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8/17/2019 K2pi.net.Vn TichPhan

1/7 w w w

. k 2 p i

. n e tT NG H P 10 BÀI TOÁN TÍCH PHÂN Bài 1 Tính tích phân

I = π

4

0

1 + tan 2 x x −(x −tan x )cos2 x

3 + cos 2 x dx

L i gi i

1 + tan 2 x x −(x −tan x )cos2 x

3 + cos 2x =

x + x tan 2 x −x cos2 x + tan x cos2 x

3 + cos 2x

= x sin2 x + x tan 2 x

3 + cos 2x +

sin x cos x3 + cos 2x

= x tan 2 x (cos2 x + 1)

3 + 2cos 2 x −1 +

sin2x2(3 + cos2 x )

= x tan 2 x

2 +

sin2x

2(3 + cos2 x )

Do đó I = 12 −

12

x 2 + x tan x + ln |cos x | − 12

ln |3 + cos2 x |π

4

0=

π8 −

π 2

64 +

14

ln 2 − 14

ln 3

Bài 2 Tính tích phân

I = e

1

x 2 −2 ln x + 1x 2 .√ x + ln x dx

L i gi i 1

Ta có:I =

e

1

x 2 + 2 x + 1 −2(x + ln x )x 2 .√ x + ln x dx =

e

1

(x + 1) 2

x 2 .√ x + ln x dx −2 e

1

√ x + ln xx 2

dx

Xét I 1 = 2 e

1

√ x + ln xx 2

dx = e

1

√ x + ln x. −1x

dx = −√ x + ln xx

+ e

1

x + 12x 2 .√ x + ln x dx

T đó: I = −2√ x + ln xx

+ e

1

1 + 1x√ x + ln x dx

Xét I 2 = e

1

1 + 1x√ x + ln x dx Đ ý: (x + ln x ) = 1 +

1x

. Dài quá, mong m i ngư i gi i ti p

L i gi i 2

I = e

1

x 2 + 2 x + 1 −2(x + ln x )x 2 .√ x + ln x dx =

e

1

x + 1x ·

x + 1x.√ x + ln x dx −2

e

1

√ x + ln xx 2

dx

Mà d x + 1x

= −1x 2

dx và x + 1

x.√ x + ln x dx = d 2.√ x + ln x . Nên

I = 2x + 1

x√ x + ln x

e

1+ 2

e

1

√ x + ln xx 2

dx −2 e

1

√ x + ln xx 2

dx = 2e + 1

e√ e + 1 −4

L i gi i 3

Ta có:√ x + ln x

x=

1 + 1x

2√ x + ln x x −√ x + ln xx 2

= 1−x −2 ln x2x 2 .√ x + ln x

và √ x + ln x =1x

+ 1 . 1

2√ x + ln x = x2 + x

2x 2 .√ x + ln x1


2/7 w w w

. k 2 p i

. n e tNên I = 2 e1 1 −x −2 ln x2x 2 .√ x + ln x dx + 2 e1 x 2 + x2x 2 .√ x + ln x dx = 2 √ x + ln xx e1 + 2 √ x + ln x e1V y I = 2 √ e + 1e + 2√ e + 1 −4Bài 3 Tính tích phân

I =

π

2

0

3e2 x + sin x (4ex + 3 sin x ) −1(ex + sin x )2

dx

L i gi i

I = π

2

02 dx +

π

2

0

e2 x + sin 2 x −1(ex + sin x )2

dx = 2xπ

2

0+

π

2

0

e2 x −cos2 x

(ex + sin x )2 dx = π +

π

2

0

e2 x −cos2 x

(ex + sin x )2 dx

Đ t A = π

2

0

e2 x −cos2 x

(ex + sin x )2 dx =

π

2

0

(ex −cos x )(ex + cos x )(ex + sin x )2

dx

Đ t u = ex −cos x⇒ du = ( ex + sin x ) dx , dv =

ex + cos x(ex + sin x )2

dx ch n v = − 1

ex + sin x

⇒A = −

ex −cos xex + sin x

π

2

0+

π

2

0dx = −

eπ

2

eπ

2 + 1+ x

π

2

0= −

eπ

2

eπ

2 + 1+

π2

.⇒

I = − e

π

2

eπ

2 + 1+

3π2

.


