# substitution

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```					MATH 1441                           Technical Mathematics for Biological Sciences

Integration: The Substitution Method
This short document illustrates the so-called method of substitution for determining antiderivatives or
indefinite integrals in its simplest form: the integration of powers of functions. As developed here, the
method of substitution effectively amounts to reversing the special form

d                  d
u n   nu n 1 u                                             (1)
dx               dx

of the chain rule, which is a familiar basic method of finding derivatives of powers of functions. In the next
document in this series, we extend this method of substitution to the more general form of the chain rule,

d               d           d
dx           du f  u   dx u 
f  u   
                                                              (2)

which handles any function of u, not just simple powers. This will make it possible to extend the method of
substitution to any integral which fits a pattern you can find in one of many published tables of integral
formulas -- some of which contain thousands of entries. As a result, the method of substitution is the basis
of the most generally useful method for finding integrals.

Essentially, all that the method of substitution involves is checking (perhaps by trial and failure, or better, trial
and success) to recast the integral

 f  x  dx                                                        (3)

into the form

 Au
n
du                                                       (4)

by introducing some appropriate substitution, u = u(x). Then, using the elementary methods described in the
previous document introducing integral calculus, we would get

A n 1
 f  x  dx   Au       du          u C
n
(5)
n 1

Of course, before declaring the final result for such an indefinite integral, (3), the symbol u must be replaced
by its equivalent in terms of x, since u itself will be introduced as an aid for organizing our thoughts and is
not a part of the original problem at all. When definite integrals are involved, this last step is not strictly
necessary, as we'll demonstrate near the end of this document.

To avoid one of the most common sources of error in applying the method of substitution, you must note that
in the proposed transformation of the indefinite integral

 f  x  dx                     Au
n
du                      (6a)

two parts of the original integral are changed:

f x                     Au n                                      (6b)
and
dx                  du                                              (6c)

Neither of these changes is simply cosmetic -- it is not simply a matter of replacing one symbol by another
symbol. Both changes are the result of a deliberate choice of u, and then taking the exact mathematical

David W. Sabo (2000)                         Integration: The Method of Substitution                   Page 1 of 10
form of u into account. In fact, carrying out the transformation of "dx" into "du" correctly is absolutely crucial
to the success of the method, and requires an explicit, detailed calculation and substitution sequence of
operations.

We can most easily describe this method by demonstrating its use in a number of examples. In the first
example just below, we will formalize the general procedure as a sequence of simple steps.

Example 1: Determine:             x   3x 2  7 dx .

Solution

It shouldn't take you long to realize that none of the three elementary methods discussed in the course so far
will work here. Recall that these three methods could be summarized as

1 n 1
x       dx         x C
n
(i)
n 1

A n 1
Ax          dx  A  x n dx         x C
n
(ii)
n 1
and
 f  x   g  x   dx   f  x  dx   g  x  dx
                                                                          (iii)

Clearly, the integrand here is not a simple power of x, as in (i), nor is it a constant multiple of a simple power
of x, as in (ii). It is a product, but both factors involve the variable x, so it just doesn't fit form (ii) at all.
Finally, although there is a sum inside the square root, the application of form (iii) requires that the entire
integrand be the sum of two or more terms, and this is clearly not the case.

The strategy for implementing (6a - c) above consists of four simple steps, which may occasionally (though
very rarely) have to be repeated more than once:

step (i): Generally, start by speculating that the best choice of u is the most complicated expression that is
raised to a power in the integrand. In this example, since

          
1

x       3 x 2  7 dx   x 3 x 2  7        2
dx

he best initial attempt defines

1
u  3x 2  7             since then                    3x 2  7  u  u    2

step (ii): Once u is chosen, write out the definition of u in differential form, and solve for x in terms of u.
For the present example, we get

d           d
u              u  x  3x 2  7 x  6x x
dx          dx       
so that
1
x              u
6x

Without going deeply into the theory, we will just state that the formula you get approximating x in terms of
u gives the exact relationship that should be used to substitute for dx in terms of du in transforming the
integrand. Although it is a bit of a misuse of the notation (recall our discussion of this in the document on
differentials), we can write here formally

1
dx           du
6x

Page 2 of 10                              Integration: The Method of Substitution                         David W. Sabo (2000)
step (iii): Make the substitutions suggested by steps (i) and (ii) into the integral. Here, this gives us

1  1     
 x  3x                 
1
2
7        2
dx   x u 2    du                                                          (*)
 6x 

Note: we've done only the exact substitutions that arose in steps (i) and (ii). In step (i), we speculated: let

              
1
2
u = 3x + 7, since then the ugly power,                                  3x 2  7  3x 2  7                     2
can be replaced by the relatively simpler
1/2                                                                                                 2
looking power of u: u . Similarly, in step (ii), we determined that if u = 3x + 7, then it makes sense to write
1
dx       du , and this substitution for dx was made to get (*). Nothing more complicated has been done to
6x
arrive at (*).

step (iv): Finally, simplify the integrand as much as possible, to verify that all reference to x disappears,
leaving just the right-hand-side of (6a). Here we get from (*)

1         1          1 12
 xu
     2
 6x du    6 u du
      

It is absolutely necessary that all reference to x in the integrand disappears in this step. If that does not
happen when the normal rules of algebra are applied to simplify the integrand, then the method of
substitution with the choice of u that was made has failed.

