INTEGRALS
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.
INTRODUCTION
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
2
x
1

x
dx

2
Equation 1
INTRODUCTION
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
Equation 2
So, formally, without justifying our calculation, we
could write:
 2x
1  x dx   1  x 2 x dx
2
2
  udu
 u
2
3
3/ 2
C
 ( x  1)
2
3
2
3/ 2
C
INTRODUCTION
However, now we can check that we have the
correct answer by using the Chain Rule to
differentiate the final function of Equation 2:
d 2 2
32
2
12
2 3
 3 ( x  1)  C   3  2 ( x  1)  2 x
dx
 2x x 1
2
INTRODUCTION
In general, this method works whenever we have
an integral that we can write in the form

f ( g ( x)) g ( x)dx
INTRODUCTION
Equation 3
Observe that, if F’ = f, then


F
(
g
(
x
))
g
( x)dx  F ( g ( x))  C

because, by the Chain Rule,
d
 F ( g ( x))  F '( g ( x)) g '( x)
dx
INTRODUCTION
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:
 f ( g ( x)) g '( x) dx   f (u ) du
 Thus, we have proved the following rule.
SUBSTITUTION RULE
Equation 4
If u = g(x) is a differentiable function whose range
is an interval I and f is continuous on I, then
 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
Example 1
Find
x
3
cos( x  2) dx
4
 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.
SUBSTITUTION RULE
Example 1
Thus, using x3 dx = du/4 and the Substitution Rule,
we have:
1
1
x
cos(
x

2)
dx

cos
u

du

4
4  cos u  du


3
4
 14 sin u  C
 14 sin( x 4  2)  C
 Notice that, at the final stage, we had to return to the original
variable x.
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
3
4
x
cos(
x
 2) dx by the simpler integral

1
cos u du

4
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.
SUBSTITUTION RULE
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
E. g. 2—Solution 1
Evaluate

 Let u = 2x + 1.
 Then, du = 2 dx.
 So, dx = du/2.
2 x  1 dx
SUBSTITUTION RULE
E. g. 2—Solution 1
Thus, the rule gives:

2 x  1 dx  

1
2
du
u
2
u
12
du
32
u
 
C
3/ 2
1 32
 3u C
1
2
 (2 x  1)
1
3
32
C
SUBSTITUTION RULE
Another possible substitution is
E. g. 2—Solution 2
u  2x 1
dx
Then, du 
2x 1
So, dx  2 x  1du  udu
 Alternatively, observe that u2 = 2x + 1.
 So, 2u du = 2 dx.
SUBSTITUTION RULE
E. g. 2—Solution 2
Thus,

2 x  1 dx   u  u du
  u du
2
3
u
 C
3
32
1
 3 (2 x  1)  C
SUBSTITUTION RULE
Find

x
1 4x
2
Example 3
dx
 Let u = 1 – 4x2.
 Then, du = -8x dx.
 So, xdx = -1/8 du and

x
1 4x
2
dx  
1
8

1
1 2
1
du   8  u du
u
  18 (2 u )  C   14 1  4 x 2  C
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 f ( x)  x / 1  4 x and its indefinite
2
integral g ( x)   14 1  4 x 2
 We take the case
C = 0.
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
SUBSTITUTION RULE
So, it seems reasonable, from the graphical
evidence, that g is an antiderivative of f.
SUBSTITUTION RULE
Calculate
e
5x
Example 4
dx
 If we let u = 5x, then du = 5 dx.
 So, dx = 1/5 du.
 Therefore,
e
5x
dx 
1
5
 e du
u
 e C
1
5
u
 e C
1
5
5x
SUBSTITUTION RULE
Find
Example 5
x
1

x
dx

5
2
 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.
SUBSTITUTION RULE
Example 5
Also, x2 = u – 1; so, x4 = (u – 1)2:

1  x x dx   1  x x  x dx  
2
5
2

1
2

1
2
4
du
u (u  1)
2
2
 u (u  2u  1) du
 (u  2u  u ) du
2
5/ 2
3/ 2
1/ 2
 12 ( 72 u 7 / 2  2  52 u 5/ 2  32 u 3/ 2 )  C
 (1  x )
1
7
2 7/2
 (1  x )
2 5/ 2
2
5
 (1  x )
1
3
2 3/ 2
C
SUBSTITUTION RULE
Example 6

Calculate tan x dx
 First, we write tangent in terms of sine and cosine:
sin x
 tan x dx   cos x dx
 This suggests that we should substitute u = cos x,
since then du = – sin x dx, and so sin x dx = – du:
sin x
du
 tan x dx   cos x dx   u   ln | u |  C
  ln | cos x |  C
SUBSTITUTION RULE
Since
Equation 5
–ln|cos x| = ln(|cos x|-1)
= ln(1/|cos x|)
= ln|sec x|,
the result can also be written as
tan
x
dx

ln
|
sec
x
|

C

DEFINITE INTEGRALS
When evaluating a definite integral by substitution,
two methods are possible.
DEFINITE INTEGRALS
One method is to evaluate the indefinite integral
first and then use the FTC.
 For instance, using the result of Example 2, we have:

