5.5 the substitution rule in this section, we will learn: to substitute a new variable in place of...
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5.5
The Substitution Rule
In this section, we will learn:
To substitute a new variable in place of an existing
expression in a function, making integration easier.
INTEGRALS
Due to the Fundamental Theorem of Calculus, it is
important to be able to find antiderivatives.
However, our antidifferentiation formulas do not
tell us how to evaluate integrals such as
INTRODUCTION
22 1x x dx Equation 1
To find this integral, we use the problem solving
strategy of introducing something extra.
The ‘something extra’ is a new variable.
We change from the variable x to a new variable u.
INTRODUCTION
Suppose we let u be the quantity under the root
sign in Equation 1, u = 1 + x2.
Then, the differential of u is du = 2x dx. Notice that, if the dx in the notation for an integral were
to be interpreted as a differential, then the differential 2x dx would occur in Equation 1.
INTRODUCTION
So, formally, without justifying our calculation, we
could write:2 2
3/ 223
2 3/ 223
2 1 1 2
( 1)
x x dx x x dx
udu
u C
x C
Equation 2INTRODUCTION
However, now we can check that we have the
correct answer by using the Chain Rule to
differentiate the final function of Equation 2:
2 3 2 2 1 232 23 3 2
2
( 1) ( 1) 2
2 1
dx C x x
dx
x x
INTRODUCTION
In general, this method works whenever we have
an integral that we can write in the form
INTRODUCTION
( ( )) ( )f g x g x dx
Observe that, if F’ = f, then
because, by the Chain Rule,
Equation 3INTRODUCTION
( ( )) '( ( )) '( )d
F g x F g x g xdx
( ( )) ( ) ( ( ))F g x g x dx F g x C
That is, if we make the ‘change of variable’ or
‘substitution’ u = g(x), from Equation 3, we have:
'( ( )) '( ) ( ( ))
( )
'( )
F g x g x dx F g x C
F u C
F u du
INTRODUCTION
Writing F’ = f, we get:
Thus, we have proved the following rule.
INTRODUCTION
( ( )) '( ) ( )f g x g x dx f u du
SUBSTITUTION RULE
If u = g(x) is a differentiable function whose range
is an interval I and f is continuous on I, then
Equation 4
( ( )) '( ) ( )f g x g x dx f u du
SUBSTITUTION RULE
Notice that the Substitution Rule was proved using
the Chain Rule for differentiation.
Notice also that, if u = g(x), then du = g’(x)dx.
So, a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.
SUBSTITUTION RULE
Thus, the Substitution Rule says:
It is permissible to operate with dx and du after
integral signs as if they were differentials.
SUBSTITUTION RULE
Find
We make the substitution u = x4 + 2.
This is because its differential is du = 4x3 dx, which,
apart from the constant factor 4, occurs in the integral.
Example 1
3 4cos( 2)x x dx
SUBSTITUTION RULE
Thus, using x3 dx = du/4 and the Substitution Rule,
we have:
Notice that, at the final stage, we had to return to the original variable x.
3 4 1 14 4
14
414
cos( 2) cos cos
sin
sin( 2)
x x dx u du u du
u C
x C
Example 1
SUBSTITUTION RULE
The idea behind the Substitution Rule is to replace
a relatively complicated integral by a simpler
integral. This is accomplished by changing from the original
variable x to a new variable u that is a function of x. Thus, in Example 1, we replaced the integral
by the simpler integral3 4cos( 2)x x dx1
cos4
u du
SUBSTITUTION RULE
The main challenge in using the rule is to think of
an appropriate substitution.
You should try to choose u to be some function in the integrand whose differential also occurs, except for a constant factor.
This was the case in Example 1.
SUBSTITUTION RULE
If that is not possible, try choosing u to be some
complicated part of the integrand, perhaps the
inner function in a composite function.
Finding the right substitution is a bit of an art.
It is not unusual to guess wrong.
If your first guess does not work, try another substitution.
SUBSTITUTION RULE
SUBSTITUTION RULE
Evaluate
Let u = 2x + 1.
Then, du = 2 dx.
So, dx = du/2.
2 1x dx
E. g. 2—Solution 1
Thus, the rule gives:
1 212
3 212
3 213
3 213
2 12
3/ 2
(2 1)
dux dx u
u du
uC
u C
x C
SUBSTITUTION RULE E. g. 2—Solution 1
Another possible substitution is
Then,
So,
Alternatively, observe that u2 = 2x + 1. So, 2u du = 2 dx.
