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Integration by Parts Objective: To integrate problems without a u- substitution
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 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
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
dxu
xev
dxedv x
Example 4(repeated)
• Use integration by parts to evaluate
xev 2xu
dxedv xxdxdu 2
dxex x2
dxxeexdxex xxx 222xu
dxu
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
dxu
xev
dxedv x dxexeexdxex xxxx 222
Cexeexdxex xxxx 2222
Example 5
• Use integration by parts to evaluate xdxex cos
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
• Page 520
• 3-33 multiples of 3
• Section 7.2
• 3-30 multiples of 3
