# Integration by substitution

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.

# Integration by Parts

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

# Table of Integrals for

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

# Table of Integrals for ,

a. Compute for

b. Write down the reduction formula for and

c. Compute the polynomials appeared in and in for

# Chebyshev Polynomials

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

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