Let and .
Then and .
Thus
We will compute integrals using the integration by parts technique.
INTRODUCTION
In this section, we are interested in integrating products. We have some experience with this from our exposure to -substitution problems such as In this scenario, the factor of in the integrand is of great significance as it is the derivative of which is the inside of the composition . We will now turn to problems where the factors in the product making up the integrand are unrelated in this or any other significant way, i.e., they are functionally independent of one another. We will effectively create a product rule for integrals and interestingly, we will use the product rule for derivatives to help us. One stark difference between this product rule for integrals and the product rule for derivatives, known as integration by parts (IBP), is that IBP does not give us a final answer. Instead, integration by parts simply transforms our problem into another, hopefully easier one, which we then have to solve. In this last respect, IBP is similar to -substitution. The following integrals can be computed using IBP:
IBP Formula
Recall the product rule: (where and are functions of ). Integrating both sides of this equation with respect to gives where the has intentionally been suppressed from the notation for simplicity. Now the integral on the left hand side is simply by the anti-differentiation definition of the integral symbol. So, Subtracting the second integral from both sides gives which can be condensed as the famous integration by parts formula.
- 1.
- To use the IBP formula, the original problem must be written in the form . This means that and must be declared (similar to declaring in -substitution), and then we compute by differentiating , and by anti-differentiating .
- 2.
- The constant of integration does not appear on the right hand side of the IBP formula (it mysteriously disappeared in the above derivation) because it was taken into account by the constant of integration implied in the integral on the right hand side, .
- 3.
- The IBP formula does not give a final answer to a problem. It replaces one integral with another another integral (which still has to be computed).
We will now use the integration by parts (IBP) formula to compute integrals involving products of unrelated factors, like the ones shown above.
example 1 Compute the integral using IBP:
The integrand is the product of two factors, and . We consider one factor to differentiate and the other factor to anti-differentiate. The factor we prefer to differentiate is since the derivative of is a constant and constants are easy to deal with in integrals. As for the factor, whether we differentiate it or anti-differentiate it makes little difference. In fact, only a change in sign occurs, as Thus to apply the IBP formula, we let Next, we compute by differentiating and we compute by anti-differentiating :
We do not need to add the constant of integration to v, as any
anti-derivative will suffice for our purpose and we choose the constant of integration
to be zero.
Now we use the Integration by Parts formula, :
We have found an answer, but at this stage we should prove that it is the correct answer by differentiating it. The result should be . Note that the differentiation requires the product rule on the term: . We have which proves that our answer was correct.
Tabular Integration
Integrals of the form require IBP times when is a sine, cosine or exponential function. A notational shortcut called tabular integration can be used to streamline the process of using IBP multiple times. As an example, let’s revisit the previous example:
example 4 Compute the integral using tabular integration:
We will construct a table that lists the derivatives of and the anti-derivatives of . We can then obtain the final answer by a series of multiplications and additions or subtractions, as indicated by the arrows in the table.
Thus,
(problem 4) Compute the integral using Tabular Integration: Complete the table and
then enter the final answer.
Thus,
example 5 Compute the integral using Tabular Integration: We construct a table
with the derivatives of and the anti-derivatives of . We add arrows which indicate
multiplication of an element of the column with elements in the column one row
lower. We adorn the arrows with alternating and signs to indicate either addition or
subtraction of terms. 
Thus,
(problem 5) Compute the integral using Tabular Integration: Complete the table and
then enter the final answer.
Thus,
Inverse Functions
We will now look at some examples the involving inverse functions and . We do not (yet) know anti-derivatives for these functions, so we will have to let equal the inverse function.
example 6 Compute the integral: In our previous examples, we would have
let and . But notice that,in this case, we do not know an anti-derivative
of , so we would be stumped in finding . Instead, we will let then The
power of has increased, rather than decreased, as in previous examples,
but this is offset by the fact that the derivative of the natural log has a
completely different form which is easier to work with. Now, by IBP, we
have,
This method can be used to anti-differentiate as well.
Special Examples
example 9 Compute the indefinite integral: This classic problem requires IBP twice
with the interesting twist that we will not get a final answer, but rather a full
circle back to the original integral, which we will call . The final answer will
then be obtained by solving an equation of the form where is a constant
(not equal to 1). We begin by letting We will now use IBP twice with
being the trig function both times. For the first application which gives
For the second application of IBP we continue with At this point in the
computation we must observe that this last integral is the original integral,
(with a coefficient of 1/9). Thus we can substitute in to get from which
we can solve for : and since , We will now check our answer. It will be
slightly easier if we factor out the common factors first: The derivative of the
RHS should yield the integrand: which is the integrand, so our answer is
correct.
(problem 9a) Compute the integral:
Let and .
Then and .
Thus Let and .
Then and .
Thus
Then and .
Thus Let and .
Then and .
Thus
Solving algebraically for the integral gives:
(problem 9b) Compute the integral:
Let and .
Then and .
Thus Let and .
Then and .
Thus
Then and .
Thus Let and .
Then and .
Thus
Solving algebraically for the integral gives:
Our next example involves the function and arises in the calculation of arc length of the parabola .
At the end of our computation, we will need to recall the following integral
example 10 Compute the indefinite integral:
We will call the original integral and use IBP, arriving back at the original integral (as in the previous example). Then, Using the trig identity , gives
and so and finally,
example 11 Compute the integral:
We will call the original integral and use IBP, arriving back at the original integral (as in the previous example). Let Then, Using the trig identity , gives and so and finally,
REDUCTION OF POWERS
(problem 12) Create a reduction of powers formula for the integral: To create the reduction formula, we will use IBP with This gives the reduction
formula:
(problem 13) Create a reduction of powers formula for the integral: Use IBP with
This gives: Using , we can write this last integral as Combining the terms involving
the original integral gives the reduction formula:
(problem 14) Create a reduction of powers formula for the integral Use IBP with
This gives the reduction formula:
APPLICATIONS
example 15 Find the volume of the solid of revolution obtained by revolving the
region between the curves and about the -axis.
We use the method of shells and the formula with We get which we compute using integration by parts with This leads to
(problem 15) Find the volume of the solid of revolution obtained by revolving the
region between the curves and about the -axis.
We use the method of shells and the formula with We get which we compute using integration by parts with This leads to
Here is a detailed, lecture style video on integration by parts:
_