4

3

)1(tan2

3

3 6

3 6

2

2 3

dt t

x

x B x

A x

B x

A x

2 1 1

2 )

(

B A B

A B A B

A x B

3

11

2ln3

11ln3

12ln3

113

123

1)1)(

2(

5 3

5 3 5

3 5

3 5

3 5

1(

13

4

1

2 2

2 2

2

2 2

2

2 2

2 2

2 4

x

x x

x

x x

x x

x

)3(2

1)

1(2

1

2 2

2 2

1(2

13

0

12

11

12

1

t

t x

4tan

1

)tan1(1

4 0

4

0

2

2 1

34

2

134

Trang 6

Câu f: dx

x x

x x x dx x

x

x x

x dx

x x

x x

x x

2

2 4

1

2

2 4

1 2

2 2

4

1

)1)(

1()

1(

)1)(

1()

1(

1)

1(

1 2 4

1

4 1 2 4

1

4 1 2 4

1)

1(

)1)(

1()

1

dx x

xdx x

x d dx x

x x

dx dx

x x

x x dx

31

ln1ln1

1 4

3 2

1

4 1) ( 1)

dx x x

1

t

t x

11

ln4

11

114

1)1(4

1)1( 4

16 1

16 1

16 1

)1)(

1(

31

2 2

x x

C Bx x

A x

x x

Bx Cx Bx A Ax

0 0

C B C

A C B A B A

0

1 0 2

1 0

1 0

2 1

0

21

ln1

21

1

21

11

x x

x x

dx x x

x x

dx dx

x x

x x

2 2

1 0 2

1 0 1

0 2 1

0 2

2 1

0

2

3 2

1 2

3 1 ln

2

1 1 ln 1 2

3 1

) 1 (

x x

x x

dx dx

x x

x x x

2

3 2

2

3 2

x

Suy ra

33

23

2)

1(tan43

)1(tan23

2

32

1

6 6 6

6

2

2 1

0

2 2

dt x x

dx

Vậy

32ln33

2.2

31ln

2

11ln1

0

2 1

0 1

2 2

2

1

0

2 4

3

3 2

1 1

x

xdx dx

x x

x I

2

3 2

tan 2

3 2

x

.

Trang 7

Suy ra 33 183

)1(tan43

)1(tan43

3

32

11

3 6 3

6

2

2 1

0

2 2

2

1

0

2 4

xdx dx

x x

0 2 2

0 2 2

0

224

2

1224

2

32

x x

dx dx

x x

x dx

x x

x dx

x x

x I

2 2

2 0

2 2

2 0 2

2 0

2 2

0

2

2

3)1(3ln3)1(42ln3)1(42

)42

(

x

dx x

dx x

x x

dx x

33

3)

1(tan3

)1(tan33

)1(

3 6

3 6

3 6

2

2 2

0

2 2

3ln21

ln22ln71

22

7)

2)(

1(

352

3

4 3

4 3

2 2

1 0 2 1

0 2 1

0

)65(2656

5

522652

152265

114

x x

x x d x

x

dx dx

x x

x dx

x x

x dx

x x

x

2

9ln3ln2ln2ln232

65ln2)3

)(

2

(

1 0

1 0 1

0

1 0

1 0 2

dx x

x

1 0

3

)1

3 2

3

)1

5 1)

(x x

3

)12

2 1 2 212

7x dx x

x

1 0

x

j.  

2 1 4

21

1

dx x

x I

2 4

21

1

dx x

1 0

1 0 2 1

0 2

2 1

0

2

2

)2)(

2(

44

44

44

44

x x

dx dx

x

dx dx

x

x dx x

x I

(

x x

x x

Trang 8

ln2ln12

2

12

2

1 0

1 0

1 0

5 3

5 3

5 3

5 3

5

102

)2(32

10)2(32

43

x

dx dx

x

dx x

dx x

dx x

x dx

x

x I

3ln1061ln103ln103.35.32ln103

210

x

2 3

2

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

B x

A x

C x

B x

A x

1 2

0 )

2

(

2

C B C

B A B A C

B A x B A Ax

1)

1(

1)

1 0 2 3

1 0

2 1

0

3 1

0

2 1

0

1 )

1 ( 2

1 1

1 )

1 (

) 1 ( )

1 (

) 1 ( )

1 ( ) 1 ( )

1

x d x

x d x

dx x

dx dx

x

x I

Câu d:  

1

0

3 2

3

)1

1

t

t x

0

2 2 3

3 4

0

2 2

3 4

0

3 2

2 3

cossincos

1.cos

sin)

1(tan

tan)

