april 2009 纪光 - 北京景山学校 integration by parts – p.1 how to integrate some special...

April 2009 纪光 - 北京景山学校Integration by parts – p.

1

How to integrate some special products of two functions

x sin x dx0

2

1. Not all functions can be integrated by using simple derivative formulas backwards …

2. Some functions look like simple products but cannot be integrated directly.

Ex. f(x) = x .sin x

but … u’.v’≠ (u.v)’ So what ?!?

April 2009 纪光 - 北京景山学校 2

x x 2

2

sin x cos x

April 2009 纪光 - 北京景山学校 Integration by parts – p.3

Reminder 1Derivative of composite functions

if g(x) = f[u(x)]and u and f have derivatives,

theng’(x) = f’[u(x)] . u’(x)

hense

f [u(x)]u'(x) dx f [u(x)]


Examples of integrals products of composite functions

cos(x 2) (k)

1

2sin2x (k)

cos x sin x dx

1

2sin2 x (k)

1

(1 x 2)(k)

cos2x dx

2x

(1 x 2)2 dx

2x sin(x 2) dx

x

1 x 2 dx

(1 x 2 (k)


Formulas of integrals of products made of composite functions

dxuu n

un1

n 1(n 1)

dxuu .cos

sinu

ln u dx

u

u

dxeu u

eu


Reminder 2Derivative of the Product of 2

functionsIf u and v have derivatives u’ and v’,then

(u.v)’ = u’.v + u.v’u.v’ = (u.v)’ - u’.v

hence by integration of both sides

u. v dx uv a

ba

b

u .v dxa

b

u v uv u v


x sin x dx0

2

u = 1st part(to derive)

u = 1st part(to derive)

v’ = 2nd part(to integrate)

u’ =(x)’ = 1u’ =(x)’ = 1

u = x

sin x = (- cos x)’

v’ = sin x

u’v = 1.(- cos x)

v = - cos x


x sin x dx0

2

x sin x dx x( cos x)' dx x( cos x) ( cos x) dx x sin x dx

0

2 sin x x cos x 0

2 1

u v uv u v

u = x u’ = 1 v’ = sin x v = - cos x


x 2 cos x dx0

2

1st part(to derive)

1st part(to derive)

2nd part(to integrate)

u’ =(x2)’ = 2xu’ =(x2)’ = 2x

u = x2

cos x = (sin x)’

v’ = cos x

u’v = 2x.sin x

v = sin x


10

x 2 cos x dx0

2

u = x2 u’ = 2x v’ = cos x v = sin x

x 2 cos x dx x 2(sin x ) dx x 2 sin x 2x.sin x dx x 2 cos x dx

0

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

0

2

2

4 2

u v uv u v


11

(ln x) x dx1

e1st part

(to derive)1st part

(to derive)2nd part

(to integrate)

u’ =(ln x)’ =u’ =(ln x)’ =

u = ln x

1

x

u’v =

x

2

x = =>v = .

v’ = x

x 2

2

x 2

2


12

(ln x)x dx ln x.x 2

2

dx ln xx 2

2

1

x.x 2

2dx

(ln x)x dx1

e ln xx 2

2 x 2

4

1

e

e2 1

4

u v uv u v

(ln x) x dx1

e u = ln x u’ = v’ = x v =

1

x

x 2

2


13

ln x dx1

e1st part

(to derive)1st part

(to derive)2nd part

(to integrate)

u’ =(ln x)’ =u’ =(ln x)’ =

u = ln x

1

x 1 = (x)’=> v = x

v’ = 1

u’v = 1


14

ln x dx ln x.(x ) dx (ln x)x 1

xx dx

(ln x)x dx1

e x ln x x 1e 1

u v uv u v

u = ln x u’ = v’ = 1 v = x

1

x

ln x dx1

e ln x.1 dx1

e


15

ln x 2dx

1

e1st part

(to derive)1st part

(to derive)2nd part

(to integrate)

u’ =(ln x)’ =u’ =(ln x)’ =

u = ln x

1

x ln x = (x ln x – x)’

v’ = ln x

u’v = ln x - 1

v = x ln x – x


16

(ln x)2 dx ln x.(x ln x x ) dx (ln x)(x ln x x) (ln x 1) dx (ln x)x dx

1

e x(ln x)2 2x(1 ln x) 1

ee 2

u v uv u v

u = ln x u’ = v’ = ln x v = x.ln x - x

1

x

ln x 2dx

1

e ln x ln x dx1

e


17

Use the IBP formula to calculate :

€

I = x. exdx1

e

∫


18

Use the IBP formula to calculate :

€

I =ln x

x2dx

1

e 2

∫


19

Use the IBP formula to prove that :

€

In+1 = −2n+1

e2+(n +1)In€

In =ln x( )

n

x 2dx

1

e 2

∫

Let , for any n > 0 :

Then find I2, I3, I4


20

祝好云谢谢再见

april 2009 纪光 - 北京景山学校 integration by parts – p.1 how to integrate some special...

Documents

x ln x x v

x ln x x slide

x sin x

ln x x slide

ln x ln x

ln x v

cos x v

sin x v

april 2009 纪光 - 北京 景山学校 integration by parts – p.1 how to integrate some special...

april 2009 纪光 - 北京景山学校 integration by parts – p.1 how to integrate some special...