2.
Integration by Parts
There are two techniques for integrations.
3.
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rule.
There are two techniques for integrations.
4.
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rule.
Integration by parts finds antiderivatives of
derivatives that are obtained by the product
rule,
There are two techniques for integrations.
5.
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rule.
Integration by parts finds antiderivatives of
derivatives that are obtained by the product
rule, i.e.
There are two techniques for integrations.
∫uv' + u'v dx = uv + c
6.
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rule.
Integration by parts finds antiderivatives of
derivatives that are obtained by the product
rule, i.e.
There are two techniques for integrations.
∫uv' + u'v dx = uv + c
However, this version is not useful in practice
because its difficult to match the integrand of
an integration problem to uv' + u'v.
7.
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rule.
Integration by parts finds antiderivatives of
derivatives that are obtained by the product
rule, i.e.
There are two techniques for integrations.
However, this version is not useful in practice
because its difficult to match the integrand of
an integration problem to uv' + u'v.
The following is the workable version.
∫uv' + u'v dx = uv + c
8.
Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
∫ uv'dx = uv – ∫ vu'dx
9.
Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
∫ uv'dx = uv – ∫ vu'dx
An simpler way to describe it is to use the
notation of differentials:
10.
Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
∫ uv'dx = uv – ∫ vu'dx
where dv = v'dx and du = u'dx.
An simpler way to describe it is to use the
notation of differentials:
∫ udv = uv – ∫ vdu
11.
Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
∫ uv'dx = uv – ∫ vu'dx
where dv = v'dx and du = u'dx.
An simpler way to describe it is to use the
notation of differentials:
∫ udv = uv – ∫ vdu
When employed, we matched the integrand to
the left hand side of the formula,
12.
Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
∫ uv'dx = uv – ∫ vu'dx
where dv = v'dx and du = u'dx.
An simpler way to describe it is to use the
notation of differentials:
∫ udv = uv – ∫ vdu
When employed, we matched the integrand to
the left hand side of the formula, then convert
it to the right hand side and hope that the
integral on the right hand side will be easier.
13.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
14.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
Match the integrand xexdx to udv, specifically
15.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
Match the integrand xexdx to udv, specifically
let u = x and dv = exdx
16.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
Match the integrand xexdx to udv, specifically
17.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
Match the integrand xexdx to udv, specifically
18.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
du
dx
= 1
Match the integrand xexdx to udv, specifically
19.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
du
dx
= 1
du = 1dx
Match the integrand xexdx to udv, specifically
20.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
du
dx
= 1
du = 1dx
integrate
Match the integrand xexdx to udv, specifically
21.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
du
dx
= 1
du = 1dx
integrate
v = ∫ ex dx
Match the integrand xexdx to udv, specifically
22.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
du
dx
= 1
du = 1dx
integrate
v = ∫ ex dx
v = ex
Match the integrand xexdx to udv, specifically
23.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
du
dx
= 1
du = 1dx
integrate
v = ∫ ex dx
v = ex
Thus ∫ xexdx =
∫ u dv = uv – ∫ v du
Match the integrand xexdx to udv, specifically
24.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
du
dx
= 1
du = 1dx
integrate
v = ∫ ex dx
v = ex
Thus ∫ xexdx = xex
∫ u dv = uv – ∫ v du
Match the integrand xexdx to udv, specifically
25.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
du
dx
= 1
du = 1dx
integrate
v = ∫ ex dx
v = ex
Thus ∫ xexdx = xex – ∫ exdx
∫ u dv = uv – ∫ v du
Match the integrand xexdx to udv, specifically
26.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
let u = x and dv = exdx
To convert, we
need du = u'dx
and v = ∫ dv
differentiate
du
dx
= 1
du = 1dx
integrate
v = ∫ ex dx
v = ex
Thus ∫ xexdx = xex – ∫ exdx = xex – ex + c
∫ u dv = uv – ∫ v du
Match the integrand xexdx to udv, specifically
27.
Integration by Parts
Steps for Integration by parts:
28.
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the integrand to be u and
set the rest as dv. It's possible that dv = dx.
29.
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the integrand to be u and
set the rest as dv. It's possible that dv = dx.
2. Take the derivative of u to find du.
Take the antiderivative of dv to find v.
30.
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the integrand to be u and
set the rest as dv. It's possible that dv = dx.
2. Take the derivative of u to find du.
Take the antiderivative of dv to find v.
∫ udv as uv – ∫ vdu3. Rewrite the integral
31.
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the integrand to be u and
set the rest as dv. It's possible that dv = dx.
2. Take the derivative of u to find du.
Take the antiderivative of dv to find v.
∫ udv as uv – ∫ vdu3. Rewrite the integral
4. Try to find ∫ vdu
32.
