(c) 2014 joan s kessler distancemath.com 11 Integration By Parts.

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Integration By Parts

sin( )x x dx

2 xx e dxln x dx

cos xe x dx

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Suppose we want to integrate this function.

Up until now we have no way of doing this.

x, and sin(x) seem totally unrelated.

If u(x) is a function and v(x) is another function we seem to have

It almost seems like the reverse of the product rule. Let’s explore the product rule.

sin( )x x dx

u v dx

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Integration By Parts

Start with the product rule:

d dv duuv u v

dx dx dx

d uv u dv v du

u dv d uv v du

u dv d uv v du

u dv d uv v du

u dv uv v du

This is the Integration by Parts formula.

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u dv uv v du

The Integration by Parts formula is a “product The Integration by Parts formula is a “product rule” for integration.rule” for integration.

u differentiates to zero (usually).

dv is easy to integrate.

Choose u in this order: LIPET

Logs, Inverse trig, Polynomial, Exponential, Trig

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Example 1:

cos x x dxpolynomial factor u x

du dx

cos dv x dx

sinv x

u dv uv v du u dv uv v du LIPETLIPET

sin cosx x x C

u v v du

sin sin x x x dx

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Example 2:

ln x dxlogarithmic factor lnu x

1du dx

x

dv dx

v x

u dv uv v du LIPET

lnx x x C

1ln x x x dx

x

u v v du

ln x dx

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This is still a product, so we need to use integration by parts againagain.

Example 3:2 xx e dx

u dv uv v du LIPET

2u x xdv e dx

2 du x dx xv e u v v du

2 2 x xx e e x dx 2 2 x xx e xe dx u x xdv e dx

du dx xv e 2 2x x xx e xe e dx

2 2 2x x xx e xe e C 2 xx e dx

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Example 4:

cos xe x dxLIPET

xu e sin dv x dx xdu e dx cosv x

u v v du sin sinx xe x x e dx

sin cos cos x x xe x e x x e dx

xu e cos dv x dx xdu e dx sinv x

sin cos cos x x xe x e x e x dx

This is the expression we started with!

uv v du

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Example 5:

cos xe x dxLIPET

u v v du

cos xe x dx 2 cos sin cosx x xe x dx e x e x

sin coscos

2

x xx e x e x

e x dx C

sin sinx xe x x e dx

xu e sin dv x dxxdu e dx cosv x

xu e cos dv x dx xdu e dx sinv x

sin cos cos x x xe x e x e x dx

sin cos cos x x xe x e x x e dx

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Example 5 :

cos xe x dx u v v du

This is called “solving for the unknown integral.”

It works when both factors integrate and differentiate forever.

cos xe x dx 2 cos sin cosx x xe x dx e x e x

sin coscos

2

x xx e x e x

e x dx C

sin sinx xe x x e dx

sin cos cos x x xe x e x e x dx

sin cos cos x x xe x e x x e dx

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More integration by Parts Ex 6.More integration by Parts Ex 6.

u v v du

2

3arcsin 3

1- 9x x x dx

x

2 -1/2arcsin 3 3 (1 9 ) x x x x dx

arcsin 3x dx Let u = arcsin3x dv = dx

v = x2

3

1 9du dx

x

2 1/ 21arcsin 3 (1 9 )

3x x x C

-18 -118

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More integration by Parts Ex. 7More integration by Parts Ex. 7

u v v du

22 2

1 1(2x)

2( 5) 2( 5)x dx

x x

22

2

1ln( 5)

2( 5) 2

xx C

x

3

2 2( 5)

xdx

x Let u = x2 du = 2x dx

2 2( 5)

xdv dx

x

2

1

2( 5)v

x

Formdu

u

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A Shortcut: Tabular Integration

Tabular integration works for integrals of the form:

f x g x dxwhere:

Differentiates to zero in several steps.

Integrates repeatedly.

u dv

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2 xx e dx de .& v rif x integ & ralsg x

2x

2x

20

xexe

xexe

2 xx e dx 2 xx e 2 xxe 2 xe C

Compare this with the same problem done the other way:

Alternate signs

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Same Example :2 xx e dx

u dv uv v du LIPET

2u x xdv e dx

2 du x dx xv e u v v du

2 2 x xx e e x dx 2 2 x xx e xe dx u x xdv e dx

du dx xv e 2 2x x xx e xe e dx

2 2 2x x xx e xe e C This is easier and quicker to do with tabular integration!

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2 2 2x x xx e xe e C

2( 2 2)xe x x C

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3 sin x x dx

3x

23x

6x

6

sin x

cos x

sin xcos x

0

sin x

3 cosx x 2 3 sinx x 6 cosx x 6sin x + C

de .& v rif x integ & ralsg x

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5 sinx dx5

4

3

2

sin

5 cos

20 sin

60 cos

120 sin

120 cos

0 sin

x x

x x

x x

x x

x x

x

x

5 5 4 3 2sin cos 5 sin 20 cos 60 sin 120 cos 120sinx dx x x x x x x x x x x x C

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5(3 2) xx e dx

3 2x

3

5xe5

5

xe

5

25

xe

0

5(3 2)

5

xx e 53

25

xe

5 53 7

5 25

x xxe eC + C

de .& v rif x integ & ralsg x

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5 53 7

5 25

x xxe eC

51 (15 7)

25xe x C

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Try

3 cos 2x xdx

3 2xx e dx

32 2( 2)x x dx

2 3 214 6 6 3

8xe x x x C

3 214 sin 2 6 cos 2 6 sin 2 3cos 2

8x x x x x x x C

5

222

( 2) 35 40 32315

x x x C

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3 2xx e dx3 2

2 2

2

2

2

13

21

641

681

016

x

x

x

x

x

x e

x e

x e

e

e

++

--

++

--

++

2 3 214 6 6 3

8xe x x x C

3 2 2 2 2 22 3 3 3

2 4 4 8

x x xxx x e xe x e

e C

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3 cos 2x xdx de .& v rif x integ & ralsg x

3 214 sin 2 6 cos 2 6 sin 2 3cos 2

8x x x x x x x C

3

2

cos 2

13 sin 2

21

6 cos 241

6 sin 281

0 cos 216

x x

x x

x x

x

x

3 2sin 2 3 cos 2 3 sin 2 3cos 2

2 4 4 8

x x x x x x xC ++

--

++

--

++

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32 2( 2)x x dx

5

222

( 2) 35 40 32315

x x x C

32 2

5

2

7

2

9

2

( 2)

22 ( 2)

5

42 ( 2)

35

80 ( 2)

315

x x

x x

x

x

++

--

++

--

Homework

Assignment

25