Integration by Parts

Objective: To integrate problems without a u-substitution

Integration by Parts

• When integrating the product of two functions, we often use a u-substitution to make the problem easier to integrate. Sometimes this is not possible. We need another way to solve such problems.

)()( xgxf

Integration by Parts

• As a first step, we will take the derivative of )()( xgxf

Integration by Parts

• As a first step, we will take the derivative of

)()()()()()( // xfxgxgxfxgxfdx

d

)()( xgxf

Integration by Parts

• As a first step, we will take the derivative of

)()()()()()( // xfxgxgxfxgxfdx

d

)()( xgxf

)()()()()()( // xfxgxgxfxgxfdx

d

Integration by Parts

• As a first step, we will take the derivative of

)()()()()()( // xfxgxgxfxgxfdx

d

)()( xgxf

)()()()()()( // xfxgxgxfxgxfdx

d

)()()()()()( // xfxgxgxfxgxf

Integration by Parts

• As a first step, we will take the derivative of

)()()()()()( // xfxgxgxfxgxfdx

d

)()( xgxf

)()()()()()( // xfxgxgxfxgxfdx

d

)()()()()()( // xfxgxgxfxgxf

)()()()()()( // xgxfxfxgxgxf

Integration by Parts

• Now lets make some substitutions to make this easier to apply.

)(xgv )(xfu

)()()()()()( // xgxfxfxgxgxf

)(/ xgdv )(/ xfdu

udvvduuv

Integration by Parts

• This is the way we will look at these problems.

• The two functions in the original problem we are integrating are u and dv. The first thing we will do is to choose one function for u and the other function will be dv.

)(xgv )(xfu

)(/ xgdv )(/ xfdu udvvduuv

Example 1

• Use integration by parts to evaluate xdxx cos

Example 1

• Use integration by parts to evaluate

xv sinxu

xdxdv cosdxdu

xdxx cos

Example 1

• Use integration by parts to evaluate

xv sinxu

xdxdv cosdxdu

xdxx cos

xdxxxxdxx sinsincos

Example 1

• Use integration by parts to evaluate

xv sinxu

xdxdv cosdxdu

xdxx cos

xdxxxxdxx sinsincos

Cxxxxdxx cossincos

Guidelines

• The first step in integration by parts is to choose u and dv to obtain a new integral that is easier to evaluate than the original. In general, there are no hard and fast rules for doing this; it is mainly a matter of experience that comes from lots of practice.

Guidelines

• There is a useful strategy that may help when choosing u and dv. When the integrand is a product of two functions from different categories in the following list , you should make u the function whose category occurs earlier in the list.

• Logarithmic, Inverse Trig, Algebraic, Trig, Exponential

• The acronym LIATE may help you remember the order.

Guidelines

• If the new integral is harder than the original, you made the wrong choice. Look at what happens when we make different choices for u and dv in example 1.

Guidelines

• If the new integral is harder that the original, you made the wrong choice. Look at what happens when we make different choices for u and dv in example 1.

xdxx cosxu cos

xdxdu sin

2

2xv

xdxdv

xdxx

xx

xdxx sin2

cos2

cos22

Guidelines

• Since the new integral is harder than the original, we made the wrong choice.

xdxx cosxu cos

xdxdu sin

2

2xv

xdxdv

xdxx

xx

xdxx sin2

cos2

cos22

Example 2

• Use integration by parts to evaluate dxxex

Example 2

• Use integration by parts to evaluate

xev xu

dxedv xdxdu

dxxex

Example 2

• Use integration by parts to evaluate

xev xu

dxedv xdxdu

dxxex

dxexedxxe xxx

Example 2

• Use integration by parts to evaluate

xev xu

dxedv xdxdu

dxxex

dxexedxxe xxx

Cexedxxe xxx

Example 3

• Use integration by parts to evaluate xdxln

Example 3

• Use integration by parts to evaluate

xv xu ln

dxdv dxx

du1

xdxln

Example 3

• Use integration by parts to evaluate

xv xu ln

dxdv dxx

du1

xdxln

dxxxxdx lnln

Example 3

• Use integration by parts to evaluate

xv xu ln

dxdv dxx

du1

xdxln

dxxxxdx lnln

MEMORIZE

Cxxxxdx lnln

Example 4(repeated)

