﻿ Integration By Parts

Integration by Parts

When I was first introduced to the formula for integration by parts, I was never really told where it came from. Rather, I was just given the formula and told when to use it. The origins, however, are useful in not only understanding but also remembering the formula.

There is a power rule for derivatives; there is a power rule for integrals. There is a chain rule for derivatives; there is a chain rule for integrals. There is a product rule for derivatives; what is there for integrals? There is integration by parts. To see how the two are related, let us recall the product rule for derivatives.

The product rule states: .

Let us solve for . We have . Now, go through and integrate both sides. This gives us .

However, by the Fundamental Theorem of Calculus, the first integral on the right side is just equal to f(x)g(x), and so we have the formula for integration by parts. It is:

.

However, to keep things simpler, let us replace f(x) and g(x) with u and v, respectively. Doing this, we can rewrite the expression as , which is the form that we will use from this point on.

#### Integration by Parts

One should spend some time looking at the formula and practice doing problems that use the formula. That way it will stay in memory. Of course, it is preferable if one can remember the derivation of the formula, not just the formula itself.

Now that we established the formula, a few questions remain: when do I use integration by parts? What is u and dv in the first integral?

We use integration by parts when we see two distinct functions that don’t seem to be related to each other through a u-substitution. For example,  does not require integration by parts since the 2x is the derivative of x2.

Typically, one will see a function that involves two distinct functions. There are five major classes of functions that we will consider. The first is Logarithmic Functions. The second is Inverse Functions, such as tan-1 and so forth. The third is Polynomial Functions, those are the functions that we first integrated. The fourth is Exponential Functions. The last is Trigonometric Functions.

This order also helps to answer the second question: what is u and dv. Think of the classes of functions as a list of desirability for choosing u. That is, we would really like to have a function with logs in it to be u. The last thing we would want to be u is a function with sine and cosine in it. The following pneumonic may help. It is called LIPET (lie-pet).

A common mistake of those using integration by parts is neglected to put a dx with the term in both dv and du. This is a minor point, but, for the sake of completeness, it needs to be included on intermediate steps. Just remember, whenever we use integration by parts, we use everything inside of the integral for u and dv, that includes the dx.

Let us take a look at an example now. Doing so should help the above paragraphs make more sense.

Example 1:

Evaluate .

Solution:

First we check to see if this integral can be solved with a u-substitution. We see that it cannot. Now, we take a look to see what we have. We notice that we have a Polynomial Function and an Exponential Function. Following the convention of LIPET, we set u = x and dv = exdx.

Now, we need to find du and v. To find du we just take the derivative of u. In order to find v, though, we have to integrate dv. That is why we want to choose logs and inverses for u. It is much easier to take their derivative than to find their anti-derivative. (Conversely, it is not too hard to integrate a trigonometric function or an exponential function.)

Well, simple calculations reveal that du = dx and v = ex. (Notice that we implicitly differentiated the equation u = x.) Now, we follow the formula given: .

So, we have: . The only thing left is to do the last integral: . But that is just ex. And so, .

It is tempting to want to select the polynomial for u. In fact, many students tell themselves that is what they need to do. To reiterate: that is wrong. You should follow LIPET when choosing u and dv. Let us take a look at another example in which the polynomial is not selected to be u.

Example 2:

Evaluate .

Solution:

We start by looking at the functions we have. We have x3 and ln(x). Next we look at LIPET to decide what to choose as u. We notice that we do have a logarithmic function, so we set u = ln(x). That means that dv = x3dx.

Now we need du and v. Implicitly differentiating u = ln(x), we get . Integrating dv = x3dx, we get . Now we just plug what we have into the formula. Doing so, we have:

All that remains is to evaluate the last integral. Using the power rule, we see that it is just . And so, we see that .

There is another way to do integration by parts, made popular in the 1983 movie Stand and Deliver. It is called the tabular method for integration by parts. It greatly reduces the amount of time and work required to solve some integrals. Often times it is not discussed because it only works when a polynomial function is chosen to be the u.

