Calculus: Organizing techniques of integration

January 25, 2010 — rip

introduction & overview

The purpose of this notebook is to organize the useful techniques of integration which are taught in freshman calculus and then presumed known (ha!) at the beginnings of a course in ordinary differential equations.

Heads up: I’m going to mention hyperbolic trig functions, but until you meet them, they are not relevant and you should ignore them. I’m just trying to be thorough, but I fear that I might be confusing. So I’m going to mention them in the details, but omit them from the summary.

First off, there are three categories of integrals:

known
special techniques
general techniques

Here it is in a nutshell: If an integral is on the “known” list, you’re wasting time trying to use special or general techniques. if an integral can be done using special techniques, you’re wasting time using general techniques.

So the primary general guideline is: try the three categories in order.

This is usually the same order in which the methods are taught. What happens, however, is that because the general methods are the freshest in memory when students take the last test, all but the brightest inevitably — and unsuccessfully — try the general methods first, instead of the older, specific, methods. And all but the brightest fail to recognize known integrals, and even they miss them under stress.

There’s no reason for anyone who has mastered the individual techniques to lose out on organizing them into a coherent whole; and no reason to just hope you will somehow recognize known integrals before you’ve had much experience.

The general methods are not universal can-openers. They are the last resort, not the first.

It is no fun to get back a test and discover that you spent a whole lot of time trying to do integration by parts when it was a known integral! Or that you tried to do trig substitution when it was a known integral. Or that you tried general substitution instead of partial fractions, or whatever.

Furthermore, most of the special techniques are used to turn the given integral into a known integral. You have to know when you’ve solved the problem in principle. If you don’t recognize an integral, at least check it against the few that are “known”.

I constructed this algorithm when I was a first-year graduate student. I was TAing the sophomore math course, and I had to explain to my students what they needed to know from their freshman year in order to solve first-order differential equations. I started out, of course, with a list of techniques culled from their first-year calculus book, and then discovered that they needed a list of known integrals, because they simply didn’t recognize them.

Some faculty and grad students argued that anyone good enough to understand my organization was too good to need it. I want to believe they are wrong, and that the market for this may be small, but it’s not zero.

This organization (or algorithm or checklist) does assume that you have learned the specific integrals and the special and general techniques. If you’re struggling to get these details right, you’re not ready for the algorithm yet.

This presentation is sketchy about all the details. One, I merely list the techniques. You may have already written out that much for yourself. But I leave it to you to write out the details of each of the special techniques – and there are details!

Two, I explicitly advocate that you try general techniques last, known integrals first, and special techniques in between.

Three, and perhaps most importantly, this provides a crutch for people who do not yet recognize the known integrals by inspection. If you do enough calculus, you will come to recognize the known integrals automatically — no checklist will be necessary. But that takes time — and this checklist will work until practice pays off and provides you with automatic recognition.

Now I’m going to present the final guideline: check your answer by taking the derivative! Almost all of us can differentiate better than we integrate.

1. known integrals

Just what do I mean by known integrals? They’re the ones that all math faculty, and most grad students, recognize on sight — but beginners don’t. The challenge is to get these integrals before you have learned to recognize them. But if you don’t get that “ah ha!” of recognition, what can you do?

We introduce a beginning step for beginners. For motivation, let us consider

$\int (1+x^2)^{17}\ x \, dx\$ .

which Mathematica could write out for us as…

$1/36+x^2/2+(17 x^4)/4+(68 x^6)/3+85 x^8+238 x^{10}$
$+(1547 x^{12})/3+884 x^{14}+(2431 x^{16})/2+(12155 x^{18})/9$
$+(2431 x^{20})/2+884 x^{22}+(1547 x^{24})/3+238 x^{26}+85 x^{28}$
$+(68 x^{30})/3+(17 x^{32})/4+x^{34}/2+x^{36}/36\$ .