• Use integration by parts to evaluate dxex x2

Example 4(repeated)

• Use integration by parts to evaluate

xev 2xu

dxedv xxdxdu 2

dxex x2

Example 4(repeated)

• Use integration by parts to evaluate

xev 2xu

dxedv xxdxdu 2

dxex x2

dxxeexdxex xxx 222

Example 4(repeated)

• Use integration by parts to evaluate

xev 2xu

dxedv xxdxdu 2

dxex x2

dxxeexdxex xxx 222xu xev

dxedv xdxdu

Example 4(repeated)

• Use integration by parts to evaluate

xev 2xu

dxedv xxdxdu 2

dxex x2

dxxeexdxex xxx 222xu

dxdu

xev

dxedv x

dxexeexdxex xxxx 222

Example 4(repeated)

• Use integration by parts to evaluatexev 2xu

dxedv xxdxdu 2

dxex x2

dxxeexdxex xxx 222

xu xev

dxedv x dxexeexdxex xxxx 222

Cexeexdxex xxxx 2222

dxdu

Example 5

• Use integration by parts to evaluate xdxex cos

Example 5

• Use integration by parts to evaluatexev xu cos

dxedv xxdxdu sin

xdxex cos

Example 5

• Use integration by parts to evaluatexev xu cos

dxedv xxdxdu sin

xdxex cos

xdxexexdxe xxx sincoscos

Example 5

• Use integration by parts to evaluatexev xu cos

dxedv xxdxdu sin

xdxex cos

xdxexexdxe xxx sincoscos xu sin

xdxdu cos

xev

dxedv x

Example 5

• Use integration by parts to evaluatexev xu cos

dxedv xxdxdu sin

xdxex cos

xdxexexdxe xxx sincoscos xu sin

xdxdu cos

xev

dxedv x

xdxexexexdxe xxxx cossincoscos

Example 5

• Use integration by parts to evaluatexev xu cos

dxedv xxdxdu sin

xdxex cos

xdxexexdxe xxx sincoscos xu sin

xdxdu cos

xev

dxedv x

xdxexexexdxe xxxx cossincoscos

xexexdxe xxx cossincos2

Cxexe

xdxexx

x

2

cossincos

Example 5

• Use integration by parts to evaluate

xv sinxeu

xdxdv cosdxedu x

xdxex cos

Example 5

• Use integration by parts to evaluate

xv sinxeu

xdxdv cosdxedu x

xdxex cos

xdxexexdxe xxx sinsincos

Example 5

• Use integration by parts to evaluate

xv sinxeu

xdxdv cosdxedu x

xdxex cos

xdxexexdxe xxx sinsincos xeu

dxedu x xdxdv sin

xv cos

Example 5

• Use integration by parts to evaluate

xv sinxeu

xdxdv cosdxedu x

xdxex cos

xdxexexdxe xxx sinsincos xeu

dxedu x xdxdv sin

xv cos

xdxexexexdxe xxxx coscossincos

Example 5

• Use integration by parts to evaluate xdxex cos

xdxexexexdxe xxxx coscossincos

xdxexexexdxe xxxx coscossincos

Example 5

• Use integration by parts to evaluate

xdxexexexdxe xxxx coscossincos

xdxex cos

xdxexexexdxe xxxx coscossincos

xexexdxe xxx cossincos2

Example 5

• Use integration by parts to evaluate

xdxexexexdxe xxxx coscossincos

xdxex cos

xdxexexexdxe xxxx coscossincos

xexexdxe xxx cossincos2

Cxexe

xdxexx

x

2

cossincos

Tabular Integration

• Integrals of the form where p(x) is a polynomial can sometimes be evaluated using a method called Tabular Integration.

dxxfxp )()(

Tabular Integration

• Integrals of the form where p(x) is a polynomial can sometimes be evaluated using a method called Tabular Integration.