I = π

4

0

sin x −2x. cos xex (1 + sin 2 x )

dx

L i gi i 1

Ta có : I = π

4

0sin x −x. (cos x + sin x ) + x(sin x −cos x )ex (1 + sin2 x ) dx

= π

4

0

sin xex (1 + sin2 x )

dx − π

4

0

x (sin x + cos x )ex (1 + sin2 x )

dx + π

4

0

xex

sin x −cos x(sin x + cos x )2

dx = I 1 −I 2 + I 3

Tính I 3 : Đ tu =

xex

dv = sin x −cos x(sin x + cos x )2

→du =

1 −xex

v = 1

sin x + cos x

I 3 = x

ex (sin x + cos x )

π

4

0 − π

4

0

1 −xex (sin x + cos x )

dx

Khi đó I = π

4

0sin x

ex (1 + sin 2 x )dx −

π

4

0

dxex (sin x + cos x )

+ xex (sin x + cos x )

π

4

0

Đ tu =

1sin x + cos x

dv = 1ex

→du = −

cosx −sin x(sin x + cos x )2

v = − 1ex

Khi đó :

I = π

4

0

sin xex (1 + sin 2 x )

dx + π

4

0

cos x −sin x(sin x + cos x )2

dx + x + 1

ex (sin x + cos x )

π

4

0

= π

4

0

cos xex (1 + sin 2 x )

dx + x + 1

ex (sin x + cos x )

π

4

0

Đ t x = π4 −t →dx = −dt khi đó

I = π

4

0

cos( π4 −t )e

π

4 −t (1 + cos2 t)dt +

x + 1ex (sin x + cos x )

π

4

0

2


3/7 w w w

. k 2 p i

. n e t= 12√ 2e π4 π40 e t (cos t + sin t )cos2 t dt + x + 1ex (sin x + cos x ) π40V y I = 12√ 2e π4 e tcos t π40 + x + 1ex (sin x + cos x ) π40 = π2 −√ 2e π4 + 12√ 2e π4L i gi i 2

I =

π

4

0sin x + cos x

−2 x + 12 . cos x

ex (1 + sin 2 x ) dx .

Đ t t = x + 12

ex (sin x + cos x )

=⇒ dt = ex (sin x + cos x ) − x +

12 . [e

x (cos x −sin x ) + ex (cos x + sin x )]

e2 x (cos x + sin x )2 dx

= sin x + cos x −2 x +

12 . cos x

ex (1 + sin2 x ) dx

Đ i c n: x = 0

→t =

1

2, x =

π

4 →t =

π + 2

4√ 2.eπ

4

.

=⇒ I = tπ +2

4 √ 2 .eπ

4

12

= π + 24√ 2.e π4 −

12

= π + 2 −2√ 2.e

π

4

4√ 2.e π4L i gi i 3

2I =

π

4

0

2sin x −4x. cos xex (1 + sin 2 x )

dx =

π

4

0

(sin x −cos x )ex (1 + sin2 x )

dx −

π

4

0

4x. cos xex (1 + sin 2 x )

dx +

π

4

0

sinx + cos xex (1 + sin 2 x )

dx

Xét :

J =

π

4

0

(sin x −cos x )ex (sinx + cos x )2

dx = 1

ex (sinx + cos x )

π

4

0+

π

4

0

1ex (sinx + cos x )

dx

Suy ra :

2I = 1

ex (sinx + cos x )

π

4

0+

π

4

0

2ex (sinx + cos x )

dx −

π

4

0

4x. cos xex (1 + sin 2 x )

dx

L i có :

K =

π

4

01

ex (sinx + cos x )dx =

x

ex (sinx + cos x )

π

4

0+

π

4

02x cos x

ex (sinx + cos x )2dx

Nên :π

4

0

2ex (sinx + cos x )

dx −

π

4

0

4x. cos xex (1 + sin2 x )

dx = 2x

ex (sinx + cos x )

π

4

0

V y :

2I = 1

ex (sinx + cos x )

π

4

0+

2xex (sinx + cos x )

π

4

0=⇒ I =

12√ 2.e π4 −1 +

π

4√ 2.e π4


I = π

4

0

x + tan xcos2 x (1 + tan x )2

dx

3


4/7 w w w

. k 2 p i

. n e tL i gi i Ta có I = π40 tan xdxcos2 x (1 + tan x )2 + π40 xdxcos2 x (1 + tan x )2 = I 1 + I 2I 1 = π40 tan xd (tan x + 1)(1 + tan x )2 = π40 d (1 + tan x )1 + tan x − π40 d (1 + tan x )(1 + tan x )2

= ln |1 + tan x |

π

4

0 + 1

1 + tan x

π

4

0 = ln2 + 11 + π4 −1

I 2 = π

4

0

xd (1 + tan x )(1 + tan x )2

= − π

4

0xd

11 + tan x

= −x1 + tan x

π

4

0+

π

4

0

dx1 + tan x

= −π8

+ π

4

0

cos xdx1 + sin x

= −π8

+ π

4

0

d (1 + sin x )1 + sin x

= −π8

+ ln |1 + sin x |π

4

0= −π

8 + ln

2 + √ 22

V y I = ln 2 + √ 2 − π4 + π −

π8

Bài 6 Tính tích phânI =

π

6

0

tan x + x tan2 xcos2 2x

dx

L i gi i

Ta có: I = π

6

0

tan xcos2 2x

dx + π

6

0

x tan2 xcos2 2x

dx = I 1 + I 2

I 1 = π

6

0

sin xcos x (2cos2 x −1)