1(tan

)1.(tantan

t t

t dt

t

t t

dt t t I

3 1

1 1 1

1 0

2 3 1

0 2 1

dx dx

x x

x dx

x x

x dx

x x

x x

x x

4 2

1

2 1 5

5 2

1 5

5 2

1

5

5 5

2

1

)1()

1(

)1()1(

5 5

5 2

33ln5

12ln1

ln5

1ln1

)1(5

1

x x

x

x d

3 1

0

2 1

0

3 1

0

3 1

0

3 1

0

1)12(2

1)12(2

1)

12(

122

1)

12(

1122

1)

1

2

dx x

dx x

dx dx

x

x dx

x

x dx

1 2 ( 2

1 1

2

1 4

1 ) 1 2 (

) 1 2 ( 4

1 )

1

2

(

) 1

0

3 1

2 1

2 1

2 1

2 2

1612

7

127112

x dx

x x

x I

x I

2 2

2 2

x C x

B Ax x

C x

B Ax x

x

x x

C A x

x C x

B Ax

Trang 9

4 2

0

C B C

x x

x x x

1 0 2 1

0 2 1

0

2

9)1(5

29)

2(5

9)

1(5

292

2

14

x

dx dx

x

x dx

x x

x dx

x x x

x I

0

1 0 2 1

0

1 0 2 1

0

2

2

15

22ln5

91ln10

925

915

21

)1(

10

9

x

dx x

x x

dx x

dx x

1

t

t x

Suy ra

41

tan

)1(tan1

4 0

4 0 2

2 1

272

2

14

1 0

2 3

2 2 2

2

2 2

1

4 2

21

111

11

1

1

dx x

x

x dx

x x

x dx x

x

x dt dx x dt

2 2

1

t

t x

12

2ln22

12

2

2 5 2

2 5 2

2 2

2 2

5 1

1

2 2

2 2

5 1

1

2 2

2

2

2 2

5 1

1

2 4 2

11

11

1

111

1

111

1

x x

x x d dx

x x

x dx

x x

x

x

x dx

x x

x I

x x d t x

5 1

1

t

t x

x

11

1

4 4

1

4 1 2

2 2

dt t

x x

x x d

Bài 4 Tính các tích phân sau: (sử dụng tích phân hàm vô tỉ)

a. I x xdx

2 1

2 1 x dx x

2

1 x

dx x

2

3 1 x dx x

Trang 10

j.  

1 0

2

2 4 x dx x

2 2

3

2)1

x I

1

ln2

x

dx x

Hướng dẫn:

Câu a:

)47(3

1)3(3

1)3()3(2

1)3(32

1

2 1

3 2 2

2 1

2

1 2 2

2 1 2 2

16 0

16 0

16 0

16

0

99

19

)9()9)(

9(

)9(

dx x x

x x

x x

dx x x

x x

9(27

2)

9()

3 3 16

0 2 1 16

1(4

1)1()1(3

1)1(1

3

11

42 0

42 0

3 3

1 3 42

2

t

t x

x

Vậy

3 2

3 2

3 2

3 2

3

2

t t t

dt dt

t

dt dt

t

t t

dt t

0 2 3 2 3 1

0 2 1 1

0

2 1 1

0 1

0

x x

dx x dx x

dx x x

x x

x

Trang 11

4 0 2 4

0

2 4

0

2 4

2 2

2

2

)2cos1(2

1sin

cos

cossincos

cossinsin

1

cossin1

t

tdt t t

tdt t t

tdt t x

sin4

12

12cos2

2

1)1(21

.2

)1(1

2

3 0

3 3

0 2 3

0

2 1

dx x

4 2 1

0

2 2 0

1 2 1

0

2 2

1

0

2 3

15

2)

()

1()

()1(1

1

t

t x

2 0 2 2

0

2 0 2 2

0

2 1

t tdt

t tdt

t dx

sin4

12

12cos2

12

0

2 0

2 0

2 0

2

t

t x

Vậy

dt t tdt

t tdt

t tdt

t t dx

x x

I 4 4sin 2cos 2cos 4 4sin cos 4 sin 2 2 (1 cos4)

2 0

2 0

2 2

2 0 2 2

0 2 2

0

2 2

x

.