1. Differentiate p(x) repeatedly until you obtain 0, and list the results in the first column.

dxxfxp )()(

Tabular Integration

• Integrals of the form where p(x) is a polynomial can sometimes be evaluated using a method called Tabular Integration.

1. Differentiate p(x) repeatedly until you obtain 0, and list the results in the first column.

2. Integrate f(x) repeatedly until you have the same number of terms as in the first column. List these in the second column.

dxxfxp )()(

Tabular Integration

• Integrals of the form where p(x) is a polynomial can sometimes be evaluated using a method called Tabular Integration.

1. Differentiate p(x) repeatedly until you obtain 0, and list the results in the first column.

2. Integrate f(x) repeatedly until you have the same number of terms as in the first column. List these in the second column.

3. Draw diagonal arrows from term n in column 1 to term n+1 in column two with alternating signs starting with +. This is your answer.

dxxfxp )()(

Example 6

• Use tabular integration to find dxxx 12

Example 6

• Use tabular integration to find

Column 1

dxxx 12

2x

x2

2

0

Example 6

• Use tabular integration to find

Column 1 Column 2

dxxx 12

2x

x2

2

0

1x

2/3)1)(3/2( x

2/5)1)(15/4( x

2/7)1)(105/8( x

Example 6

• Use tabular integration to find

Column 1 Column 2

dxxx 12

2x

x2

2

0

1x

2/3)1)(3/2( x

2/5)1)(15/4( x

2/7)1)(105/8( x

Example 6

• Use tabular integration to find

Column 1 Column 2

dxxx 12

2x

x2

2

0

1x

2/3)1)(3/2( x

2/5)1)(15/4( x

2/7)1)(105/8( x

Cxxxxx 2/72/52/32 )1)(105/16()1()15/8()1()3/2(

Example 7

• Evaluate the following definite integral 1

0

1tan xdx

Example 7

• Evaluate the following definite integral

xu 1tan

1

0

1tan xdx

21

1

xdu

dxdv

xv

Example 7

• Evaluate the following definite integral

xu 1tan

1

0

1tan xdx

21

1

xdu

dxdv xv

21

1

0

1

1tantan

x

xdxxxdx

Example 7

• Evaluate the following definite integral

xu 1tan

1

0

1tan xdx

21

1

xdu

dxdv xv

21

1

0

1

1tantan

x

xdxxxdx

21 xu

dxx

du

2

xdxdu 2

Example 7

• Evaluate the following definite integral

xu 1tan

1

0

1tan xdx

21

1

xdu

dxdv xv

21

1

0

1

1tantan

x

xdxxxdx

21 xu

dxx

du

2

xdxdu 2

u

duxxdx2

1tantan 1

1

0

1

Example 7

• Evaluate the following definite integral

xu 1tan

1

0

1tan xdx

21

1

xdu

dxdv xv

21

1

0

1

1tantan

x

xdxxxdx

21 xu

dxx

du

2

xdxdu 2

u

duxxdx2

1tantan 1

1

0

1

)1ln(2

1tantan 21

1

0

1 xxxdx

Example 7

• Evaluate the following definite integral 1

0

1tan xdx

)1ln(2

1tantan 21

1

0

1 xxxdx

)01ln(2

10tan0)11ln(

2

11tan1tan 2121

1

0

1 dx

Example 7

• Evaluate the following definite integral 1

0

1tan xdx

)1ln(2

1tantan 21

1

0

1 xxxdx

)01ln(2

10tan0)11ln(

2

11tan1tan 2121

1

0

1 dx

2ln4

002ln2

1

4tan

1

0

1 dx

Homework

• Section 7.2• Page 498• 3-30 multiples of 3