# Integration ```Objectives
 Use pattern recognition to find an indefinite
integral.
 Use a change of variables to find an indefinite
integral.
 Use the General Power Rule for Integration to
find an indefinite integral.
ANTIDIFFERENTIATION OF A
COMPOSITE FUNCTION
Let g be a function whose range is an interval I and let
f be a function that is continuous on I. If g is
differentiable on its domain and F is an antiderivative
of f on I, then
 f ( g ( x)) g '( x)dx  F ( g ( x))  c
If u = g(x), then du = g’(x)dx and
 f (u )du  F (u )  c
PATTERN RECOGNITION
In this section you will study techniques for
integrating composite functions.
This is split into two parts, pattern recognition
and change of variables.
u-substitution is similar to the techniques used
for the chain rule in differentiation.
PATTERN RECOGNITION FOR
FINDING THE ANTIDERIVATIVE
Find
 5cos 5xdx
Let g(x)=5x and we have g’(x)=5dx
So we have f(g(x))=f(5x)=cos5x
From this, you can recognize that the integrand
follows the f(g(x))g’(x) pattern. Using the trig
integration rule, we get
  cos 5x  5dx  sin 5x  c
You can check this by differentiating the answer to
obtain the original integrand.
RECOGNIZING PATTERNS
Look at the more complex part of the function- the stuff ‘inside’.
Does the stuff outside look like the derivative of the stuff inside?
2
4
2
2
x
(
x

1)
dx

u

x
 1  du  2 x

x
2
 1
5
5
c
RECOGNIZING PATTERNS


3x 2 x3  1 dx  u  x3  1  du  3x 2
3/ 2
2 3
  x  1  c
3
RECOGNIZING PATTERNS
2
2
sec
x
tan
x

3
dx

u

tan
x

3

du

sec
x



tan x  3


2
2
c
CHANGE OF VARIABLES
Find
 5cos 5xdx
Let u=5x and we have du=5dx
 cos udu  sin u  c
  cos 5x  5dx  sin 5x  c
You can check this by differentiating the answer to
obtain the original integrand.
CHANGE OF VARIABLES
A.K.A. U-SUBSTITUTION
 2x  x
Example 3
2
 1 dx
2
u  x  1  du  2 xdx
2
 2x  x
 u  du
2

x


2
 1 dx
2
u 
2 1
2 1
2
 1
3
c 
3
c
u 
3
3
c
PATTERN RECOGNITION
The integrands in Example 1 fit the f(g(x))g'(x) pattern exactly—you only had to recognize the
pattern.
You can extend this technique considerably with the Constant Multiple Rule
Many integrands contain the essential part (the variable part) of g'(x) but are missing a constant
multiple.
In such cases, you can multiply and divide by the necessary constant multiple, as shown in
Example 3.
MULTIPLYING AND DIVIDING BY A
CONSTANT
Example 3
 xx
2
 1 dx
2
u  x  1  du  2 xdx
2
2
1
2
   2 x  x  1 dx
2

1 2
1 3
u
du

u c

2
6
3
1 2
now back substitute :  x  1  c
6
MULTIPLYING AND DIVIDING BY A
CONSTANT ALTERNATE METHOD
Example 3
 xx
2
 1 dx
2
let u  x 2  1  du  2 xdx
but we have only x, not 2 x...so we do algebra
1
du  xdx  now substitute :
2
1
 u 2 du
2

1 2
1 3
u
du

u c

2
6
3
1 2
now back substitute :  x  1  c
6
CHANGE OF VARIABLES
EXAMPLE 4

2 x  1dx
EXAMPLE 4 – CHANGE OF
VARIABLES
Find
Solution:
First, let u be the inner function, u = 2x – 1.
Then calculate the differential du to be du = 2dx.
Now, using
and
substitute to obtain
EXAMPLE 4 – SOLUTION
cont’d
CHANGE OF VARIABLES
Example 5
x
2 x  1dx
CHANGE OF VARIABLES
Example 5
x
2 x  1dx
let _ u  2 x  1  du  2dx
solve for x : (u  1) / 2
1
 u  1  2 du
Substitute :  x 2 x  1dx   
u
 2  2
1
3
1


1
1
1  2 5/ 2 2 3/ 2 
2
2
2
   u  1u du    u  u du   u  u   c
4
4 
45
3


5
3
1
1
Back _ Substitute :  2 x  1 2   2 x  1 2  c
10
6
CHANGE OF VARIABLES EXAMPLE
Example 6
2
sin
 3x cos 3xdx
EXAMPLE 7 – SUBSTITUTION AND THE GENERAL POWER
RULE
EXAMPLE 7 – SUBSTITUTION AND THE GENERAL POWER
RULE
cont’d
GUIDELINES FOR MAKING A
CHANGE OF VARIABLES
1.
2.
3.
4.
5.
6.
Choose a substitution u=g(x). Usually it is best
to choose the INNER part of a composite
function, such as a quantity raised to a power.
Compute du=g’(x)dx
Rewrite the integral in terms of the variable u
Find the resulting integral in terms of u
Replace u by g(x) to obtain an antiderivative in
terms of x.
GENERAL POWER RULE FOR
INTEGRATION
Theorem: The General Power Rule for Integration
If g is a differentiable function of x, then
 g  x  
  g  x  g   x  dx  n  1
n 1
n
Equivalently, if u=g(x), then
u n 1
 u du  n  1  c, n  1
n
 c, n  1
CHANGE OF VARIABLES FOR
DEFINITE INTEGRALS
If the function u=g(x) has a continuous derivative on the
closed interval [a,b] and f is continuous on the range of g,
then
b
g b
a
g a
 f  g  x   g   x  dx     f  u  du
Basically, this gives us permission to find the area under the
curve with these types of integration problems.
DEFINITE INTEGRALS AND
CHANGE OF VARIABLES
Example 8
 xx
1
0
2
 1 dx
3
EXAMPLE 8 – CHANGE OF
VARIABLES
Evaluate
Solution:
To evaluate this integral, let u = x2 + 1.
Then, you obtain
Before substituting, determine the new upper and lower limits of
integration.
EXAMPLE 8 – SOLUTION
Now, you can substitute to obtain
cont’d
EXAMPLE 8 – SOLUTION
cont’d
Try rewriting the antiderivative
in terms of the variable x and evaluate the definite
integral at the original limits of integration, as shown.
Notice that you obtain the same result.
DEFINITE INTEGRALS AND
CHANGE OF VARIABLES
Example 9:
This problem requires a little more work:
5
x
1 2 x  1 dx
INTEGRATION OF EVEN AND ODD
FUNCTIONS
Even with a change of variables, integration can be difficult.
Occasionally, you can simplify the evaluation of a definite
integral over an interval that is symmetric about the y-axis or
about the origin by recognizing the integrand to be an even or
odd function (see Figure 4.40).
Figure 4.40
INTEGRATION OF EVEN AND ODD
FUNCTIONS
EXAMPLE 10 – INTEGRATION OF AN ODD
FUNCTION
Evaluate
Solution:
Letting f(x) = sin3x cos x + sin x cos x produces
f(–x) = sin3(–x) cos (–x) + sin (–x) cos (–x)
= –sin3x cos x – sin x cos x = –f(x).
EXAMPLE 10 – SOLUTION
cont’d
So, f is an odd function, and because f is symmetric about the origin over [–π/2, π/2], you can
apply Theorem 4.16 to conclude that
Figure 4.41
HOMEWORK:
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1,3,9,14,16,20,25,27,31,33,34,41,45,51,54,59,69,
79,81,95,103