Trang 12

2 4

2

2

2 4

2

2cot

sin2sin22cos2

sincos1

cos1

t dt

t x

t dt

t t

t

4 2

.sin2cot

dt t t

4 2

2 4

2

)

cos1(

2cos2

2

cos2sin

t dt

t t t

t

2

214

sin 4 2

0 2

1

0 2

1 0 2 1

0

2

1

21

.21

.1

1

x

dx x x

xdx x x

xdx x x

x dx x

2

1

211

21

112

x

dx I

x

dx dx

x dx

x x

21

22

x

dx I

x

dx I

4 0

2 4

0 2 4

0

4 0

2 4

0

2

2 4

2 1

)(sinsin

1

)(sincos

coscos

cos1coscos

1

cos1

tan

)tan1(1

t d t

t d t

tdt t

dt

t t dt

t

t dt

t

dt t x

dx

)(sin)sin1)(

sin1(

sin12

1)(sin)sin1)(

sin1(

sin12

1)(sin)

sin1)(

sin

1

(

)sin1()sin

4

0

t d t t

t t

d t t

t t

d t t

t t

4 0

4 0

4 0

4

sin1ln2

1sin1

)(sin2

11sin

)(sin2

1sin

1

)(sin2

1sin

11ln2

1223ln2

110sin

0sin1ln2

114

sin

4sin

212

12

1

t

t x

x

2 0

2 2

0 4 2

0 6 2

0

3 2 1

sinsinsin

)cos1()

tdt t tdt

x dx

x I

2 0

34

14

cos12

12cos214

12

cos2

12cos

dt t t

dt t t

3 1

Trang 13

t

t x

3 2 0 2 2

15124

)32(

)32(5

124

)32

dt t

t

tdt x

x x

dx x x

x x

14

1)1(tan4

)tan1(22

142

0

3 0

3 0

2

2 3

tdt dx dx e tdt e

t e

ln

0

t

t x

0 2 1

0

1 0 2

2

ln

2.1

t

dt t

dt dt

t

dt t dx e

Lại đặt t tanu dt ( 1 tan 2u)du

0

u

u t

t

.

Suy ra:

4)

tan1(

)tan1(1

4 0

4 0

2

2 1

2313

0

t

t x

2 1

0

7 8 8

1

0

8 15

12

1.3

13

13

5 36

1 ) (

36

1

3 5 2

1 ) ln 2 ( 3 2 2

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

2

1 2 3 1

2 3 1

2 1 1

e e

x x

x d x x

d x

21

2

11

2 1

2 2

Trang 14

Đặt 3 9 2 1 6 9 9 2 1 1 (6 2) 2(3 11)

2 2

2 2 2

x t

x x x tx t x

x x

1 1

0

t

t x

x

Vậy

2

126ln3

113

2 2 1

1sin

28

.cossin

217

.cos2

cos7

.cos

dx x x

dx x x

dx x

0

t

t x

2

1

t

dt t

0

u

u t

t

Suy ra

262

12

1cos

2

cos2

2cos

4

cos22

12

2

0

6 0

6 0

2sin

22cos

2tan

dx x

4 3cos1

dx x

3 4 6

2

cossin

dx x

x

4 0 4cos

x

x x

6 sin sin 6

x x

dx

3 4sin

3cos

dx x

x

3 4

2

2 cossin

4cos

4sin

xdx

Trang 15

q. 

2 0

2cos.4cos

xdx x

r. I 2cos2x(sin x cos x)dx

0

4 4

3sin

5

4 cossin

xdx x

2 0

3

2 cossin

xdx x

xdx I

44

sin4

12

1)4cos1(2

12

0

2 0

sin

3 4

3

4 3

d x

dx x

x xdx

I

3

4 3 4

3

4 3

2 4

3

2 4

dx x dx

x xdx

I

cos1

)(coscos

1

)(cossin

sinsin

2

2 2

3

2 2

3 2 2

x d dx

x

x dx

x I

1sin

)(sin2

11sin

)(sin2

11sin

)(sinsin

1

)(sincos

coscos

3

4 3

4 3 2 4

3

2 4

3 2 4

x d x

x d x

x d dx x

x dx

x I

dx x

3

t

t x

5)

1(cos

.cos

1tancos

.cos

1.cos

1

3 1

3 5 2

2 3

4

2 2

2 3

4

2 2

)tan1(cos

.coscos

4 0

2 4

0

2 2

4 0

2 2 4

dx x x

x

dx x

dx I

Câu i: 