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the integrand to be u and
set the rest as dv. It's possible that dv = dx.
2. Take the derivative of u to find du.
Take the antiderivative of dv to find v.
∫ udv as uv – ∫ vdu3. Rewrite the integral
4. Try to find ∫ vdu
Remarks:
* Try integration by parts if substitution doesn’t apply
33.
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the integrand to be u and
set the rest as dv. It's possible that dv = dx.
2. Take the derivative of u to find du.
Take the antiderivative of dv to find v.
∫ udv as uv – ∫ vdu3. Rewrite the integral
4. Try to find ∫ vdu
Remarks:
* Try integration by parts if substitution doesn’t apply
* dv must be integrable so we may find v
34.
Integration by Parts
Example: Find ∫udv = uv – ∫vduLn(x) dx∫
35.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
36.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dx
37.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dx
differentiate
du
dx = 1
x
38.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dx
differentiate
du
dx =
du =
1
x
x
dx
39.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dx
differentiate
du
dx =
du =
integrate
v = ∫ dx1
x
x
dx
40.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dx
differentiate
du
dx =
du =
integrate
v = ∫ dx
v = x
1
x
x
dx
41.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dx
Hence ∫ Ln(x) dx = xLn(x) – ∫ x
differentiate
du
dx =
du =
integrate
v = ∫ dx
v = x
1
x
x
dx
x
dx
42.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dx
Hence ∫ Ln(x) dx = xLn(x) – ∫ x
differentiate
du
dx =
du =
integrate
v = ∫ dx
v = x
1
x
x
dx
x
dx
= xLn(x) – x + c
43.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
44.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
Match the integrand x3Ln(x)dx to udv,
45.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Match the integrand x3Ln(x)dx to udv,
46.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
differentiate
du
dx = 1
x
Match the integrand x3Ln(x)dx to udv,
47.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
differentiate
du
dx =
du =
1
x
x
dx
Match the integrand x3Ln(x)dx to udv,
48.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
differentiate
du
dx =
du =
integrate
v = ∫ x3 dx1
x
x
dx
Match the integrand x3Ln(x)dx to udv,
49.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
differentiate
du
dx =
du =
integrate
v = ∫ x3 dx
v =
1
x
x
dx x4
4
Match the integrand x3Ln(x)dx to udv,
50.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Hence ∫ Ln(x) dx = Ln(x)
differentiate
du
dx =
du =
integrate
v = ∫ x3 dx
v =
1
x
x
dx x4
4
x4
4
Match the integrand x3Ln(x)dx to udv,
51.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Hence ∫ Ln(x) dx = Ln(x) – ∫
differentiate
du
dx =
du =
integrate
v = ∫ x3 dx
v =
1
x
x
dx
x
dx
x4
4
x4
4
x4
4
Match the integrand x3Ln(x)dx to udv,
52.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Hence ∫ Ln(x) dx = Ln(x) – ∫
differentiate
du
dx =
du =
integrate
v = ∫ x3 dx
v =
1
x
x
dx
x
dx
x4
4
x4
4
x4
4
= Ln(x) – ∫ x3dx
x4
4
1
4
Match the integrand x3Ln(x)dx to udv,
53.
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Hence ∫ Ln(x) dx = Ln(x) – ∫
differentiate
du
dx =
du =
integrate
v = ∫ x3 dx
v =
1
x
x
dx
x
dx
= Ln(x) – + c
x4
4
x4
4
x4
4
= Ln(x) – ∫ x3dx
x4
4
1
4
x4
4
x4
16
Match the integrand x3Ln(x)dx to udv,
54.
Integration by Parts
∫udv = uv – ∫vdu
Sometime we have to use the integration by
parts more than once.
55.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Sometime we have to use the integration by
parts more than once.
56.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
Sometime we have to use the integration by
parts more than once.
57.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
differentiate
du
dx = 2x
Sometime we have to use the integration by
parts more than once.
58.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
differentiate
du
dx = 2x
du = 2xdx
Sometime we have to use the integration by
parts more than once.
59.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
differentiate
du
dx = 2x
du = 2xdx
integrate
v = ∫ ex dx
Sometime we have to use the integration by
parts more than once.
60.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
differentiate
du
dx = 2x
du = 2xdx
integrate
v = ∫ ex dx
v = ex
Sometime we have to use the integration by
parts more than once.
61.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
Hence ∫ x2ex dx = x2ex – ∫ 2xexdx
differentiate
du
dx = 2x
du = 2xdx
integrate
v = ∫ ex dx
v = ex
Sometime we have to use the integration by
parts more than once.