This technique is best illustrated with an example. Let us consider a function that would be incredibly difficult to do with our usual technique for integration by parts.

Example 3:

Evaluate .

Solution:

We could start this problem by setting u = x5 and dv = sin(x)dx, but we would have to do integration by parts five times. Thankfully, the tabular method will make short order of this problem.

We begin by making two columns. On the left column, we write the polynomial. Below it, we write each derivative until we reach zero. (That is why this one works for a polynomial function. No other function will eventually reach zero.)

Now, on the right column, we write the dv. Then below it we list each anti-derivative. We continue until we are lined up with the zero in the first column. For this example, our table would like the following:

Now, we go through and draw arrows diagonally down and to the right, starting at the top left.

And off to the left of the columns, we alternate writing a positive or negative sign, starting with a positive sign. We continue in this manner until we are level with the zero in the first column. Doing this, we get a table that looks the like the next one.

Now, we just follow the arrows. Along each arrow, we multiple terms. Then we add or subtract terms depending on the sign on the far left.

So, for the first row, we get x5cos(x). (We multiplied +x5 with cos(x). We continue in this manner until we reach the last arrow.

After multiplying all of the diagonals together (while being mindful of the signs), we get the following:

The above answer is precisely the answer we would get if we were to do integration by parts the long way. (This can be verified, but would take a couple of pages so it is left as an exercise for the reader.)

This technique is quite useful, but can take a while to master. And remember, this technique only works if a polynomial function is going to be selected as u. Just because an integral contains a polynomial does not mean that the tabular method can be applied.

For example, one could not apply this method to the integral in Example 2. In that instance, LIPET told us to select ln(x) as u, not x3. (In fact, only one use of integration by parts was required to solve that integral.)

There is one more technique for integration by parts. These integrals are sometimes referred to as “party trick” integrals because they involve a trick that if not seen, will make the integral impossible.

Typically, a u is chosen such that after one or two uses of integration by parts, the new integral to emerge will be easy to evaluate, and thus the problem will be solved. There are a few instances where this is not the case. If after a few rounds of integration by parts, it seems that the integral is just repeating itself, then it is very likely a “party trick” integral.

Example 4:

Evaluate .

Solution:

Following LIPET, we set u = ex and dv = cos(x)dx. That means that du = exdx and also that v = sin(x). Following the formula for integration by parts, we have:

.                                                    (1)

The integral on the right side does not look much better than what we started with. In fact, the only difference is that we have a sin(x) instead of a cos(x). To evaluate this integral, we will have to use integration by parts again.

Again, following LIPET, we set u = ex and dv = sin(x)dx. That means that du = exdx and v = cos(x). By the formula for integration by parts, we have:

.                                                  (2)

Now we are back to where we started. If we continued in this manner, we would be repeating ourselves, again and again, with no end in sight.

So, how shall we proceed? Let us plug equation (2) back into equation (1). Doing so, we have:

This is where the trick comes in. Notice that we can add  to both sides of the equation. Doing so yields:

.

Now, we divide both sides by two and add a +C. This gives us our final answer. Thus, after much work and an insightful trick, we have that

The following is a checklist to determine if we want to use integration by parts. As is seen from above, it is not a quick procedure (with the exception of the tabular method, which does not always apply.) Consequently, it is not the first technique that one should try.

1. Can the integral be solved directly? If not, proceed to 2.

2. Are there two distinct functions being multiplied together? If so, proceed to 3.

3. Can a u substitution be used? If so, is there a du? If not, proceed to 4.

4. Pick a u and a dv according to LIPET and evaluate using integration by parts.

Hopefully these questions come naturally when an integral is presented. If not, over time such a mindset will develop. Solving integrals is not easy and there is not always directions on which technique to use. Therefore, it is critical to learn rules for when integration by parts is used and when it is not.

Ultimately, the best teacher will not come from a textbook, but will come from the reader. By asking questions and seeing similarities in questions that use integration by parts, the reader can see when to use the technique.