Integration By Parts

Section 1.2 Integration By Parts

In this section, we introduce an essential method for the evaluation of integrals, called integration by parts. This method often comes in handy when the substitution method does not work. It sometimes is needed in combination with substitution (demonstrated and practiced in the lecture).

Subsection 1.2.1 Motivation

Consider the following three integrals:

$\displaystyle \displaystyle \int \sin(x) \ dx$
$\displaystyle \displaystyle \int x \sin(x^2) \ dx$
$\displaystyle \displaystyle \int x \sin(x) \ dx$

We readily evaluate the first integral since we recognize that the integrand, $\sin(x)\text{,}$ is the derivative of $-\cos(x)\text{.}$ That is, $$\displaystyle \int \sin(x) \ dx = -\cos(x) + C.$$

We can evaluate the second integral by applying the substitution $u = x^2$ (implying $du = 2 x \ dx\text{,}$ or $x \ dx = \frac{1}{2} \ du$). Go ahead, and give it a try. You should obtain $$\int x \sin(x^2) \ dx = -\frac{1}{2} \cos(x^2) + C.$$

But what about the third integral introduced above? We do not recognize it as the derivative of a known function. Further, no substitution will allow us to make progress. It appears that we are stuck with the methods that we know. We need a new method to evaluate the third integral.

Subsection 1.2.2 Preview: Establishing the Need for a Method that Reverses the Product Rule

After the dust clears, we will see that the third integral will evaluate as follows: $$\int x \sin(x) \ dx = - x \cos(x) + \sin(x) + C.$$

We can of course check this by differentiating the function on the right-hand side. Since this function involves a product, $x \cos(x)\text{,}$ we need to apply the product rule, as follows:

\begin{alignat*}{1} \frac{d}{dx} \left[ -x \cos(x) + \sin(x) \right] &= \frac{d}{dx} \left[ -x \cos(x) \right] + \frac{d}{dx} \sin(x) \\ &= \left[ (-x) ( -\sin(x) ) + (-1)( \cos(x) )\right] + \cos(x) \\ &= x \sin(x) - \cos(x) + \cos(x) \\ &= x \sin(x). \end{alignat*}

Based on the above, it appears that we need a method to reverse the product rule in a systematic way.

Subsection 1.2.3 Introduction to Integration By Parts

In the next video, we will introduce you to integration by parts, show how this method reverses the product rule in a systematic way, and work through the example introduced above.

Figure 1.2.1. Video demonstrating how Integration By Parts works. NOTE: The video covers the motivation and preview, so you may skip through the first 6 minutes or so and start viewing at 6:09. We will have this type of repetition for a few more pre-classes, after which the videos will fit seamlessly into the text.

Subsection 1.2.4 Summary

Integration by parts is a method to reverse the product rule in the evaluation of integrals: $$ \int f(x) g'(x) \ dx = f(x) g(x) - \int g(x) f'(x) \ dx.$$
In differential form, the 'formula' or pattern for integration by parts is $$\int u \ dv = u \ v - \int v \ du.$$
In general, choose $u$ and $dv$ such that the integral on the right-hand side is simpler than the original integral on the left-hand side. In a few cases, it suffices for the integral on the right-hand side to be no more complicated than the original integral on the left-hand side.
Rule of Thumb: Let $dv$ be the most complicated part of the integrand plus the differential (e.g., $dx$) for which you can find the antiderivative mentally.

Subsection 1.2.5 Don't Forget

Don't forget to return to eClass to complete the pre-class quiz.

Subsection 1.2.6 Further Study

Remember that the notes presented above only serve as an introduction to the topic. Further study of the topic will be required. This includes working through the pre-class quizzes, reviewing the lecture notes, and diligently working through the homework problems.

As you study, you should reflect on the following learning outcomes, and critically assess where you are on the path to achieving these learning outcomes:

Learning Outcomes

Apply integration by parts to evaluate indefinite and definite integrals.
Apply integration by parts repeatedly to evaluate an integral.
Apply substitution and integration by parts in combination to evaluate an integral.
Apply integration by parts to develop a reduction formula.

The following references provide a good start for review and further study:

Learning Outcome	Video	Textbook Section
1	1.E1	5.5
2	1.E1	5.5 (Examples 3 and 4)
3	1.E1 (plus 10.E1 and 12.E4 from MATH 134)	5.5 (Exercises 21 - 26)
4	1.E1	5.5 (Exercises 27 - 30)