April 2011 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 2011 纪光 - 北京 景山学校 2

x x 2

2

sin x cos x

April 2011 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)

hence

f [u(x)]u '(x) dx∫ = f [u(x)]

April 2011 纪光 - 北京 景山学校 Integration by parts – p.4

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+ x2 (+k)

April 2011 纪光 - 北京 景山学校 Integration by parts – p.5

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

April 2011 纪光 - 北京 景山学校 Integration by parts – p.6

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

April 2011 纪光 - 北京 景山学校 Integration by parts – p.7

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

April 2011 纪光 - 北京 景山学校 Integration by parts – p.8

x sin x dx0

2

x sin x dx = x(−cosx)' dx =x(−cosx)− (−cosx) dx⇒ xsinx dx

0

2 = sinx−xcosx[ ]0

2 =1

u v uv− uv

u = x u’ = 1 v’ = sin x v = - cos x

April 2011 纪光 - 北京 景山学校 Integration by parts – p.9

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

April 2011 纪光 - 北京 景山学校Integration by parts – p.

10

x 2 cos x dx0

2

u = x2 u’ = 2x v’ = cos x v = sin x

x2 cos x dx = x2 (sinx ) dx =x2 sinx− 2x.sinx dx⇒ x2 cosx dx

0

2 = x2 sinx−2(sinx−xcosx)⎡⎣ ⎤⎦0

2 =2

4−2

u v uv− uv

April 2011 纪光 - 北京 景山学校Integration by parts – p.

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

April 2011 纪光 - 北京 景山学校Integration by parts – p.

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− uv

(ln x) x dx1

e u = ln x u’ = v’ = x v =

1

x

x 2

2

April 2011 纪光 - 北京 景山学校Integration by parts – p.

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 =

v’ = 1

u’v = 1

x

April 2011 纪光 - 北京 景山学校Integration by parts – p.

14

ln x dx = lnx.(x ) dx =(lnx)x−1x

x dx =xlnx−x

⇒ lnx dx1

e

= xlnx−x[ ]1e =1

u v =uv− uv

u = ln x u’ = v’ = 1 v = x

1

x

ln x dx1

e ln x.1 dx1

e

April 2011 纪光 - 北京 景山学校Integration by parts – p.

15

ln x 2 dx1

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

April 2011 纪光 - 北京 景山学校Integration by parts – p.

16

(ln x)2 dx = lnx.(xlnx−x ) dx =(lnx)(xlnx−x) − (lnx−1) dx⇒ (lnx)2 dx

1

e

= x(lnx)2 + 2x(1−lnx)⎡⎣ ⎤⎦1e=e−2

u v =uv− uv

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

1

x

ln x 2 dx1

e ln x ln x dx1

e

April 2011 纪光 - 北京 景山学校Integration by parts – p.

17

Use the IBP formula to calculate :

€

I = x. exdx1

e

∫

April 2011 纪光 - 北京 景山学校Integration by parts – p.

18

Use the IBP formula to calculate :

€

I =ln x

x2dx

1

e 2

∫

April 2011 纪光 - 北京 景山学校Integration by parts – p.

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 Integer n ≥ 0 :

Then find I0, I1, I2

April 2011 纪光 - 北京 景山学校Integration by parts – p.

20

In = cosx( )n dx0

2

Let , for any Integer n ≥ 0 : [Wallis Integral]

1. Use the IBP formula to prove that :

2. Calculate I0 and I1

3. Find a short formula for I2n 4. Find a short formula for I2n+1

In+2 =n+1n+ 2

In

April 2011 纪光 - 北京 景山学校Integration by parts – p.

21

F(x) = sin(lnt)dt1

x

Use twice the IBP formula to calculate

April 2011 纪光 - 北京 景山学校Integration by parts – p.

22

I = x2e−x dx0

1

Use twice the IBP formula to calculate

April 2011 纪光 - 北京 景山学校Integration by parts – p.

23

I = x(cosx)2 dx0

2 &J = x(sinx)2 dx

0

2

1.Calculate I + J

2.Use IBP to calculate I – J

3.Find I and J

April 2011 纪光 - 北京 景山学校Integration by parts – p.

24

In =1n!(1−x)nex dx0

1

1. Calculate I1

2. Find a relationship between In and In-1 (n ≥ 1)

3. Show that

4. Show that

5. Show that

In =e−1k!k=0

k=n

∑

0 ≤In ≤en!

1

n!0

∞

∑ =e