Download 1.4.2 : Integration by parts Managing this process An antiderivative

1.4.2 : Integration by parts
Managing this process
In this section we discuss the technique of integration by parts, which is
essentially a reversal of the product rule of differentiation.
Example 22
�
Find
x cos x dx.
(uv )� = u � v + uv � .
Suppose we integrate both sides here with respect to x. We obtain
�
�
�
�
�
(uv )� dx = u � v dx + uv � dx =⇒ uv = u � v dx + uv � dx.
Solution How could x cos x arise as a derivative?
Well, cos x is the derivative of sin x. So, if you were differentiating
x sin x, you would get x cos x but according to the product rule you
would also get
� another term, namely sin x.
Conclusion:
This can be rearranged to give the Integration by Parts Formula :
�
�
�
uv dx = uv − u � v dx.
x cos x dx = x sin x + cos x + C .
Dr Rachel Quinlan
What happened in this example was basically that the product rule was
reversed. This process can be managed in general as follows. Recall from
differential calculus that if u and v are expressions involving x, then
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Integration by Parts
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Integration by Parts
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An antiderivative for ln x
Here is the first example again, handled according to this scheme.
Example 23
Use the integration by parts technique to determine
Solution: Write
�
Example 24
�
ln x dx.
Determine
x cos x dx.
Solution: Let u = ln x, v � = 1.
Then u � = x1 , v = x.
�
�
�
ln x dx =
uv � dx = uv − u � v dx
�
1
x dx
= x ln x −
x
= x ln x − x + C .
u = x v � = cos x
u � = 1 v = sin x
Then
�
x cos x dx
�
�
uv dx = uv − u � v dx
�
= x sin x − 1 sin x dx
=
�
Note: This example shows that sometimes problems which are not
obvious candidates for integration by parts can be attacked using this
technique.
= x sin x + cos x + C .
Dr Rachel Quinlan
MA180/MA186/MA190 Calculus
Integration by Parts
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Dr Rachel Quinlan
MA180/MA186/MA190 Calculus
Integration by Parts
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Two Rounds of Integration by Parts
Two rounds (continued)
Sometimes two applications of the integration by parts formula are
needed, as in the following example.
Let I =
Example 25
�
x 2 e x dx.
Evaluate
x 2,
�
xe x dx
To evaluate I apply the integration by parts formula a second time.
v�
u�
ex .
u = x v � = ex
u� = 1 v = e x .
�
�
x
x
Then I = xe dx = xe − e x dx = xe x − e x + C . Finally
ex .
Solution: Let u =
=
Then = 2x, v =
�
�
�
uv � dx = uv − u � v dx
x 2 e x dx =
�
= x 2 e x − 2xe x dx
�
2 x
= x e − 2 xe x dx.
Let I =
�
�
x 2 e x dx = x 2 e x − 2xe x + 2e x + C .
xe x dx.
Dr Rachel Quinlan
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Integration by Parts
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Dr Rachel Quinlan
�
An Example of Another Type
Solution Let
Then
Dr Rachel Quinlan
u=
u� = e x
u = ex
u� = e x
= cos x
v = sin x.
e x cos x dx = e x sin x −
=⇒ 2
e x sin x dx.
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Integration by Parts
=⇒
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�
�
v�
�
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e x sin x dx: Let
and
ex
�
�
Then
Example 26
�
Determine
e x cos x dx.
Integration by Parts
e x cos x dx (continued)
For
The next example shows another mechanism by which a second
application of the integration by parts formula can succeed where the
first is not enough.
MA180/MA186/MA190 Calculus
�
�
x
v � = sin x
v = − cos x.
x
e sin x dx = −e cos x +
x
x
�
�
e x cos x dx,
x
e cos x dx
= e sin x − −e cos x +
e x cos x dx
= e x sin x + e x cos x + C
e x cos x dx
=
Dr Rachel Quinlan
�
x
e cos x dx
�
1 x
(e sin x + e x cos x) + C
2
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A Definite Integral
Example 27
� 1
Evaluate
(x + 3)e 2x dx.
0
Solution: Write u = x + 3, v � = e 2x ;
�
1
(x + 3)e 2x dx
=
0
=
=
=
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�
u � = 1, v = 12 e 2x
uv � dx = (uv )|10 −
�
1
u � v dx
0
�
�
x + 3 2x ��1 1 1 2x
e dx
e � −
2
2 0
0
�
�
x + 3 2x ��1 1 1 2x ��1
e � − × e �
2
2 2
0
0
4 2 3 0 1 2 1 0 7 2 5
e − e − e + e = e − .
2
2
4
4
4
4
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Integration by Parts
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