In Version 7, I had to force it to write it all out — in Version 5, that mess was the default answer. It’s a lousy way to write the answer! it should be simply…

$\frac{1}{36} \left(x^2+1\right)^{18}+C$

and we can confirm that by differentiating, which gets us:

$x \left(x^2+1\right)^{17}\$ .

There are two issues. One, a human being could have done the integral the way Mathematica did in Version 5, but it would take time to expand that polynomial and careto get it right — and it’s an ugly answer. Two, if we recognize that this integral is of the form

$\int u^n \, du\$ ,

then we know the answer is simply $\frac{u^{n+1}}{n+1}+ C\$ . Version 7 of Mathematica was smart enough to know this.

But what if we didn’t recognize it? What then? That, after all, is the problem a beginner faces.

Instead of just hoping to recognize it, we can check any given integral against a very short list:

known integrals (column 1, plus 2 special cases and 6 alternatives)

$\int u^n \, du$ $\int u^{-1} \, du$

$\int e^u \, du$ $\int a^u \, du$

$\int \sin (u) \, du$ $\int \cos (u) \, du$ ( $\int \sinh (u) \, du\$ , $\int \cosh (u) \, du\$ )

$\int \sec ^2 (u) \, du$ $\int \csc ^2 (u) \, du\$ , ( $\int \mathrm{sech} ^2 (u) \, du\$ , $\int \mathrm{csch} ^2 (u) \, du\$ )

That’s too big a list. What I actually remember is two things; column 1:

$\int u^n \, du$

$\int e^u \, du$

$\int \sin (u) \, du$

$\int \sec ^2 (u) \, du$

and “there are alternatives and special cases”. I count on the first column to jog my memory if I see one of the additional cases.

Before we start, let me point out that the fourth line is there because it’s worth remembering in general, and it is, frankly, essential when you’re embroiled in integration. And the second column is there to elaborate on the first.

If you have not seen the hyperbolic trig functions (cosh x, sinh x, etc. — trig functions with an “h” appended), then they shouldn’t be on the techniques-of-integration tests. In fact, I don’t generally include them in a face-to-face presentation of this list.

Oh, what is that last integral in column 1?

$\int \sec ^2 (u) \, du = \tan (u) + C\$ .

i.e.

$\frac{d}{du} \tan (u) = \sec ^2(u)\$ .

In other words, instead of remembering the derivative of the tangent, we remember the integral of the secant squared.

(While I’m sure that the integral of csc squared is ±cotangent, I’d have to work out the sign. In other words, because I know the integral of $\sec ^2 (u)\$ , I know I can work out the integral of $\csc ^2 (u)\$ by differentiating the cotangent; it’s just a matter of finding the correct sign. Similarly for the hyperbolic tangent and hyperbolic cotangent: I know their derivatives are ±sech^2 and ±csch^2, and I would quickly work out whatever I needed.)

The additional columns are there because they are sort of redundant. In a fundamental sense, this entire list has 4 entries, not 8 or 12. How can we say we know how to integrate $u^n\$ if we don’t know the special case n = -1? How can we know the integral of the sine without knowing the integral of the cosine? On the other hand, getting the integral of $a^u\$ from the integral of $e^u\$ may not be obvious, but it’s just a special case and we need to know it, or how to work it out. Find the trick in your calculus book. Or look at the post about e.

To say all that another way, the single entry for $\int \sin (u) \, du\$ , for example, prompts me to ask: do I even see a sine, or cosine, or hyperbolic sine, or hyperbolic cosine? If not, move on.

Back to our given problem:

$\int (1+x^2)^{17}\ x \, dx\$ .

If — not that we know it is, but if it is — any one of the 4 known integrals, then it can only be the first,

$\int u^n \, du\$ ,

and then the only possibility is n = 17 and u = 1 + x^2. That’s a big “if”, but try it. That’s the point: to try this one possibility before we move on to consider special techniques. We compute that

du = 2 x dx,

and we discover that, because of the given x dx, our integral is, in fact,

$\frac{1}{2} \int u^{17} \, du\$

i.e.