3

0

3cos

x dx

Trang 16

x xdx du x

2 3

0

3 0

sintancos

tancos

x

x x

dx I

3 0

3 0

3 0 3 3

2 3

0

3

2

cos3

2coscos

32cos

cos132cos

x

dx x

dx dx

x

x x

xdx

23

23ln2

13

21sin

1sinln2

13

2sin1

cos3

2cos

cos3

0

3 0

2 3

3 2 3 ln 2

1 3 2 2 2 3 2 3 ln 2

1 3

0 2 4

1cos

22

cos1

dx x

x dx

x

x dx

dx du

4 0 4

0

4 0

4 0

sintan

tantan

x x

x xdx x

x dx x x

4

18cos

lntancos

)(cos

0 4

x x

x d x

2 6

3 2 6

2)

sin(cos

cos

2

4coscos

dx x

x x

dx dx

x x I

.2

3ln23

3ln2tan

1ln2tan

1

)tan1

(

3 6

x d

3 6

3

6

cossin

3sin26sincos6cossinsin6

sinsin

x x

dx x

x

dx I

3

2lncot

3ln2cot

3

)cot3(2cot

3

sin2cossinsin

3

6 3

6

3 6

2 3

x d

x x dx

x x x

2 3

4

2 3

4

3 3

4

sin

3)sin1(4cossin

)3cos4(cossin

cos3cos4sin

x x

dx x

x x

dx x

x x

3 4

2 3

)sin1(4)(sinsin

3)sin

d x

x x

d x

x

3

32cot

tancos

sincos

sin

)cos(sin

cos

3 4 2 3

4 2 3

4

2 2

2 2

x

dx x

dx x

x

dx x x

x x dx

Trang 17

“Note”:

cossin

coscos

sin

sincos

sin

)cos(sin

cossin

2 2

2 2

x

xdx x

x

dx x x

x x

dx

2 0

2 2

0

2 2

0

4

1)

2cos1(4

1cos

dx x x

dx x xdx

2

0 16

34

sin32

12sin4

18

34

cos2

12cos22

34

1

x x

x dx x

2 0

2 2

0

4

12cos22

34

1)

2cos1(4

1sin

dx x x

dx x xdx

I

16

3 4

sin 32

1 2 sin 4

1 8

2 0

2 0

cos2

1)2cos1(4cos2

1cos

.4cos

xdx dx

x x

xdx x

2 0

2 0

2 0

2

0

2cos4

16cos4

14

cos2

12

cos6

cos2

12

14

xdx dx

x x

xdx

0 2

sin 4

1 6 sin 24

4 4

Câu r: I 2cos2x(sin x cos x)dx

0

4 4

34

4cos112sin2

11cossin2)sin(cos

cos

x x

x x

x x

2 0

2

0

4 4

4

2cos4cos4

2cos34

4cos4

32cos)

cos(sin

2cos

dx x dx

x dx

x x

dx x dx

x x

2 0

2 0

2 0

2 0

2

0

6cos8

12

cos8

7)2cos6

(cos8

12

cos4

34

2cos4cos4

sin 48

1 2

2 2

0

2 2

x x

.

2 0

2 2 4

2 0

4 4 2

x xdx

x x xdx

x I

8 6

4 2

0

4 2

4 2

0

2 2

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

x d x x

x x

d x x

x x

d x x

315

8 9

sin 7

x

.

Trang 18

2 0

4 2

2 2

2 0

2 2 2

x d x x

xdx x

x xdx

x

I

15

2 5

sin 3

0

5 3

3cos1

sin4

dx x

x

3 6

4 cossin

x x

dx

2 3

2

)cos1(cos

cos

2 2

x x

dx x x

x x

4 0

2)cos2(sin

x x

dx

3 4

3tan

sin4cos52 0

x x

N

Hướng dẫn giải Câu a: 

2

0

3cos1

sin4

dx x

0

t

t x

x

22

4)

1(41

)1)(

1(41

)1(4cos

1

sin.sin

1

2 0

1

0 1

0 1

2 2

6

3

t

t x

t t t

t

dt x

x

xdx x

x

xdx x

2

4 4 2

3

2

3 6

2 4

3 6

2 4 3

6

)1()1()

sin1(sin

coscos

sin

coscos

2

2 3

2

2 2

3

2

2 3

2

2 2

2 3

2

4 2

)1()1()

1(

)1)(

1()

1()

t

t t

dt dt

t t

t t

dt t t

t dt

t

t

t

Trang 19

)23(3ln2

1327

261

1ln2

113

1)

1()

1(

2 3

2

2 3

2

1 2

2 3

2

1 4

2 3

x x

0

t

t x

2 2

1 0

2 1

0 2 1

0 2 4

0

2 2

)2()1(2)1(2121

21

tan2

dt t

dt t

t

dt t

t

dt x

dx C

2 2 2 2 ln 2 2

1 2

1 2 1 ln 2

Câu d:  

2 3

2

)cos1(cos

12

1

22

3

t

t x

x

Vậy

3

2 2 4

2 2

2 1

3 3

2 2 2

2 2

2 1

3 3

2 2

2 2

2 2

2 1

3 3

2 2 2

2 2

2 2

3

2

)1(4

)1(

121

2

)1(

121

)1()1(

)1(

12

1

111

2.11

)cos

1

(

cos

t t

dt t t

t t

dt t t

t

t t

dt t t

t t t

dt t t

22

11

3

12

11

12

11

2

14

)1(.)

