4.5 Integration By Pattern Recognition
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Integration by Pattern Recognition:
The first basic type of integration
problem is in the form: n
u n 1
 u du  n  1  C
Integrate by recognizing the Pattern
 (2 x
3
u
n
du
 5) 6 x dx
4
2
Note: If u  (2 x  5)
3
Therefore, this integral
is of the type:  u 4 du
But, u  (2x 3  5)
Then
du  6 x dx
2
Integrating we get:
5
u
4
u
 du  5  C
3
u

(
2
x
 5)
Substitute,
3
5
(
2
x

5
)
Henceforth, (2 x 3  5) 4 6 x 2 dx 
C

5

2 x  3  2dx   (2 x  3) 2dx
12
Note: If u  (2 x  3)
Then

du  2dx
n
Note: This is exactly in the u du form!
Therefore, this integral
1
is of the type: u 2 du
Integrating we get:
2
u
2
u
 3
2
Substitute, u  (2 x  3)

But,
3
1
u  (2 x  3)
3
2
Henceforth,  (2 x  3) 2dx  2 x  3 2  C
3
1
2
Multiplying by a Form of 1 to integrate:
 (x
Note: If
2
 5) xdx
u  ( x  5)
2
2
Then du
 2 xdx
Note: This is not exactly in the  u du form!
n
The 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.
Now multiply by a form of 1 to integrate:
1 2
2
( x  5) xdx   ( x  5) 2 xdx
2
2
Note: If u  ( x  5) Then

2
2
1
Note: This is exactly in the
2
du  2 xdx
2
u
 du
3
1
u
Integrate this form to get 
C
2 3
3
u
Simplifying to get
C
6
Substitute u  ( x  5) get
2


3
1 2
x 5 C
6
form!
Integrate
 3x  x
2
3
 5 dx
pick u  x +5, then du  3x dx
3

10
u du
u

C
10
Sub to get
Integrate
9
2
x


3
5


9
10
C
10
Back Substitute
Ex. Evaluate

x 5 x 2  7dx
u  5x  7 du  10 xdx
2
1
1
2
2
(
5
x

7
)
xdx

(
5
x

7
)
 10xdx




 10 
2
1
2
1 u
 
C
 10   3 / 2 
Pick u,
compute du
Sub in
3/ 2
5x


2
7
15

Integrate
3/ 2
C
Sub in
Trig Integrals in the  u du form:
n
sin
x
cos
xdx

2
Let
u  sin x  sin x  Then du  cos xdx

Note: This is exactly in the u 2 du form!
3
Integrate this form to get u  C
3
3
3
(sin x)
sin x
C 
C
3
3
Sub in
9
Basic Trig Integrals
 cos u du  sin u  C
 sin u du   cos u  C
 sec u du  tan u  C
 csc u du   cot u  C
 sec u  tan u du  sec u  C
 csc u  cot u du   csc u  C
2
2
10
The key to each basic Trig Integral is that:
n
First make sure you do not have a  u du problem.
Let u = The angle
While du = The derivative of the angle
You need to know the 6 trig. Derivatives,
so that you can work backwards and find
their Anti-derivatives!
Using the Trig Integrals
• The technique is often to find a u which is the
angle, the argument of the trig function
• Consider
 cos 3x dx
• What is the u, the du?
• Substitute, integrate
u  3x du  3dx
1
cos u du  sin u  C

3
1
 sin 3 x  C
3
12
2
3
3
Let
u
=
x
x
cosx
dx

1
  cosx3 (3x2 dx)
3
1
  cos udu
3
1
 sin u  C
3
1
 sin x3  C
3
; du = 3x2dx
; C.F. 1/3
Symmetry in Definite Integral
Integrals of Symmetric Functions