$\frac{1}{2} \frac{u^{18}}{18}+C$

$= \frac{(1+x^2)^{18}}{36} + C\$ .

(corrected: I had written 2x instead of x^2)

Thus, the guideline for known integrals is: instead of hoping for a flash of insight, we do a little pattern matching against a very small list, and see if we have a match.

And, I emphasize again, we do that before we try special techniques or general techniques.

2. special techniques

This gets us into some gory detail, but just as we can use pattern matching to check a given problem against one possible known integral, the special techniques are go / no-go. (And with pretty much only one exception, if one technique is a go, the other three are no-go.)

Let’s summarize them.

special techniques:

partial fractions

powers of trig functions

trig substitution

complete the square

I’m not going to discuss these in detail: see your calculus book. We can integrate any rational function of (any ratio of) polynomials — we just have to write the ratio in partial fractions, which requires that we factor the denominator. But, partial fractions works only for the the ratio of two polynomials. Period. Until we are looking at a rational function of polynomials, partial fractions is a no-go. Cross it off.

Powers of trig functions works for trig functions (or hyperbolic trig functions). Period. There are a few different cases, and one should be able to handle all of them. Until we are looking at powers of trig functions, the techniques for powers of trig functions are no-go. Cross it off.

Completing the square works for a general quadratic. Period. Until we are looking at a general quadratic, completing the square is a no-go. Cross it off.

Finally, trig substitution (or hyperbolic trig substitution) works for things of the form

$\left(\pm a^2 \pm x^2 \right)\$ .

Period. Until we are looking at the sum or difference of squares, trig substitution is a no-go. Cross it off.

(The one exception I refer to is things like

$\int \frac{1}{1-x^2} \, dx\$ .

We can do it by trig substituion and get an answer involving arc cos (the inverse cosine), or we can do it by partial fractions, since it is the ratio of two polynomials, and get an answer involving the natural logarithm. Either answer is correct, even though they look wildly different.

Now, completing the square usually leads to either partial fractions or trig substitution; and trig substitution may lead to powers of trig functions. It is important to understand that any substitution may lead back to a known integral or to another special technique.

So the first guideline for special techniques is: having decided that it is not a known integral, we ask if any of the special techniques can be applied. If so, use it. In the rare case that more than 1 special technique applies, take your pick.

Thus the second guideline for special techniques is: a special technique may lead back to a known integral or to another special technique.

3. general techniques

There are two general techniques:

integration by parts

substitution

I cannot emphasize too strongly that these are tools of last resort, to be used after we decide that the given problem is not a known integral in disguise, and that no special techniques can be applied.

(OK, by the time we recognize the known integrals automatically, we probably recognize when integration by parts is the way to go, and it is no longer a tool of last resort but the automatic choice in preference to a special technique. But this good judgment will come with experience.)

I distinguish substitution as a general technique from “known integrals” because for the so-called known integrals, we are checking against a short list; for substitution in general, we’re just hoping that something simplifies. “General substitution” is basically an act of desperation. A good or clever choice often works, and we are hoping to make an inspired guess. The purpose of the “known integrals” list is to eliminate the hope for inspiration when it isn’t necessary.

The first guideline for general techniques is: try integration by parts first,

$\int u \, dv = u\ v - \int v \, du$

remembering that dv = dx is a possible choice; for example:

$\int \log (x) \, dx = x\ \log (x) - \int 1 \, dx = x\ \log (x) - x + C\$ .

(You should say something like, “Oh, wow! I can integrate a logarithm using integration by parts!)

The second guideline for general techniqes is: when guessing a substitution, leave some play; think of the integral as

$\int \text{(a function of garbage)} \, d{\text{(garbage)}}$

and make choices for both function and garbage.