If step (iv) is successful in eliminating x from the integrand, we then proceed as indicated in formula (5)
above. Here, we get

1        1                 1 1    1 1
 6u        2
du 
1 1
6 2
u 2 C

1 1 32
         u C
6 32

1 2 32
       u C
63

1 32
      u C
9

Finally, substitute the definition of u in terms of x, to get the final answer

1
           
3

x     3x 2  7 dx                             3x 2  7                C
2
(**)
9

Note that we can verify that this result is correct by differentiation:

d 1                                              1 d                                         0
                                                                    
3                                                3
3x 2  7                           2
 C       3x 2  7                         2

dx  9
                                              9 dx 
                                       


1 3
                     6x 
1
        3x 2  7                   2
(use the chain rule)
9 2

           
1
 x 3x 2  7               2
 x 3x 2  7

David W. Sabo (2000)                                             Integration: The Method of Substitution                                           Page 3 of 10
This is precisely the integrand we started with, and so our final result is verified to be the correct answer.


 5  x  1 3x                    
4
Example 2: Find                           2
 6x  17 dx .

Solution

At first glance this appears to be a rather fearsome integral. However, we cannot know that the substitution
method will not work until we try it out in detail. So, as in the first example above, we start by choosing the
define u to be the most complicated thing in the integrand that is raised to a power. There is only one
obvious choice here:
2                                                          2          4       4
u = 3x - 6x + 17                         since then                (3x - 6x + 17) = u

From this choice, we get

u   6 x  6  x

From this, we get

du
dx 
 6x  6
as the formula we need to substitute for dx in the integrand. Substituting these two pieces into the original
integral then gives

 5  x  1  3x                                                     du
 6 x  17 dx   5  x  1 u 4
4
2

 6x  6
du
  5  x  1 u 4
6  x  1

5 4
6
      u du

5 1 5
       u C
6 5

1
                    
5
     3x 2  6x  17              C
6

as the final answer. You can easily verify that this is the correct result for the original integral by
differentiating this final expression with respect to x, to confirm that the original integrand is obtained.


5x 2
Example 3: Find         4
2x 3  7
dx .

Solution

In an attempt to use the method of substitution, we focus on the fourth root in the denominator of the
integrand. As a first try, we propose

Page 4 of 10                                  Integration: The Method of Substitution                           David W. Sabo (2000)
1
u = 2x + 7
3
so that                                     4
2x3  7  4 u  u                        4

This gives

u  6x x
2

so that we can use
2
du = 6x dx
or
du
dx 
6x 2

in eliminating dx from the integrand. Substituting these two things into the integral then gives us

5x 2
                 
1
                 dx   5 x 2 2x 3  7                      4
dx
4
2x  7
3

1    du
  5 x2 u              4

6 x2

5  14     5 1 3

6  u du  6 3 4 u 4  C
10
                 
3
          2x 3  7                 4
C
9

as our final result. We note that in the second line here, all reference to x cancelled out completely in the
integrand, indicating that the substitution method with this choice of u was successful. Since we may have
made algebraic errors in our work though, we should verify this solution by differentiation. We get

d 10                                  10 d                                                 0
                                                                 
3                                                          3
2x 3  7                4
 C       2x 3  7                                   4

dx  9
                                   9 dx 
                                                


10 3
                       6x 
1
            2x 3  7                     4        2
(using the chain rule)
9 4

5x 2                          5x 2
                 
1
 5 x 2 2x 3  7                      4
                               
 2x               
1
3
7           4       4
2x 3  7

which is exactly the same as the integrand we began with. This verifies that the indefinite integral obtained
just above is indeed correct.


