Section 4.5 Integration By Substitution
The Reverse Chain Rule
Find the following Derivative
To differentiate this function requires the chain rule. Suppose
we had the task of anti-differentiating the result of f ( x) . Is there and Antidifferentiation “Chain Rule”. The answer is not exactly. We will however
have to utilize two speecialized techniques known as
f ( x)  (2 x3  5) 4
Pattern Recognition
and
u - du substitution.
To use Pattern Recognition:
Is there an “inner” function whose derivative exists as a factor that is
attached to this “inner” function?
Examples of what I mean:
 2 x( x
 3x
2
 sec
2
2
 1) 4 dx Note that the inner function of ( x 2  1) 4 has a derivative of 2x which is attached
x3  1 dx Note that the inner function of
x3  1 has a derivative of 3x 2 which is attached
x(tan x  1) dx Note that the inner function of  tan x  1 has a derivative of sec 2 x which is attached
When this is the case and we recognize it we can simply integrate by
integrating the composite function only and we can disregard the attached
function.
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To Integrate functions of this form we will use the following process known
informally as “u” du substitution.
Note: Outer function (f) is normally a power function of some type or a trig.
Function of some type.
 f ( g ( x)) g ( x)dx  F ( g ( x))  C
Examples:
1.
 3(3x  1) dx
4.
 3x
3
2
x3  1 dx
2.
 (1  2 x )(4 x)dx
2
5.
4 x
 (1  2 x )
2 2
3.
 5cos 5 x dx
dx
2
We will now try the previous examples using what is known as the u-du
substitution strategy.
Examples:
1.
 3(3x  1) dx
4.
 3x
3
2
x3  1 dx
2.
 (1  2 x )(4 x)dx
2
5.
4 x
 (1  2 x )
2 2
3.
 5cos 5 x dx
dx
3
What happens if the derivative of the inner function is only off by a constant?
It turns out we can manage this problem using algebra rules. Here are some
examples.
In these cases we must use a slightly different process using a simple
algebraic technique.
Here are some examples
6.
 x(4 x
8.

2
 3)3 dx
4x
x2  1
(off by factor of 8)
7.
 sin
2
3 x cos 3 x dx (off by factor of 3)
dx (off by a factor of 2)
****** note :
Multiplying and dividing by a variable is not legal
 off by a factor of 2x oops! no can do 
Example:
x
2
 1 dx 
2
1
2x
 (x
2
 1) 2 (2 x) dx but you can expand ( x 2  1) 2  x 4  2 x 2  1
and then integrate it.
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To summarize the look for a pattern approach and the “u du” substitution method we can use the
following 2 general power rules:
 g ( x) 
g ( x) dx 
n 1
 ( g ( x))
n
n 1
C
Or
n
 u du 
u n 1
C
n 1
Finding a Definite Integral using these techniques:
Suppose you had to evaluate the definite integral
1
 (x
2
 1)3 ( x)dx
0
Using the pattern recognition method you get
1
1
( x 2  1)3 (2 x)dx

20
4
1
2


1  ( x 2  1) 4 
1  (12  1) 4  0  1  1 16 1  1 15  15



 

 2  4 4  2  4  8
2  4  0 2  4
4


Alternative Method using “u du” Substitution Method
Suppose you wanted to use substitution to integrate the function. Is it possible to evaluate the
integral. The answer is yes but you must first change your limits of integration with their equivalent
“u” values using the “u” equation:
1
 (x
2
 1)3 ( x)dx
0
Let u  x 2  1
du  2 x  dx
du
1
1 u4 
 x  dx   u 3du    you must find equivalent values of your limits using u  x 2  1
2
20
2 4 
1
This changes your "x" limits of integration from 0  1 and from 1  2. Evaluating with these limits
results in the following:
2
1  u4 
1 16 1  1 15  15
     


2  4 1 2  4 4  2  4  8
Note: You could use the power rule and still use the “x” limits of integration.
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Example 2: Evaluate the following definite integral using both approaches
2

3
x 2  4( x)dx  3.619
0
More Trig. Examples
sin x
 cos 2 x dx
csc 2 x
 cot 2 x dx
 sec
2
x(tan 4 x) dx
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