2 dx = 1

√ 32

dt

t (2t 2 −1)2 =

1

√ 32

− 2t

2t 2 −1 +

2t(2t 2 −1)

2 + 1t

dt

= −12 ln 2t2 −1 − 12 (2t 2 −1)

+ ln |t |1

√ 32

= 12 − 12 ln 32

I 2 = 12

π

6

0x tan2 xd (tan2 x ) =

12

x tan 2 2xπ

6

0 − 12

π

6

0tan2 xd (x tan2 x )

= π4 −

12

π

6

0tan 2 2xdx −

12

π

6

0

x tan2 xcos2 2x

dx = π4 −

12

π

6

0(tan 2 2x + 1) dx +

12

π

6

0dx −I 2

⇒I 2 =

π8

+ π24 −

18

π

6

0d(tan2 x ) =

π6 −

√ 38

V y I = 12 − 12 ln 32 + π6 −√ 38


I = e

1

(1 −x ln x ) ln x 4√ ln x

xe 2 x 4√ ex dx

L i gi i

I = e

1

1 −x ln xxe x

ln xex

4 ln xex dxĐ t t = 4 ln xex . Suy ra: t4 = ln xex =⇒ 4t 3 dt =

1x

ex −ex ln xe2 x

dx =⇒ 4t4 dt =

1 −x ln xxe x

4 ln xex dxDo đó, I = 4

e

1t 8 dt =

49

t 9e

1=

49

(e9 −1).

4


5/7 w w w

. k 2 p i

. n e tBài 8 Tính tích phân I = π30 1 + 2cos x(2 + cos x )2 dx

L i gi i

Ta có: 1 + 2cos x(2 + cos x )2 =

u(x )2 + cos x =

u (x )(2 + cos x ) + u(x )sin x(2 + cos x )2 suy ra u(x ) = sin x

V y I = sinx

2 + cos x

π

3

0=

√ 35


I = 2

1

1(2x 2 + 3) √ x 2 + 2 dx

L i gi i

Đ t : √ x 2 + 2 = x + t⇒2xt = 2 −t 2 ⇒x = 2 −t2

2t = 1

t − t2 ⇒dx = −1t 2 − 12 dt,√ x 2 + 2 = x + t = 1

t +

t2

. Đ i c n : x = 1⇒

t = √ 3 −1, x = 2⇒t = √ 6 −2 T đó :

I = − √ 6 −2

√ 3 −12 + t 2 .2t

2t 2 (2 + t2 ) 2 2 −t 22 t2

+ 3dt = −

√ 6 −2

√ 3 −12t

(2 −t 2 )2 + 6 t 2

dt

= − √ 6 −2

√ 3 −12t

(t 4 + 2 t 2 + 4)dt = −

√ 6 −2

√ 3 −12t

(t 2 + 1) 2 + 3dt

Đ n đây ta l i đ t : t2

+ 1 = u⇒2tdt = du

Đ i c n : t = √ 3 −1⇒u = 4 −2√ 3, t = √ 6 −2⇒u = 10 −4√ 6Và ta có : ⇒I = −

10 −4 √ 6

4 −2 √ 3du

u 2 + 3


I = 1

0

a x (x ln a + 1)(a x + x ln a + 2) 2

dx

L i gi i Đ t t = x ln a =⇒ a

x = e t

I = 1ln a

ln a

0

e t (t + 1)(e t + t + 2) 2

dt = − 1ln a

ln a

0

e t (t + 1)e t + 1

d 1e t + t + 2

I = − 1ln a

e t (t + 1)e t + 1

1e t + t + 2

ln a

0+

1ln a

ln a

0

1e t + t + 2

d et (t + 1)e t + 1


I =

2

1

2

x3

+ 2 x dx

L i gi i Ta có:

5


6/7 w w w

. k 2 p i

. n e tBài 2 Tính tích phân I = 21 2x 3 + 2 x dx

L i gi i

Ta có:Bài 3 Tính tích phân

I = 2

1

2x 3 + 2 x

dx

L i gi i Ta có:


I = 2

12x 3 + 2 x dx

L i gi i Ta có:


I = 2

1

2x 3 + 2 x

dx

L i gi i Ta có:


I = 2

1

2x 3 + 2 x

dx

L i gi i Ta có:


I = 2

1

2x 3 + 2 x

dx

L i gi i Ta có:


I =

2

1

2

x3

+ 2 x dx

L i gi i Ta có:

6


7/7w w w

. k 2 p i

. n e tBài 9 Tính tích phân I = 21 2x 3 + 2 x dx

L i gi i

Ta có:Bài 10 Tính tích phân

I = 2

1

2x 3 + 2 x

dx

L i gi i Ta có:

7

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