4
0
2 x  1 dx 

 (2 x  1)
32 4
 (9)
1
3
1
3
1
3

2 x  1 dx
32
4
0
0
 (1)
 13 (27  1) 
32
26
3
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.
SUB. RULE FOR DEF. INTEGRALS
Equation 6
If g’ is continuous on [a, b] and f is continuous on
the range of u = g(x), then

b
a
f ( g ( x)) g '( x)dx  
g (b )
g (a)
f (u )du
SUB. RULE FOR DEF. INTEGRALS
Proof
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
a
f ( g ( x)) g '( x)dx  F ( g ( x)) a
b
 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 (a)
f (u ) du  F (u ) g ( a )
g (b )
 F ( g (b))  F ( g (a))
SUB. RULE FOR DEF. INTEGRALS Example 7
Evaluate

4
0
2 x  1 dx using Equation 6.
 Using the substitution
from Solution 1 of
Example 2, we have:
u = 2x + 1 and dx = du/2
SUB. RULE FOR DEF. INTEGRALS Example 7
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
SUB. RULE FOR DEF. INTEGRALS Example 7
Thus,

4
0
2 x  1 dx  
9
1
1 2
u du
 12  23 u
 (9
1
3

26
3
32 9
32
1
1 )
32
SUB. RULE FOR DEF. INTEGRALS Example 7
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 8
Evaluate

2
1
dx
2
(3  5 x)
 Let u = 3 - 5x.
 Then, du = – 5 dx, so dx = – du/5.
 When x = 1, u = – 2, and when x = 2, u = – 7.
SUB. RULE FOR DEF. INTEGRALS Example 8
Thus,

2
1
dx
1 7 du
 
2
2

2
(3  5 x)
5
u
7
1 1
   
5  u  2
11 1 1
   
5  7 2  14
SUB. RULE FOR DEF. INTEGRALS Example 9

Calculate
e
1
ln x
dx
x
 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,

e
1
1
1
ln x
u 
1
dx   u du   
0
x
2 0 2
2
SUB. RULE FOR DEF. INTEGRALS Example 9
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.
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
Theorem 7
Suppose f is continuous on [–a , a].
a. If f is even, [f(–x) = f(x)], then

a
a
a
f ( x) dx  2 f ( x) dx
0
b. If f is odd, [f(-x) = -f(x)], then

a
a
f ( x) dx  0
INTEGS. OF SYMM. FUNCTIONS
Proof - Equation 8
We split the integral in two:

a
a
0
a
a
0
f ( x) dx   f ( x) dx   f ( x) dx
 
a
0
a
f ( x) dx   f ( x) dx
0
INTEGS. OF SYMM. FUNCTIONS

a
a
Proof
0
a
a
0
f ( x) dx   f ( x) dx   f ( x) dx
 
a
0
a
f ( x) dx   f ( x) dx
0
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
Proof
Therefore,

a
0
a
f ( x) dx    f (u )(du )
0
a
  f (u ) du
0
INTEGS. OF SYMM. FUNCTIONS
Proof - Equation 9
So, Equation 8 becomes:

a
a
f ( x) dx
a
a
0
0
  f (u ) du   f ( x) dx
INTEGS. OF SYMM. FUNCTIONS
Proof a
If f is even, then f(–u) = f(u).
So, Equation 9 gives:

a
a
f ( x) dx
a
a
0
0
  f (u ) du   f ( x ) dx
a
 2  f ( x) dx
0
INTEGS. OF SYMM. FUNCTIONS
Proof b
If f is odd, then f(–u) = –f(u).
So, Equation 9 gives:

a
a
f ( x ) dx
a
a
0
0
   f (u ) du   f ( x ) dx
0
INTEGS. OF SYMM. FUNCTIONS
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

b
a
f ( x) dx 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
Therefore, part (b) says the integral is 0 because
the areas cancel.
INTEGS. OF SYMM. FUNCTIONS
Example 10
As f(x) = x6 + 1 satisfies f(–x) = f(x), it is even. So,

2
2
2
( x  1) dx  2  ( x  1) dx
6
6
0
2
 2  17 x  x 
0
7
 2  128
7  2

284
7
INTEGS. OF SYMM. FUNCTIONS
Example 11
As f(x) = (tan x)/ (1 + x2 + x4) satisfies f(–x) = –f(x),
thus f(x) is odd and,
tan x
dx

0
2
4
1 1  x  x
1