Integration by Substitution
BY
DR. JULIA ARNOLD AND MS. KAREN
OVERMAN
USING TAN’S 5TH EDITION APPLIED
CALCULUS FOR THE MANAGERIAL ,
LIFE, AND SOCIAL SCIENCES TEXT
Recall from the previous section that integration is the opposite of
differentiation.
Recall the Chain Rule for Derivatives:
d
dx
F h x   F' h x   h' x 
which is the derivative of the outside multiplied by
the derivative of the inside
So it stands to reason if integration is the opposite of differentiation,
that

F' h x   h' x dx
 F h x 
Since identifying this pattern can be difficult, we use a method
called substitution to integrate expressions with this form or to
reverse the Chain Rule.
The Method of Substitution and How it Works
Consider

5
2(2x  4) dx
The more complicated expression in the integral is (2x + 4)5.
Using a different variable like u, for instance, we will make a
substitution for the inside of the exponential function.
Let u = 2x + 4, then (2x + 4)5 becomes u5.
Now du may be different from dx, so let’s find du.
If u = 2x + 4, then du = 2 dx.
You will note that we did not write
du
dx
differentials on their respective sides.
but wrote the
Now by substitution we can rewrite the problem from the x
variable to the u variable.

5
2(2x  4) dx 

5
(2x  4) 2dx 

5
u du
Integrating the expression in the u variable we get:
What would this be in the x variable?
2 x  4 6
6
u
6
6
C
C
The problem is solved!!
Checking by taking the derivative:
6 2 x  4   2
5
6
 2 2 x  4 
5
Be sure to check out the simpler examples in the text.
Here are some variations on those examples.
Example 1a. Find
 4x x
2
3

4
dx
Solution: Remember to pick the most complicated expression
and let the inside be u.
Which would you pick for u?
X2 +3
(X2 +3)4
4x
Example 1a. Find
 4x x
2
3

4
dx
Solution: Remember to pick the most complicated expression
and let the inside be u.
Which is more complicated, 4x or x  3  ?
2
What is the inside?
Try again.
4
Example 1a. Find
 4x x
2
3

4
dx
Solution: Remember to pick the most complicated expression
and let the inside be u.
Which would you pick for u?
u = x²+3, is the correct choice for u.
Example 1a. Find
 
4x x
2
3

4
dx 

u 4xdx
4
We need all the variables in our integral to be in terms of u, so
the next step is to find du.
u  x
2
3
du  2xdx
This time the 4x dx doesn’t exactly match the 2x dx, but it
only differs by a factor of 2, a constant.
So to adjust for this we
solve the above for dx.
u  x
2
3
Or, we could multiply by 2
u  x
2
3
du  2xdx
du  2xdx
du
2 du  4xdx
2x
 dx
 
Example 1a. Find
u  x
2
4x x
2
3
2x
 4x x
2
3

4
u
4
4x
2
3
du  2 xdx
Or
 dx
dx 

u 4xdx
u  x
2 du  4xdx
Substitute just for dx
4
dx 
3
du  2 xdx
du

4
du
2x

Substitute for 4xdx
u
4
2 du
 
As you can see, the end result is the same:
4x x
2
3

4
dx 

u 4xdx


u 2 du
4
4
To finish the problem
 4x x

2
3

4
dx 
u 4xdx 
u
4
4

4
2 du  2 u du  2
u
5
5
C 
2
5
x
2
3

5
C
Example 2a. Find
5
3x  1 dx
Solution: What did you pick for u?
u = 3x + 1
du = 3 dx
du
3
 dx
Substitute:
5
3x  1 dx 
5
u
You must change all
variables to u.
du
3
Just like with derivatives, we do a rewrite on the square root.
5
3x  1 dx 
5
u
du
3

5
3
1

3
3
5 2 2
10
2
3x  1  2  C
u du   u  C 
3 3
9
 x x
Example 3a. Find
2
Solution: Pick u.
u  x
3
3
1

