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4.5 Integration By Pattern Recognition
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Integration by Pattern Recognition:
Cnuduun
n
1
1
The first basic type of integration problem is in the form:
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Integrate by recognizing the Pattern duu n
dxxdu 26 Then
Therefore, this integral is of the type: duu 4
dxxx 243 6)52(
Note: If )52( 3 xuIntegrating we get:
Cuduu 5
54
Substitute, )52( 3 xu
Henceforth, Cxdxxx
5)52(6)52(
53243
)52( 3 xuBut,
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dxx 2)32( 21
Note: If )32( xu dxdu 2Then
Note: This is exactly in the form! duun
dxx 232
Therefore, this integral is of the type: duu 2
1Integrating we get:
23
23
21 uu
)32( xuBut, Substitute, )32( xu
Henceforth, Cxdxx 232
132
322)32(
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Multiplying by a Form of 1 to integrate:
xdxdu 2Then
Note: This is not exactly in the form! duu nThe inside of the Integral has to be multiplied by 2
Therefore the outside of the Integral has to be multiplied by ½, since( 2) (½) = 1, and as long as we multiple the entire integral by a numeric form of 1 we can proceed with integration.
xdxx 22 )5(
Note: If )5( 2 xu
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Now multiply by a form of 1 to integrate:
xdxdu 2Then
xdxx 2)5(21 22
Substitute get)5( 2 xu Cx 32 5
61
xdxx 22 )5(
Note: This is exactly in the form! duu 2
21
Integrate this form to get Cu
321 3
Simplifying to get Cu
6
3
Note: If )5( 2 xu
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Integrate
92 33 5x x dx3 2pick +5, then 3 u x du x dx
10
10u C
9u du 103 5
10
xC
Sub to get Integrate Back Substitute
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Ex. Evaluate 25 7x x dx
3/ 225 7
15
xC
Sub in
3/ 21
10 3/ 2u C
Integrate
Pick u, compute du75 2 xu xdxdu 10
Sub inxdxxxdxx 10)75(101)75( 2
12212
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Trig Integrals in the form:duun xdxx cossin 2
xxu sinsin Let Then xdxdu cosNote: This is exactly in the form! duu 2
Integrate this form to get Cu
3
3
Sub inCxCx
3sin
3)(sin 33
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Basic Trig Integrals
2
2
cos sin
sin cos
sec tan
csc cot
sec tan sec
csc cot csc
u du u C
u du u C
u du u C
u du u C
u u du u C
u u du u C
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The key to each basic Trig Integral is that:
Let u = The angle
While du = The derivative of the angle
First make sure you do not have a problem.
duun
You need to know the 6 trig. Derivatives, so that you can work backwards and find their Anti-derivatives!
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Using the Trig Integrals
• The technique is often to find a u which is the angle, the argument of the trig function
• Consider
• What is the u, the du?
• Substitute, integrate
cos3x dx3 3u x du dx
1 cos sin3
1 sin 33
u du u C
x C
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2 3x cosx dx Let u = x3 ; du = 3x2dx ; C.F. 1/3
3 21 cosx (3x dx)3 1 cosudu3 1sinu C3
31sinx C3
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Symmetry in Definite Integral
Integrals of Symmetric Functions