4x 2  9
Example 4: Find            6x 2
dx

Solution
This integral would appear to be very similar in overall appearance to the three integrals we've just
successfully completed using the method of substitution. So, proceeding as before, we notice that the
square root in the numerator of the integrand is the most complicated part of the integrand. This leads us to
propose to use

David W. Sabo (2000)                             Integration: The Method of Substitution                                                          Page 5 of 10
1
2
u = 4x + 9            since then                               4x 2  9  u  u    2

The differential form of this definition of u is

u  8x x

and so we can use

du
dx 
8x

to eliminate dx from the integral in favor of du. Carrying out these substitutions gives

       
1
4x 2  9            2             1
4x 2  9                                         u 2 du
    6x 2
dx  
6x 2
dx  
6x 2 8x
1
1 u 2
48  x 3
          du

Now we see that there is trouble. We've done the proposed substitutions, and we've done all of the
simplifications possible in the integrand, but it still contains x (as well as u). This is unacceptable. The
substitution method has failed here with this particular choice for u. We cannot do anything more with the
integral in the last line. It is a dead end -- any attempt to push further ahead will produce invalid results,
because of this mixing of two different variables.

When this happens, the first thing to consider briefly is whether an alternative choice for u might be made
that might work. To be useful, there would have to be another expression in the integrand that is raised to a
power. In this particular integrand, the only other expression raised to a power is x itself, in the denominator.
But making the choice

u=x

accomplishes nothing for us, since this really just amounts to a change of symbols. So, as far as the
method of substitution is concerned, we are stymied -- there is nothing more we can do here. This integral
cannot be determined using the simple method of substitution described in this document.

In the next document in this series, we will introduce an extension of the method of substitution which allows
us to use published tables of integral formulas. When we re-encounter this particular integral in that
document, we will find that the special table of integral formulas prepared for use in this course does indeed
contain a formula that will work. So, the problem with this example will turn out to be not that there is no
antiderivative for the integrand, but merely that this simplest of substitution methods is not adequate to find
that antiderivative.


The examples above all dealt with indefinite integrals. We conclude this document by using the method of
substitution to evaluate two definite integrals.

5
3x
Example 5: Compute         
2    x2  7
dx .

Solution
Recall that to evaluate a definite integral, you first determine the corresponding indefinite integral (or
antiderivative) and then evaluate it between the upper and lower bounds on the definite integral. So, to
obtain the required indefinite integral using the method of substitution, we propose

Page 6 of 10                             Integration: The Method of Substitution                           David W. Sabo (2000)
1
u=x +7
2
so that                              x2  7  u  u      2

The differential form of the equation for u is

u  2x x

so that to eliminate dx from the integrand, we will use

du
du = 2x dx               or                dx 
2x

Making the substitutions, we get

5                      x 5                            x 5
3x                           1    du 3       1

2      x2  7
dx     
x 2
3x u     2
  u 2 du
2x 2 x  2

Notice that we've added a bit of notation to the limits on the integral in order to remind ourselves that these
two values are values of x, not of u. We could then proceed as follows:

x 5                            x 5                          1 5
3
2 u 2 du  2 12 u 2
2 x
1      31 1
x 2

 3 x2  7             2
 3 52  7  3 22  7
2

 3 32  3 11  7.0207

Alternatively, we can avoid re-introducing x entirely in evaluating this definite integral by noting that


2
x=2                      u = 2 + 7 = 11
and

2
x=5                      u = 5 + 7 = 32

Then

x 5               u  32
3      1     3         1     3 1 1 32          32

2 u 2du  2 u 11 u 2du  2 12 u 2 11  3 u 11
2 x

 3 32  3 11  7.0207

By changing the limits of the definite integral to the corresponding values of u, it is not necessary to
substitute for u in terms of x before the final evaluation of the definite integral. You can see from this
detailed work, however, that both methods really amount to doing the same arithmetic.

y
Example 6: The equation                                                                                        4
2          2   4                                                                           3
y = 9x - x
2
produces a sort of "bow-tie" shaped graph as shown to the right.
Compute the area of the region enclosed within this shape.                                                                           x
1   2
Solution

The graph in the first and third quadrant comes from using the
formula

y  9x 2  x 4  x 9  x 2

David W. Sabo (2000)                          Integration: The Method of Substitution                                          Page 7 of 10
for -3  x  3. To get the graph in the second and fourth quadrant, we need to use the negative square root

y
y  x 9  x2                                                                   4
3
from the original formula. Notice that the graph occurs only for
the interval -3  x  3, since for |x| > 3, the square root contains a                    2
negative value and so y itself does not have a real value. The
other point to note here is that the section of the curve in each of                                          x
the four quadrants is effectively generated by the same formula,
except for changes of sign, and so the regions enclosed in each                                  1     2
quadrant are identical in shape to each other, except for being
flipped left-to-right or top-to bottom. The regions in quadrants 2,
3, and 4 have the same shape as the region in quadrant 1 (see
the figure to the right), and so each has the same area as the
region in quadrant 1. This means that the total area enclosed will
just be four times the area enclosed in the first quadrant. This is
fortunate, because the region enclosed in the first quadrant has
just the sort of shape for which we already know how to calculate areas using integration: it is bordered on
the bottom by the x-axis, on the top by a curve for which we have a formula, and then on the left and right by
specific values of x. In fact, you will recall from the earlier document introducing integration, that we can
write immediately here that the desired enclosed area is expressible simply as