62.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
Hence ∫ x2ex dx = x2ex – ∫ 2xexdx
differentiate
du
dx = 2x
du = 2xdx
integrate
v = ∫ ex dx
v = ex
Sometime we have to use the integration by
parts more than once.
Use integration by parts again for ∫ 2xexdx
63.
Integration by Parts
∫ 2xexdxIts easier to do separately.
64.
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
65.
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx
66.
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
67.
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx)
68.
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx) = 2xex – 2ex
69.
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx) = 2xex – 2ex
Therefore,
∫ x2ex dx = x2ex – ∫ 2xexdx
70.
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx) = 2xex – 2ex
Therefore,
∫ x2ex dx = x2ex – ∫ 2xexdx
= x2ex – (2xex – 2ex) + c
71.
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx) = 2xex – 2ex
Therefore,
∫ x2ex dx = x2ex – ∫ 2xexdx
= x2ex – (2xex – 2ex) + c
= x2ex – 2xex + 2ex + c
72.
Integration by Parts
∫udv = uv – ∫vdu
Another example where we have to use the
integration by parts twice.
73.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ sin(x)ex dx
Another example where we have to use the
integration by parts twice.
74.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ sin(x)ex dx
Match the integrand sin(x)exdx to udv,
let u = sin(x) dv = exdx
Another example where we have to use the
integration by parts twice.
75.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ sin(x)ex dx
let u = sin(x) dv = exdx
du
dx = cos(x)
Another example where we have to use the
integration by parts twice.
Match the integrand sin(x)exdx to udv,
76.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ sin(x)ex dx
let u = sin(x) dv = exdx
du
dx = cos(x)
du = cos(x)dx
Another example where we have to use the
integration by parts twice.
Match the integrand sin(x)exdx to udv,
77.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ sin(x)ex dx
let u = sin(x) dv = exdx
du
dx = cos(x)
du = cos(x)dx
v = ∫ ex dx
v = ex
Another example where we have to use the
integration by parts twice.
Match the integrand sin(x)exdx to udv,
78.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ sin(x)ex dx
let u = sin(x) dv = exdx
Hence ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx
du
dx = cos(x)
du = cos(x)dx
v = ∫ ex dx
v = ex
Another example where we have to use the
integration by parts twice.
Match the integrand sin(x)exdx to udv,
79.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ sin(x)ex dx
let u = sin(x) dv = exdx
Hence ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx
du
dx = cos(x)
du = cos(x)dx
v = ∫ ex dx
v = ex
Another example where we have to use the
integration by parts twice.
Use integration by parts again for ∫ cos(x)exdx
Match the integrand sin(x)exdx to udv,
80.
Integration by Parts
let u = cos(x) dv = exdx
For ∫ cos(x)exdx
81.
Integration by Parts
let u = cos(x) dv = exdx
du
dx = -sin(x)
du = -sin(x)dx
For ∫ cos(x)exdx
82.
Integration by Parts
let u = cos(x) dv = exdx
du
dx = -sin(x)
du = -sin(x)dx
v = ∫ ex dx
v = ex
For ∫ cos(x)exdx
83.
Integration by Parts
let u = cos(x) dv = exdx
Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx
du
dx = -sin(x)
du = -sin(x)dx
v = ∫ ex dx
v = ex
For ∫ cos(x)exdx
84.
Integration by Parts
let u = cos(x) dv = exdx
Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx
du
dx = -sin(x)
du = -sin(x)dx
v = ∫ ex dx
v = ex
For ∫ cos(x)exdx
Since ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx
85.
Integration by Parts
let u = cos(x) dv = exdx
Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx
du
dx = -sin(x)
du = -sin(x)dx
v = ∫ ex dx
v = ex
For ∫ cos(x)exdx
Since ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx
∫ sin(x)ex dx = sin(x)ex – cos(x)ex – ∫ sin(x)exdx
86.
Integration by Parts
let u = cos(x) dv = exdx
Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx
du
dx = -sin(x)
du = -sin(x)dx
v = ∫ ex dx
v = ex
For ∫ cos(x)exdx
Since ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx
∫ sin(x)ex dx = sin(x)ex – cos(x)ex – ∫ sin(x)exdx
2 ∫ sin(x)ex dx = sin(x)ex – cos(x)ex
87.
Integration by Parts
let u = cos(x) dv = exdx
Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx
du
dx = -sin(x)
du = -sin(x)dx
v = ∫ ex dx
v = ex
For ∫ cos(x)exdx
Since ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx
∫ sin(x)ex dx = sin(x)ex – cos(x)ex – ∫ sin(x)exdx
2 ∫ sin(x)ex dx = sin(x)ex – cos(x)ex
∫ sin(x)ex dx = ½ (sin(x)ex – cos(x)ex) + c
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