calc 7.2(10)

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Integration by Parts

Objective: To integrate problems

without a u-substitution

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When integrating the product of twofunctions, we often use a u-substitution tomake the problem easier to integrate.

Sometimes this is not possible. We needanother way to solve such problems.

)()( xgxf

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As a rst step, we will take the derivativeof

)()( xgxf

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[ ] )()()()()()( //

xfxgxgxfxgxfdx

d

+=

)()( xgxf

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[ ] )()()()()()( //

xfxgxgxfxgxfdx

d

+=

)()( xgxf

[ ] )()()()()()( // xfxgxgxfxgxfdx

d +=

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[ ] )()()()()()( //

xfxgxgxfxgxfdx

d

+=

)()( xgxf


d +=

[ ] )()()()()()( // xfxgxgxfxgxf +=

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[ ] )()()()()()( //

xfxgxgxfxgxfdx

d+=

)()( xgxf


d +=

[ ] )()()()()()( // xfxgxgxfxgxf +=

[ ] )()()()()()( // xgxfxfxgxgxf =

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Now lets make some substitutions to makethis easier to apply.

)(xgv=)(xfu=

[ ] )()()()()()( // xgxfxfxgxgxf =

)(/ xgdv=)(/ xfdu=

= udvvduuv

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his is the way we will look at theseproblems.

he two functions in the original problem weare integrating are u and dv. he rst thingwe will do is to choose one function for u andthe other function will be dv.

)(xgv=)(xfu=

)(/ xgdv=)(/ xfdu= = udvvduuv

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Example 1

!se integration by parts to evaluate xdxx cos

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Example 1

!se integration by parts to evaluate

xv sin=xu=

xdxdv cos=dxdu=

xdxx cos

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Example 1


xv sin=xu=

xdxdv cos=dxdu=

xdxx cos

= xdxxxxdxx sinsincos

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Example 1


xv sin=xu=

xdxdv cos=dxdu=

xdxx cos

= xdxxxxdxx sinsincos

Cxxxxdxx ++= cossincos

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Guielines

he rst step in integration by parts is tochoose u and dv to obtain a new integralthat is easier to evaluate than the original.

"n general, there are no hard and fastrules for doing this# it is mainly a matter ofe$perience that comes from lots of

practice.

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Guielines

here is a useful strategy that may help whenchoosing u and dv. When the integrand is aproduct of two functions from di%erent categoriesin the following list , you should make u the

function whose category occurs earlier in the list.

&ogarithmic, "nverse rig,Algebraic, rig,'$ponential

he acronym &"A'may help you remember theorder.

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Guielines

"f the new integral is harder than theoriginal, you made the wrong choice. &ookat what happens when we make di%erent

choices for u and dv in e$ample (.

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Guielines

"f the new integral is harder that theoriginal, you made the wrong choice. &ookat what happens when we make di%erent

choices for u and dv in e$ample (. xdxxcos

xu cos=

xdxdu sin=

2

2

xv=

xdxdv=

+= xdxx

xx

xdxx sin2

cos2

cos22

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Guielines

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

xdxxcosxu cos=

xdxdu sin=

2

2

xv=

xdxdv=

+= xdxx

xx

xdxx sin2

cos2

cos22

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Example !

!se integration by parts to evaluate dxxex

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Example !


xev=xu=

dxedv x=dxdu=

dxxex

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Example !


xev=xu=

dxedv x=dxdu=

dxxex

= dxexedxxe xxx

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Example !


xev=xu=

dxedv x=dxdu=

dxxex

= dxexedxxe xxx

Cexedxxe xxx +=

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Example "

!se integration by parts to evaluate xdxln

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Example "


xv=xu ln=

dxdv=dxx

du 1=

xdxln

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Example "


xv=xu ln=

dxdv=dxx

du 1=

xdxln

= dxxxxdx lnln

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Example "


