Integration By Parts 1

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This is the first in a series of posts on one of my favorite methods of antidifferentiation: integration by parts. I didn’t love it at first, but a little practice and a few tricks made me appreciate it. Teaching it, well, that is where the love affair begins.

Tricks of the Trade

(with Professor Glesser)

What are you talking about?

Let me assume that the reader is familiar with basic differentiation (including the product rule) and antidifferentiation of some basic elementary functions, i.e., the reader knows such facts as the power rule and how to antidifferentiate exponential functions as well as sine and cosine.

Integration by parts is an analogue to the product rule for derivatives (which tells you how to differentiate a product of functions). In the language of differentials, we have

d(uv) = u\ dv + v\ du

for functions u and v of some common variable, say x. Integrating both sides, we get

uv = \int d(uv) = \int u\ dv + \int v\ du.

The usual form of the integration by parts formula is now obtained by subtracting a term:

\int u\ dv = uv - \int v\ du.

Uh…What?

An example may be helpful. A canonical first example is \int x\sin(x)\ dx. The typical calculus student, fooled by the simplicity of the sum rule and not having the product rule in mind, will incorrectly assert \int x \sin(x)\ dx = (\int x\ dx)(\int \sin(x)\ dx) = (\frac{1}{2}x^2)(-\cos(x)) = -\frac{1}{2}x^2\cos(x). Of course, differentiating shows that this answer is wrong. Why? Well, because antidifferentiation is additive but isn’t multiplicative.

So let’s try the integration by parts formula. We start by noting that \sin(x) is the derivative of -\cos(x), i.e., d(-\cos(x)) = \sin(x)\ dx. Consequently, we could write

\int x \sin(x)\ dx = \int x\ d(-\cos(x)).

We may now apply the integration by parts formula where u = x and v = -\cos(x). This gives

\int x\ d(-\cos(x)) = -x\cos(x) - \int -\cos(x)\ dx = -x\cos(x) + \sin(x) + C.

Could I see one more?

Sure, here is a less obvious example. Consider \int \ln(x)\ dx.

Wait, there is no product of functions.

There is a product; it is just a bit silly. You see \ln(x) = \ln(x)\times 1. Yes, it is one of those kinds of tricks. Now, I know that 1 is the derivative of x and so I can use the integration by parts formula with u = \ln(x) and v = x. This gives:

\int \ln(x)\ dx = x\ln(x) - \int x\ d(\ln(x)) = x \ln(x) - \int x \frac{1}{x}\ dx = x\ln(x) - \int dx = x\ln(x) - x + C.

How do I keep everything straight?

A very common bookkeeping measure is to make a little table including u, v, du and dv. For our first example, you would start with:

u = x \qquad dv = \sin(x)\ dx
du = ? \qquad v = ?

You then compute du = 1\ dx = dx and v = -\cos(x) to complete the table:

u = x \qquad dv = \sin(x)\ dx
du = dx \qquad v = -\cos(x)

You can then simply plug everything into the integration by parts formula.

This isn’t so bad. Why do you need multiple posts?

For those who don’t know the punchline, I won’t spoil it here. It suffices to say that there are some harder problems out there and there are some really efficient ways of handling these difficulties. Stay tuned!

4 thoughts on “Integration By Parts 1”

  1. I had the opportunity to first teach this at Vanderbilt University. So, I switched the order of the letters and would shout “VU!” whenever I could and hold up the little hand signal they do at the football games (thumb, fore-, and middle-fingers). I am now teaching high school, but still in the Nashville area, so they still get it.

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  2. Pingback: Integration by Parts 2 « GL(s,R)

  3. Pingback: Integration by Parts 3 « GL(s,R)

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