October 1, 2010

Integration by Parts 3

Filed under: Tricks of the Trade — Adam Glesser @ 8:43 am
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Tricks of the Trade

(with Professor Glesser)

In the first two installments of this series

Integration by Parts 1
Integration by Parts 2

we introduced integration by parts as a way to compute antiderivatives of a product of functions and we saw how certain integration by parts problems are handled more efficiently with the so-called tabular method (or, in Stand and Deliver, the “tic-tac-toe” method). In this post, we will consider the following question: As integration by parts requires the making of a choice—which is your u and which is your dv—how can we make this choice so that the resulting integral is easier to compute?

From the Mailbag

Über-reader CalcDave wrote in the comments to the last post in this series that,

I usually make a show of how sometimes the order does matter…That is, I’ll let u = x^4 and dv = \sin(x)\ dx the first time and then go through it and say something like, “Well, that didn’t get us much of anywhere. What if we switch up our u and dv this time? Let’s let u = \cos(x) and dv = x^3.” Then when you work it through, everything cancels out and we’re back to the original problem.

Indeed, Dave. Let’s take a look at what happens if we switch it up.

Egad, Dave is right. Since the product of the terms in the last line of the table is what we will need to integrate, doing it this way just makes things worse.  Ah, but what if we start with the cosine on the left and then switch it up? Oh, yeah, we’ll just get back what we started with. This suggests that we should always put a polynomials on the left so that it doesn’t go up in degree. It turns out that there are several examples where this is precisely the wrong thing to do. We implicitly saw this in the first post, but let me give you a couple of more explicit examples.

\int x\sin^{-1}(x)\ dx

If we split this up using our ‘rule’ to always put the polynomial on the left, then we are forced to integrate \sin^{-1}(x). Let’s say you just happen to know the antiderivative of \sin^{-1}(x) is x\sin^{-1}(x) + \sqrt{1 - x^2} + C (I didn’t, although I can use integration by parts to figure it out!). You would now get:
and be forced to integrate the monstrosity on the right. Not for me thank you. However, if you put the \sin^{-1}(x) on the left, we get:
and at the very least we have gotten rid of the \sin^{-1}(x). In fact we have done more, but we’ll have to wait until the next post to resolve this.

Another example is \int x\ln(x)\ dx. Although we did integrate \ln(x) in our first post, it gave an answer of x\ln(x) - x + C and we don’t want to integrate that since it we don’t know how to integrate x\ln(x) (in a future post, we will resolve this last problem directly). On the other hand, if we put the \ln(x) on the left, the derivative will return \frac{1}{x} and the natural logarithm is gone. So when does it pay to put the polynomial on the right? Whenever the derivative of the other function changes it into an algebraic function, it will be right to integrate the polynomial. Otherwise, you should differentiate the polynomial. If we also include trigonometric functions and exponential functions, the rule of thumb is:
Logarithms Inverse-Trig Algebraic Trig Exponential

This list represents a good order in which to choose your u in the following sense: if you have two functions, whichever comes first in the above list should be your u. Some people enjoy a good mnemonic to memorize the order. I’ve heard the following:

LIATE rule (or alL I ATE rule)

Lions In Africa Tackle Elephants

Liberals In America Typify Elitists

Little Indians Are Tiny Engines

Lets Integrate All The Equations

This says, for example, that when confronted with \int \sin(x) e^x\ dx, differentiate the \sin(x) and integrate e^x.

Next Time

In our next segment, we will introduce the box method for handling several of the integrals left unsolved in this post.


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