I’ll remark that integration by parts is pretty much the only possibility for integrating the logarithm. It is not a known integral, and no special technique applies; and there is no plausible choice for “garbage” other than x, no plausible choice for “function” other than the logarithm. That leaves integration by parts. It happens to work. If it didn’t, we might have created and taught another “elementary function”, defined as the integral of the natural logarithm.

an additional special technique

I suspect that almost every freshman calculus text includes (what I consider) an obscure transformation for integrating any rational combination of (i.e. any ratio of) trig functions. It would be used only if and after the special technique “powers of trig functions” failed to work out.

Let me go look it up — I don’t have it committed to memory.

z = tan x/2.

I know it exists. I know where to look it up. I think it’s not worth remembering in detail, only that it exists and what it solves. I would call it a special technique, but I need it so rarely that I don’t bother to remember it. That is, I don’t remember what it is, but I do remember that it is.

And if I were taking a test that included “techniques of integration”, I would make sure I knew this one if it had been covered. There are some gory details.

Summary

The executive summary is simply, as I said at the beginning in red: If an integral is on the known list, you’re wasting time trying to use special or general techniques. if an integral can be done using special techniques, you’re wasting time using general techniques.

In more detail…. Given an integral to evaluate, assuming you do not recognize it immediately…

There are three categories of techniques

is it a known integral?
if not, can a special technique be applied to it?
if neither, will a general technique work?

you should try the categories in order.
known: instead of hoping for a flash of insight, we do a little pattern matching against a very small list, and see if we have a match.
having decided that it is not a known integral, we ask if any of the special techniques can be applied.
a special technique or a general technique may lead back to a known integral or to another special technique.
I would try integration by parts before trying a general substitution.
when guessing a general substitution, leave some play when choosing what’s what.
There is a rarely-needed very general method for integrating a rational function of sines and cosines.
Finally, check your answer by taking the derivative!

I should remark that I could easily write down integrals which cannot be done using what we know: they end up being used to define new (“not elementary”) functions. $\int x\ \tan (x) \, dx\$ is an example — that is, it can be transformed to such a non-elementary function. (Go ahead, try it!)

In more detail (but omitting hyperbolic trig functions)

known

$\int u^n \, du$

$\int e^u \, du$

$\int \sin (u) \, du$

$\int \sec ^2 (u) \, du$

and “there are alternatives and special cases”.

special

partial fractions
powers of trig functions
trig substitution
complete the square

general

by parts
general substitution

Finally, check your answer by taking the derivative! Almost all of us can differentiate better than we integrate.

appendix: my favorite “proof” that 0 = 1

I am sure that most of you have seen some algebraic “proofs” that 0 = 1, usually involving a division by zero. Here’s one “proof” using calculus. We consider

$\int \frac{1}{x} \, dx$

and integrate by parts — of course it’s a known integral, but integrate by parts anyway!! Let u=1/x, dv = dx, and then from

$\int u \, dv = u\ v - \int v \, du$

we get

$\int \frac{1}{x} \, dx = \frac{x}{x} - \int x\ \left( \frac{-1}{x^2} \right) \, dx$

i.e.

$\int \frac{1}{x} \, dx = 1 + \int \frac{1}{x} \, dx\$ .

Now subtract $\int \frac{1}{x} \, dx\$ from both sides, getting:

0 = 1.

Not good. Definitely not good. Really bad, in fact.

So what did I do wrong, huh? You know I must have done something wrong.

Posted in calculus. Tags: calculus, integration, mathematics. 23 Comments »

Links for Friday « The Number Warrior Says:
February 12, 2010 at 8:14 am

[…] Applied Mathematics Blog: Calculus: Organizing techniques of integration I constructed this algorithm when I was a first-year graduate student. I was TAing the sophomore […]

Sue VanHattum Says:
February 12, 2010 at 2:39 pm

Your first example is wrong. You ask the integral of (1+x^2)^17, when (I think) you meant to ask the integral of x*(1+x^2)^17.

rip Says:
February 12, 2010 at 8:03 pm

Thanks for the comment. I do make mistakes — and someone has already found one in here. But I got “x dx” right in the two places that jump out at me: I put the “x” with the dx….

here: “For motivation, let us consider

$\int (1+x^2)^{17}\ x \, dx\$ .”

and here:

“Back to our given problem:

Is there one I’m not seeing?