xv=xu ln=

dxdv=dxx

du 1=

xdxln

= dxxxxdx lnln

MEMORIZE

Cxxxxdx += lnln

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Example #$repeate%

!se integration by parts to evaluate dxex x2

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Example #$repeate%


xev

=

2xu=

dxedv x=xdxdu 2=

dxex x2

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Example #$repeate%


xev

=

2xu=

dxedv x=xdxdu 2=

dxex x2

+= dxxeexdxex xxx 222

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Example #$repeate%


xev

=2xu=

dxedv x=xdxdu 2=

dxex x2

+= dxxeexdxex xxx 222xu= xev =

dxedv x

=dxdu=

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Example #$repeate%


xev

=2xu=

dxedv x=xdxdu 2=

dxex x2

+= dxxeexdxex xxx 222xu=

dxdu=

xev

=

dxedv x

=

[ ] ++= dxexeexdxex xxxx 222

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Example #$repeate%


xev =2xu=

dxedv

x=xdxdu 2=

dxex x2

+= dxxeexdxex xxx 222

xu= xev =

dxedv

x

=[ ] ++= dxexeexdxex xxxx 222

Cexeexdxex xxxx +=

2222

dxdu=

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Example &

!se integration by parts to evaluate xdxex cos

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Example &


xev=xu cos=

dxedv x=xdxdu sin=

xdxex cos

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Example &


xev=xu cos=

dxedv x=xdxdu sin=

xdxex cos

xdxexexdxe xxx sincoscos +=

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Example &


xev=xu cos=

dxedv x=xdxdu sin=

xdxex cos

xdxexexdxe xxx sincoscos +=xu sin=

xdxdu cos=

xev=

dxedv x

=

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Example &


xev=xu cos=

dxedv x=xdxdu sin=

xdxex cos


xdxdu cos=

xev=

dxedv x

=xdxexexexdxe xxxx cossincoscos +=

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Example &


xev=xu cos=

dxedv x=xdxdu sin=

xdxex cos


xdxdu cos=

xev=

dxedv x

=xdxexexexdxe xxxx cossincoscos +=

xexexdxe xxx cossincos2 +=

Cxexe

xdxexx

x ++

= 2cossin

cos

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Example &


xv sin=xeu=

xdxdv cos=dxedu x

=

xdxex cos

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Example &


xv sin=xeu=

xdxdv cos=dxedu x

=

xdxex cos

= xdxexexdxe xxx sinsincos

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Example &


xv sin=xeu=

xdxdv cos=dxedu x=

xdxex cos


xeu=

dxedu x

= xdxdv

sin=

xv cos=

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Example &


xv sin=xeu=

xdxdv cos=dxedu x=

xdxex cos


xeu=

dxedu x

= xdxdv

sin=

xv cos=

[ ] = xdxexexexdxe xxxx coscossincos

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Example &

!se integration by parts to evaluate xdxex cos


+= xdxexexexdxe xxxx coscossincos

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Example &



xdxex cos


xexexdxe

xxx

cossincos2 +=

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Example &



xdxex cos


xexexdxe

xxx

cossincos2 +=

C

xexexdxe

xxx +

+= 2

cossincos

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Tabular Integration

"ntegrals of the form wherep)$* is a polynomial can sometimes beevaluated using a method called abular

"ntegration.

dxxfxp )()(

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Tabular Integration

"ntegrals of the formwhere p)$* is a polynomial can sometimesbe evaluated using a method called

abular "ntegration.(. +i%erentiate p)$* repeatedly until you

obtain , and list the results in the rstcolumn.

dxxfxp )()(

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Tabular Integration

"ntegrals of the form wherep)$* is a polynomial can sometimes beevaluated using a method called abular

"ntegration.(. +i%erentiate p)$* repeatedly until you

obtain , and list the results in the rstcolumn.

. "ntegrate f)$* repeatedly until you havethe same number of terms as in the rstcolumn. &ist these in the second column.

dxxfxp )()(

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Tabular Integration

"ntegrals of the form wherep)$* is a polynomial can sometimes beevaluated using a method called abular"ntegration.

(. +i%erentiate p)$* repeatedly until you obtain, and list the results in the rst column.

. "ntegrate f)$* repeatedly until you have the

same number of terms as in the rst column.&ist these in the second column.

. +raw diagonal arrows from term n in column (to term n/( in column two with alternating

signs starting with /. his is your answer.

dxxfxp )()(

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Example '

!se tabular integration to nd dxxx 12

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Example '

!se tabular integration to nd

0olumn (

dxxx 12

2x

x2

20

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Example '


0olumn ( 0olumn

dxxx 12

2x

x2

20

1x

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

2/5

)1)(15/4( x

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

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Example '


0olumn ( 0olumn

dxxx 12

2x

x2

20

1x

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

2/5

)1)(15/4( x

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

+

+

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Example '


0olumn ( 0olumn

dxxx 12

2x

x2

20

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(

+

+

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Example (

'valuate the following denite integral

1

0

1tan xdx

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Example (


xu 1tan=

1

0

1tan xdx

211x

du+

= dxdv=

xv=

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Example (


xu 1tan=

1

0

1tan xdx

21

1

xdu

+= dxdv= xv=

+=

2

1

1

0

1

1tantan

x

xdxxxdx

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Example (


xu 1tan=

1

0

1tan xdx

21

1

xdu

+= dxdv= xv=

+=

2

1

1

0

1

1tantan

x

xdxxxdx

21 xu +=

dxx

du=

2

xdxdu 2=

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Example (


xu 1tan=

1

0

1tan xdx

21

1

xdu

+= dxdv= xv=

+=

2

1

1

0

1

1tantan

x

xdxxxdx

21 xu +=

dxx

du=

2

xdxdu 2=

=

u

du

xxdx 2

1tantan

1

1

0

1

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Example (


xu 1tan=

1

0

1tan xdx

21

1

xdu

+= dxdv= xv=

+=

2

1

1

0

1

1tantan

x

xdxxxdx

21 xu +=

dxx

du=

2

xdxdu 2=

=

u

du

xxdx 2

1tantan

1

1

0

1

)1ln(

2

1tantan 21

1

0

1xxxdx +=

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Example (


1

0

1tan xdx

)1ln(

2

1tantan 21

1

0

1xxxdx +=

)01ln(2

10tan0)11ln(

2

11tan1tan 2121

1

0

1 +++= dx

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Example (


1

0

1tan xdx

)1ln(

2

1tantan 21

1

0

1xxxdx +=

)01ln(2

10tan0)11ln(

2

11tan1tan 2121

1

0

1 +++= dx

2ln4

002ln2

1

4tan

1

0

1 =+=

dx

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)omewor*

Section 1.

2age 345

- multiples of

calc 7.2(10)

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