SUBSTITUTION RULE E. g. 2—Solution 2
2 1u x
2 1
dxdu
x
2 1dx x du udu
Thus,
SUBSTITUTION RULE E. g. 2—Solution 2
2
3
3 213
2 1
3
(2 1)
x dx u u du
u du
uC
x C
SUBSTITUTION RULE
Find
Let u = 1 – 4x2. Then, du = -8x dx. So, xdx = -1/8 du and
21 4
xdx
x
1 21 18 82
21 18 4
1
1 4
(2 ) 1 4
xdx du u du
ux
u C x C
Example 3
SUBSTITUTION RULE
The answer to the example could be checked by
differentiation.
Instead, let us check it with a graph.
SUBSTITUTION RULE
Here, we have used a computer to graph both the
integrand and its indefinite
integral
We take the case C = 0.
2( ) / 1 4f x x x 21
4( ) 1 4g x x
SUBSTITUTION RULE
Notice that g(x): Decreases when f(x) is negative Increases when f(x) is positive Has its minimum value when f(x) = 0
So, it seems reasonable, from the graphical
evidence, that g is an antiderivative of f.
SUBSTITUTION RULE
SUBSTITUTION RULE
Calculate
If we let u = 5x, then du = 5 dx. So, dx = 1/5 du. Therefore,
5 15
15
515
x u
u
x
e dx e du
e C
e C
Example 4
5xe dx
SUBSTITUTION RULE
Find
An appropriate substitution becomes more obvious if we factor x5 as x4 . x.
Let u = 1 + x2.
Then, du = 2x dx.
So, x dx = du/2.
5 21x x dx
Example 5
SUBSTITUTION RULE
Also, x2 = u – 1; so, x4 = (u – 1)2:2 5 2 4 2
212
5/ 2 3/ 2 1/ 212
7 / 2 5/ 2 3/ 21 2 2 22 7 5 3
2 7 / 2 2 5/ 21 27 5
2 3/ 213
1 1 ( 1)2
( 2 1)
( 2 )
( 2 )
(1 ) (1 )
(1 )
dux x dx x x x dx u u
u u u du
u u u du
u u u C
x x
x C
Example 5
SUBSTITUTION RULE
Calculate
First, we write tangent in terms of sine and cosine:
This suggests that we should substitute u = cos x, since then du = – sin x dx, and so sin x dx = – du:
sintan
cos
xx dx dx
x
sintan ln | |
cosln | cos |
x dux dx dx u C
x ux C
Example 6
tan x dx
SUBSTITUTION RULE
Since –ln|cos x| = ln(|cos x|-1)
= ln(1/|cos x|)
= ln|sec x|,
the result can also be written as
Equation 5
tan ln | sec |x dx x C
DEFINITE INTEGRALS
When evaluating a definite integral by substitution,
two methods are possible.
One method is to evaluate the indefinite integral
first and then use the FTC. For instance, using the result of Example 2, we have:
DEFINITE INTEGRALS
44
0 0
43 213 0
3 2 3 21 13 3
2613 3
2 1 2 1
(2 1)
(9) (1)
(27 1)
x dx x dx
x
DEFINITE INTEGRALS
Another method, which is usually preferable, is to
change the limits of integration when the variable
is changed.
Thus, we have the substitution rule for definite
integrals.
If g’ is continuous on [a, b] and f is continuous on
the range of u = g(x), then
( )
( )( ( )) '( ) ( )
b g b
a g af g x g x dx f u du
SUB. RULE FOR DEF. INTEGRALS Equation 6
Let F be an antiderivative of f.
Then, by Equation 3, F(g(x)) is an antiderivative of f(g(x))g’(x).
So, by Part 2 of the FTC (FTC2), we have:
( ( )) '( ) ( ( ))
( ( )) ( ( ))
b b
aaf g x g x dx F g x
F g b F g a
SUB. RULE FOR DEF. INTEGRALS Proof
However, applying the FTC2 a second time, we
also have:
( ) ( )
( )( )( ) ( )
( ( )) ( ( ))
g b g b
g ag af u du F u
F g b F g a
SUB. RULE FOR DEF. INTEGRALS Proof
Evaluate using Equation 6.
