Let be a composition of two functions and , say for all in some interval If we know the derivative of , say the chain rule tells us that the derivative of is given by the formula Since , this states that . In other words,

implies

In Leibniz notation, this statement can be written as follows: If we have the
integration formula

then we have also have the more general formula

For example, if
then

holds with
so

becomes

In particular, if
this gives us

a result that is easily verified directly since the derivative of
is
.
We notice that the general formula in

is related to

by a simple mechanical process. Suppose we replace
everywhere in

by a new symbol
and replace
by
,
the Leibniz notation for derivatives. Then it becomes

At this stage the temptation is strong to replace the combination
by
.
If we do the last formula becomes

Notice that this has exactly the same form as

except that the symbol
appears everywhere instead of
.
In other words, every integration formula such as

can be yield a more general integration formula if we simply substitute
symbols. We replace in

by a new symbol
to obtain

and then we think of
as representing a new function of
,
say
.
Then we replace the symbol
by the combination
and equation

reduces to the general formula in

For example, if we replace
by
in the formula
we obtain

In this latter formula,
may be replaced by
and
by
,
and a correct integration formula,

results.

When this mechanical process is used *in reverse*, it becomes the
method of **integration by substitution.** The object of the
method is to transform an integral with a complicated integrand, such as
,
into a more familiar integral, such as
.
The method is applicable whenever the original integral can be written in the
form

since the substitution

transforms this to
If we succeed in carrying out the integration indicated by
,
we obtain a primitive, say
,
and then the original integral may be evaluated by replacing
by
in the formula for

EXAMPLE 1. Integrate

Solution. Let us keep in mind that we are trying to write
in the form
with a suitable choice of
and
.
Since
is a composition, this suggests that we take
and
so
becomes
This choice of
gives
,
and hence
.
The extra factor
is easily taken care of by multiplying and dividing the integrand by
.
Thus we have

Now, we make the substitution
,
,
and obtain

Replacing
by
in the end result, we obtain the formula

which can be verified directly by differentiation. After a little practice one
can perform some of the above steps mentally, and the entire calculation can
be given more briefly as follows: Let
.
Then
,
and we obtain

Notice that the method works in this example because the factor
has an exponent one less than the power of
which appears in
.

EXAMPLE 2. Integrate .

Solution. Let
Then
and we get

Again, the final result is easily verified by differentiation.

EXAMPLE 3. Integrate .

Solution. Let
Then
or
Hence

EXAMPLE 4. Integrate

Solution. Let
Then
so
and
we obtain

The method of substitution is, of course, also applicable to definite integrals.

EXAMPLE 5. Evaluate

Solution. Let
Then
,
so that

Now we obtain new limits of integration by noting that
when
and that
when
Then
we write

The same result is arrived at by expressing everything in terms of
.
Thus we have

Now we prove a general theorem which justifies the process used in Example 5.

THEOREM SUBSTITUTION THEOREM FOR INTEGRALS. Assume
has a continuous derivative
on an open interval
.
Let
be the set of values taken by
on
and assume that
is continuous on
.
Then for each
and
in
,
we have

Proof. Let
and define two new functions
and
as follows:

Since
and
are indefinite integrals of continuous functions they have derivatives given
by the formulas

Now, let
denote the composite function,
Using the chain rule, we find

Applying the second fundamental theorem twice, we obtain

and

This shows that the two integrals are equal.

In order to find
,
the integration by parts formula follows the steps

to obtain

A moment's reflection shows that the result was obtained by taking successive
derivatives of
and taking successive integrals of
and combine the result as indicated below:

The procedure can be justified from

where

In view of the above formula we see that if
is a polynomial of degree
then

The integration by parts formula yields

Thus if
,
then the polynomial
is obtained from the previous polynomial
by multiplying by
and adding the term

Exercise: Build up the following table of coefficients for the

Exercise: Compute
for

It is clear now that if
is a polynomial of degree
then

a. Compute for

b. Write down the reduction formula for and

c. Compute the polynomials appeared in and in for

a. Establish the identity

b. Set
,
.
Show that

c. List the coefficients of for

d. Draw the graphs of , , on the square

e. Calculate the coefficients in the expansion for

f. Find the coefficients
in the integral

for

a. Establish the identities

b. Deduce from the identities

c. Compute the coefficients
and
in the expansions

for

d. Compute for

The reduction formula follows from

Hence

For even , write to reduce it to and expand the binomial expression. For odd , set , and when appears, use the identity

Example. Find .

Write ,

,

Therefore

and
so