3
2
dx
1
2
du  3x dx
du
3x
2
 dx
Substitute, simplify and integrate:
 x x
2
3
1

3
2
dx 
x
2
3
u  2
du
3x
2

1
3
3
5
1 2
2
3
u 2 du   u 2  C 
x 1
3 5
15


5
2
C

Example 4a. Find
Solution: Pick u.
e
 3x
dx
u   3x
du   3dx
du
3
 dx
Substitute, simplify and integrate:

e
 3x
dx 

e
u
du
3

1
3

u
e du  
1
3
e
u
C  
1
3
e
 3x
C
 3x
Example 5a. Find
Solution: Pick u.
x
2
u  3x
1
2
dx
1
du  6 xdx
du
6x
 dx
Substitute and integrate:
 3x
x
2
1
dx 
x du
u
6x

1
6 
du
u

1
6
ln u  C 
1
6
ln 3x
2
1 C
Example 6a. Find
Solution: Pick u.

lnx 2
2x
dx
u  lnx
du 
dx
x
x du  dx
Substitute and integrate:

lnx 2
2x
dx 

 u2

 2x


1
xdu 

2

u
2
du 
1 u
3
2 3
C 
lnx 3
6
C
Here’s one for you to try:
3

x e
x
4
8
dx
Sometimes it is difficult to pick u. One hint given previously was to
pick the inside of the more complicated expression. Another hint is
to pick the expression with the highest power of your variable.
Using these two hints, click on the expression you would pick for u?
x
3
or
e
x
4
8
or
x
4
8
x³ is not the right choice.
x³ is not the highest power of the variable.
Try again.
e
x
4
8
is not the right choice.
Pick the inside of the complicated expression,
or in this case the exponent!!
Try again.
Good work!!
Let
u  x
4
du  4x
du
8
3
dx
3
 x dx
4
Substitute into the integral, simplify and integrate.

3

x e

u
e
x
du
4
4
8

dx 
1
4


u
e
x
4
8
e du 
3
x dx
1
4
e
u
C 
1
4
e
x
4
8
C
An Application to Business
In 1990 the head of the research and development department of
the Soloron Corp. claimed that the cost of producing solar cell panels
would drop at the rate of
58
3 t
 2
2
, 0  t  10
dollars per peak watt for the next t years, with t = 0 corresponding
to the beginning of the year 1990. ( A peak watt is the power
produced at noon on a sunny day.) In 1990 the panels, which are
used for photovoltaic power systems, cost $10 per peak watt. Find
an expression giving the cost per peak watt of producing solar cell
panels at the beginning of year t. What was the cost at the
beginning of 2000?
An Application to Business
This tells you the expression is
a derivative.
In 1990 the head of the research and development department of
the Soloron Corp. claimed that the cost of producing solar cell panels
would drop at the rate of
58
3 t
 2
2
, 0  t  10
dollars per peak watt for the next t years, with t = 0 corresponding
to the beginning of the year 1990. ( A peak watt is the power
produced at noon on a sunny day.) In 1990 the panels, which are
used for photovoltaic power systems, cost $10 per peak watt. Find
an expression giving the cost per peak watt of producing solar cell
panels at the beginning of year t. What was the cost at the
beginning of 2000?
Since the expression is a dropping rate in cost, the expression is C’(x)
or the derivative of the cost C(x) and it should be negative since it is
dropping. Thus:
C ( t ) 
 58
3 t
 58
 3 t  2 
 2
dt 
2
2

, 0  t  10
 58 du
u
2
3

 58
3
u
2
du 
 58 u
3
1
1

58
3u

58
3 3 t  2 
C
u  3t  2
du  3 dt
du
3
 dt
The Cost Function C(t) is
C (t ) 
58
3( 3 t  2 )
C
The Cost Function C(t) is
58
C (t ) 
3( 3 t  2 )
C
Use the initial condition that the cost in 1990 was $10,
or when t = 0, C(0)= $10, thus
10 
10 
58
3(3  0  2 )
58
6
 C
 C
60  58  6 C
2  6C
2
6

Hence,
1
3
 C
C (t ) 
58
3(3 t  2 )

1
3
C (t ) 
58
3(3 t  2 )

1
3
Now to find the cost per peak watt at the beginning of the year
2000 which is 10 years from 1990 and would correspond to
t = 10.
C (10 ) 
58
96

1
3
58
3 ( 3 10   2 )

1
3

58
3 32 

1
3
 . 9375  . 94
Thus the cost per peak watt of producing solar cell panels at
the beginning of 2000 is approximately $.94 per peak watt.