1 3 3

2 4 1

3 3 4

2 1

3

3

4

2 2 2

12

1

22

0

t

t x

0

2 2

2 2 2

2 3

1

2 1

) 1 (

2 3

sin cos

dt t

t t

dt x

x

dx I

2 2 1

0

24

t

dt

Đặt t 1 2 tanu dt 2 (tan 2u 1 )du

t

4 0

1

với tan 21.

4)

1(tan4

)1(tan42)1(

2

4 4

4

2

2 1

Trang 20

2 2

0

2 2

0

3 2

sin)sin441(1

cos

sin)sin43(1

cos

sin4sin31

x

xdx x

dx x

x x

2 2

0

2

1cos

sin)cos41(1

cos

sin)sin1

0

t

t x

)41(1

cos

sin)cos

4

1

1 2

0 1

0 1

2 2

cos2

x x

2)(

2 2

2

2 2

2

cossin

4

cos)

(sin4

x x

dt t

t t dt

t

t t

2 2 2

2

2 2

2

2 2

2

2 2

2

)(sinsin

4

)(sinsin

4

cossin

4

cossin

4

cos2

x d dx

x

x dx

x

x x

dx x

x x

3 ln 2

1 2

sin

2 sin

dx x x

x x

Ta có

5cos3sin45cos3sin4

sin3cos45

cos3sin

4

6cos7sin

C x

x

x x

B A x

x

x x

C x x

B x

x A x

C A x B A x B A x

5

7 4

3

1 3

4

C B C

A

B A

B A

2

15

cos3sin4

sin3cos415

cos3sin

4

6cos7sin

dx x

x x

x

x x

dx x x

x x

cos3sin4

15

cos3sin

4

sin3cos

x x

d x

dx x x

dx x x

x x

2 0 2

15

cos3sin4ln25cos

x x

dx x

Trang 21

1

22

0

t

t x

2 1

0 2 1

0 2 1

0

2

2 2

2 2

25

1

1.31

2.4

1

25

cos3sin

4

1

t

dt t

t

dt t

t dt t

t t

dt dx

x x

6

12

1)

2(

)2

0

1 0

9ln25cos3sin4

15

cos3sin4ln2

2 0

x x

x x

dx

4

;0(

1)

2(tan

)2(tan)

2(tancos)

cos2(sin

4 0

4 0

2 4

0

2 2

x d x

x

dx x

2 2

3 4

2 2 3

xdx x

xdx I

2 3

4

2 3

4

2 3

x d x xdx

xdx x

3 4

3 4 3

4

3 3

4

3

4

2 3

4

3

tan3

tan)

1(tan3

x dx

dx x

4 0

4 0

4

.coscos

cossin

1cos

sin1

dx x

x

x x

dx

x x

dx x

x x

dx x

Ta có:

4cos

sin

.cossin

cossin

.sincos

4 0

4 0

)cos(sin

cossin

sin

0 4

0

4 0

x x

x x

d dx x x

x x J

2

182

ln422ln4)(

)

Trang 22

Câu l:          

4 0

4 0

2 4

0

2 4

x x dx

x x

xdx I

.2

2ln2

1cos

ln2

tancos

)(cos)

(tan.tancos

sin)

0

4 0

4 0

x d x d x dx

x

x x

d

x

3 6

3 6

2 3

6

2 3

x x

dx x

x xdx

3 6

3 6

3 6

3 6

(cotcot

sin

cos)

(cotcot

cot)

1(cot

x

x x

xd xdx

dx x x

2 ln 1 sin

ln 2

)cos(sin

sin4cos52 0

x x

x x

x

2 0

3 1

)cos(sin

sin

x x

x

2 0

3 2

)cos(sin

0

t

t x

t t

t t

d t

t

t x

d x x

3 0

2

3 2

0

3 2

2

cos2

sin

2cos

2

cos2

sin

2cos)

cos(sin

3 2

0

3

sincos

sinsin

cos

sin

N x d x x

x t

d t t

2 2

0

3 2

0

3 1

2

1

4cos

2)cos(sin

)cos(sin

sin)

cos(sin

cos2

x

dx x

d x x

x x

d x x

x N

N

N

14

tan21

4cos

5)

cos(sin

sin4)

cos(sin

cos5)

cos(sin

sin4cos

5

1 2 2

0

3 2

0

3 2

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