Warm-up:Evaluate the integrals.

1)

2)

dx

xex

73

dxx

x )13

1(

2

Warm-up:Evaluate the integrals.

1)

2)

dx

xex

73

dxx

x )13

1(

2

Cxex ln73

Warm-up:Evaluate the integrals.

1)

2)

dx

xex

73

dxx

x )13

1(

2

Cxex ln73

Cxx

3

sin

3

2 12

3

Integration by Parts

Section 8.2

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

xu xdxdv cos

xdxx cos

Example 1• Use integration by parts to evaluate

xv sin

xu xdxdv cos

dxdu

xdxx cos

Example 1• Use integration by parts to evaluate

xv sin

xu xdxdv cos

dxdu

xdxx cos

xdxxxxdxx sinsincos

Example 1• Use integration by parts to evaluate

xv sin

xu xdxdv cos

dxdu

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 that 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

xu dxedv x

dxxex

Example 2• Use integration by parts to evaluate

xev

xu dxedv x

dxdu

dxxex

Example 2• Use integration by parts to evaluate

xev

xu dxedv x

dxdu

dxxex

dxexedxxe xxx

Example 2• Use integration by parts to evaluate

xev

xu dxedv x

dxdu

dxxex

dxexedxxe xxx

Cexedxxe xxx

Example 3 (S):• Use integration by parts to evaluate xdxln

Example 3• Use integration by parts to evaluate

xu ln dxdv

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

Cxxxxdx lnln

Example 4 (Repeated):• Use integration by parts to evaluate dxex x2

Example 4 (Repeated):• Use integration by parts to evaluate

2xu dxedv x dxex x2

Example 4 (Repeated):• Use integration by parts to evaluate

xev

2xu dxedv x

xdxdu 2

dxex x2

Example 4 (Repeated):• Use integration by parts to evaluate

xev

2xu dxedv x

xdxdu 2

dxex x2

dxxeexdxex xxx 222

Example 4 (Repeated):• Use integration by parts to evaluate

xev

2xu dxedv x

xdxdu 2

dxex x2

dxxeexdxex xxx 222xu dxedv x

Example 4 (Repeated):• Use integration by parts to evaluate

xev

2xu dxedv x

xdxdu 2

dxex x2

dxxeexdxex xxx 222xu dxdu xev

dxedv x

Example 4 (Repeated):• Use integration by parts to evaluate

xev

2xu dxedv x

xdxdu 2

dxex x2

dxxeexdxex xxx 222xu dxdu xev

dxedv x

dxexeexdxex xxxx 222

Example 4 (Repeated):• Use integration by parts to evaluate

xev

2xu dxedv x

xdxdu 2

dxex x2

dxxeexdxex xxx 222xu dxdu xev

dxedv x

dxexeexdxex xxxx 222

Cexeexdxex xxxx 2222

Example 5:• Evaluate the following definite integral

1

0

1 )(tan dxx

Example 5:• Evaluate the following definite integral

xu 1tan

1

0

1 )(tan dxx

dxdv

Example 5:• Evaluate the following definite integral

xu 1tan

1

0

1 )(tan dxx

21

1

xdu

dxdv xv

Example 5:• Evaluate the following definite integral

xu 1tan

1

0

1 )(tan dxx

21

1

xdu

dxdv xv

21

1

0

1

1tan)(tan

x

xdxxxdxx

Example 5:• Evaluate the following definite integral

xu 1tan

1

0

1 )(tan dxx

21

1

xdu

dxdv xv

21

1

0

1

1tan)(tan

x

xdxxxdxx

21 xu

Example 5:• Evaluate the following definite integral

xu 1tan

1

0

1 )(tan dxx

21

1

xdu

dxdv xv

21

1

0

1

1tan)(tan

x

xdxxxdxx

21 xu xdxdu 2

Example 5:• Evaluate the following definite integral

xu 1tan

1

0

1 )(tan dxx

21

1

xdu

dxdv xv

21

1

0

1

1tan)(tan

x

xdxxxdxx

21 xu

dxx

du

2

xdxdu 2

Example 5:• Evaluate the following definite integral

xu 1tan

1

0

1 )(tan dxx

21

1

xdu

dxdv xv

21

1

0

1

1tan)(tan

x

xdxxxdxx

21 xu

dxx

du

2

xdxdu 2

u

duxxdxx2

1tan)(tan 1

1

0

1

Example 5:• Evaluate the following definite integral

xu 1tan

1

0

1 )(tan dxx

21

1

xdu

dxdv xv

21

1

0

1

1tan)(tan

x

xdxxxdxx

21 xu

dxx

du

2

xdxdu 2

u

duxxdxx2

1tan)(tan 1

1

0

1

10211

0

1 )1ln(2

1tan)(tan xxxdxx

Example 5:• Evaluate the following definite integral

1

0

1 )(tan dxx

10211

0

1 )1ln(2

1tan)(tan xxxdxx

Example 5:• Evaluate the following definite integral

1

0

1 )(tan dxx

10211

0

1 )1ln(2

1tan)(tan xxxdxx

)01ln(2

10tan0)11ln(

2

11tan1)(tan 2121

1

0

1 dxx

Example 5:• Evaluate the following definite integral

1

0

1 )(tan dxx

)1ln(2

1tan)(tan 21

1

0

1 xxxdxx

)01ln(2

10tan0)11ln(

2

11tan1)(tan 2121

1

0

1 dxx

002ln2

1

4)(tan

1

0

1 dxx

Example 5:• Evaluate the following definite integral

1

0

1 )(tan dxx

)1ln(2

1tan)(tan 21

1

0

1 xxxdxx

)01ln(2

10tan0)11ln(

2

11tan1)(tan 2121

1

0

1 dxx

2ln4

002ln2

1

4)(tan

1

0

1 dxx

Homework:Page 520

# 3-9 odd, 15, 25, 29, 31, 37