Using the substitution from Solution 1 of Example 2, we have: u = 2x + 1 and dx = du/2
4
02 1x dx
Example 7SUB. RULE FOR DEF. INTEGRALS
To find the new limits of integration, we note that: When x = 0, u = 2(0) + 1 = 1 and when x = 4, u = 2(4)
+ 1 = 9
Example 7SUB. RULE FOR DEF. INTEGRALS
Thus,
4 9120 1
93 21 22 3 1
3 2 3 213
263
2 1
(9 1 )
x dx u du
u
Example 7SUB. RULE FOR DEF. INTEGRALS
Observe that, when using Equation 6, we do not
return to the variable x after integrating.
We simply evaluate the expression in u between the appropriate values of u.
SUB. RULE FOR DEF. INTEGRALS Example 7
Evaluate
Let u = 3 - 5x.
Then, du = – 5 dx, so dx = – du/5.
When x = 1, u = – 2, and when x = 2, u = – 7.
2
21 (3 5 )
dx
x
Example 8SUB. RULE FOR DEF. INTEGRALS
Thus,2 7
2 21 2
7
2
1
(3 5 ) 5
1 1
5
1 1 1 1
5 7 2 14
dx du
x u
u
Example 8SUB. RULE FOR DEF. INTEGRALS
Calculate
We let u = ln x because its differential du = dx/x occurs in the integral.
When x = 1, u = ln 1, and when x = e, u = ln e = 1.
Thus,
1
lne xdx
x
Example 9SUB. RULE FOR DEF. INTEGRALS
121
1 00
ln 1
2 2
e x udx u du
x
As the function f(x) = (ln x)/x in the example is
positive for x > 1, the integral represents the area
of the shaded region in this figure.
SUB. RULE FOR DEF. INTEGRALS Example 9
SYMMETRY
The next theorem uses the Substitution Rule for
Definite Integrals to simplify the calculation of
integrals of functions that possess symmetry
properties.
INTEGS. OF SYMM. FUNCTIONS
Suppose f is continuous on [–a , a].
a. If f is even, [f(–x) = f(x)], then
b. If f is odd, [f(-x) = -f(x)], then
0( ) 2 ( )
a a
af x dx f x dx
( ) 0a
af x dx
Theorem 7
We split the integral in two:
0
0
0 0
( ) ( ) ( )
( ) ( )
a a
a a
a a
f x dx f x dx f x dx
f x dx f x dx
Proof - Equation 8INTEGS. OF SYMM. FUNCTIONS
In the first integral in the second part, we make the
substitution u = –x .
Then, du = –dx, and when x = –a, u = a.
INTEGS. OF SYMM. FUNCTIONS
0
0
0 0
( ) ( ) ( )
( ) ( )
a a
a a
a a
f x dx f x dx f x dx
f x dx f x dx
Proof
Therefore,
0 0
0
( ) ( )( )
( )
a a
a
f x dx f u du
f u du
ProofINTEGS. OF SYMM. FUNCTIONS
So, Equation 8 becomes:
0 0
( )
( ) ( )
a
a
a a
f x dx
f u du f x dx
Proof - Equation 9INTEGS. OF SYMM. FUNCTIONS
If f is even, then f(–u) = f(u).
So, Equation 9 gives:
0 0
0
( )
( ) ( )
2 ( )
a
a
a a
a
f x dx
f u du f x dx
f x dx
INTEGS. OF SYMM. FUNCTIONS Proof a
If f is odd, then f(–u) = –f(u).
So, Equation 9 gives:
INTEGS. OF SYMM. FUNCTIONS Proof b
0 0
( )
( ) ( )
0
a
a
a a
f x dx
f u du f x dx
Theorem 7 is
illustrated here.
INTEGS. OF SYMM. FUNCTIONS
For the case where f is positive and even, part (a)
says that the area under y = f(x) from -a to a is
twice the area from 0 to a because of symmetry.
INTEGS. OF SYMM. FUNCTIONS
Recall that an integral can be expressed
as the area above the x-axis and below y = f(x)
minus the area below the axis and above the curve.
INTEGS. OF SYMM. FUNCTIONS
( )b
af x dx
Therefore, part (b) says the integral is 0 because
the areas cancel.
INTEGS. OF SYMM. FUNCTIONS
As f(x) = x6 + 1 satisfies f(–x) = f(x), it is even. So,
2 26 6
2 0
2717 0
1287
2847
( 1) 2 ( 1)
2
2 2
x dx x dx
x x
Example 10INTEGS. OF SYMM. FUNCTIONS
As f(x) = (tan x)/ (1 + x2 + x4) satisfies f(–x) = –f(x),
thus f(x) is odd and,
1
2 41
tan0
1
xdx
x x
Example 11INTEGS. OF SYMM. FUNCTIONS