3
Area  4  x 9  x 2 dx                                        (*)
0

This integral looks like a candidate for the method of substitution. So, let

1
u=9-x
2
then             9  x2  u         2

Further, since

u  -2x x

we have
du
du = -2x dx             or              dx 
2 x

to eliminate dx from the integrand. Making these substitutions gives

x 3                         u 0
1     du
Area  4     
x 0
x 9  x 2 dx  4       
u 9
xu    2

2 x

0                               0            0
4     1           1 3                    4 3
        u 2 du   2 3 2 u 2
2 9                               9
 u 2
3      9

  0 2  9 2   36 square units
4 3      3


3          


Note that in the first line here, we replaced the limits of the definite integral from their original form as values
of x, with the corresponding values of u, using


2
x=0                     u=9-0 =9
and

2
x=3                     u=9-3 =0

Page 8 of 10                             Integration: The Method of Substitution                        David W. Sabo (2000)
Note as well, in the last steps of the calculation of this definite integral, where we subtract the value of the
antiderivative at the lower limit of the integral from the value of the antiderivative at the upper limit of the
integral -- we use the actual lower and upper limit values of the definite integral in that order. There is no
reason why the upper limit of the definite integral need be a larger number than the lower limit. Further, the
limits of the definite integral cannot be switched between positions without resulting in a change in the value
of the definite integral itself (you should be able to convince yourself that switching the two limits will have
the effect of reversing the sign of the final result). Once the original correct expression, (*), for the area has
been developed, everything else follows by direct substitution.


A Final Note

We have presented a very formal and strict methodology in this document -- the core of the method of
substitution is organized as four very specific operations in a very specific order in Example 1. In practice,
once you select a potentially useful definition for u, the remainder of the calculation consists of simple
substitutions and routine algebraic simplifications. In particular, the differential symbol (dx in all of the
examples above) is eliminated in favor of du by direct substitution. This is the strongly recommended
approach -- it is straightforward and unlikely to result in errors due to algebraic oversight.

Many books and commentators demonstrate a somewhat different method, which goes something like as
follows for our Example 1. What we want to do is transform the integral

x    3x 2  7 dx

into an integral involving a power of some variable, u. For reasons already described above, the best choice
of u is to turn the most complicated part of the integrand into a simple power of u. Thus, the only possibly
useful choice here is, as before,
2
u = 3x + 7

Then, since the differential form of this equation is

u  6x x

we will use the formula

du = 6x dx

to eliminate dx from the original integral. The reasoning used then goes something like this: in order to
make du appear in the transformed integral, it is necessary to first make the product "6x dx" show up in that
integral. We get partway there by moving the factor x rightwards so it is adjacent to the dx:

x    3 x 2  7 dx                 
3 x 2  7  x dx 

All that is still missing is the factor 6. So, just insert it where required just to the left of the x, but then, to
compensate for this insertion, it is necessary to divide the integral as a whole by 6. This gives

x    3x 2  7 dx                  
3x 2  7  x dx  
1
6
           
3x 2  7  6x dx 

1
Now, in this last form, we can clearly see that         3x 2  7  u  u 2 , and (6x dx) = du, and so the integral
can be rewritten

x    3x 2  7 dx 
1
6
            
3x 2  7  6x dx  
1 12
6
u du

The rest of the work is exactly the same as in our Example 1.

David W. Sabo (2000)                     Integration: The Method of Substitution                            Page 9 of 10
Now, this somewhat different approach is perfectly valid from the point of view of mathematical principles.
However, it can be confusing in practice because the operations involved in creating the expression for du
inside the integral consist of doing two things at once -- inserting factors into the integrand, and at the same
time, inserting compensating factors outside of the integral. When this approach is used, errors due to
carelessness, or less than perfect concentration are much more likely than with the strict substitution
approach we used in our examples above.

We strongly recommend the strict substitution for dx illustrated in the examples in this document, followed by
routine algebraic simplification of the integrand. This approach avoids the need to keep track of two sets of
changes at the same time. Further, this approach alerts you much more clearly when the substitution
method is failing. In the alternative approach which attempts to construct the expression for du inside the
integral, the failure to do so could mean either that the substitution method is failing or that some algebraic
error is being made. With the strict substitution approach, you just carry out whatever algebraic
simplifications are possible after substituting the expression for dx, and if those simplifications fail to result in
all references to x disappearing, you know immediately that the substitution method has failed.

Page 10 of 10                        Integration: The Method of Substitution                  David W. Sabo (2000)

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