Rip

Sue Says:
February 12, 2010 at 8:51 pm

I am so embarrassed! I checked it 2 or 4 times, but I am so used to the x coming first, I just didn’t see it. Feel free to delete my comment, if you want.

I like your integral of garbage d(garbage). I call it junk when I’m teaching this stuff.

rip Says:
February 12, 2010 at 9:30 pm

You could look at it this way: if you didn’t see it, maybe some other people didn’t see it, either.

Not too long ago I asked a question about finite fields on the sci.math newsgroup, and I got politely spanked for overlooking a requirement. It happens.

That said, if you would like me to delete your comment — and this entire sequence — I will do it. I’d rather not, but I think it should be your call.

In any case, welcome to my blog.

Sue VanHattum Says:
February 13, 2010 at 7:07 am

Thank you for your generous response. (It’s fine to leave my comment.)

mekuria getaechew Says:
June 3, 2010 at 12:16 am

you can exeplained as student need thak you

rip Says:
June 12, 2010 at 10:28 am

You’re welcome. I hope it helps.

Asghar Says:
February 22, 2011 at 4:55 am

plz stewart calculus include

rip Says:
February 22, 2011 at 5:44 pm

What do you mean? How would I include a book?

Sper Says:
February 26, 2011 at 2:59 pm

I greatly enjoyed reading this post – it’s clear, concise and so well written. I wish I’d read it when I was younger – it would have made my life so much easier:)

As for the mistake in the example, I think it may have something to do with the continuity of 1/x in zero, doesn’t it?

rip Says:
February 28, 2011 at 6:46 pm

Hi Sper,

Thanks. Nice to hear from you again.

As for the false proof, let me put it this way:

SPOILER SPACE
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.
.
.

any two antiderivatives differ by a constant; they are not the same thing. If I insist on subtracting an antiderivative (an indefinite integral) from both sides, I need to introduce a constant of integration.

khursheed Says:
April 23, 2011 at 10:01 pm

how integration used in Business?

rip Says:
April 24, 2011 at 4:28 pm

That depends on how mathematical you get. The Black-Sholes equation, as I recall, is a partial differential equation. Still, that’s the study of financial derivatives, rather esoteric.

In practice, even engineers don’t use integration all that much. In my experience, third-year engineering students have forgotten the techniques of integration.

I could argue that the calculus is really just the language of science – and if you want to understand a theory, you need to know what an integral and a derivative are… and the best way to understand them, for most of us at least, is to have spent some time computing them.

Along those lines, a quantitative business degree will presuppose that you can understand equations involving integral signs.

amare setie Says:
November 21, 2011 at 7:56 am

I could argue that the calculus is really just the language of science .

mouse Says:
August 23, 2012 at 3:37 am

I’m studying apm for the first time and dont understand integration or what e has to do with it. For ex. (6=e^2k) how would you integrate this and why? the step is what I want to understand.

rip Says:
August 23, 2012 at 4:23 pm

Hi Mouse,

Sorry, but it’s not practical for me to teach you calculus via comments on a blog. If you have a teacher, talk to him. If you don’t have one, get one.

BTW, we don’t usually integrate equations – and we don’t usually use k as a variable of integration – so your example is not a good one. That is, I would _not_ integrate that.

As for e, well, e^x is a function, so it can be integrated or differentiated. It happens to be a very interesting function, because it is its own integral and its own derivative, but that’s another issue.

Good luck,

rip

Abhishek Says:
April 16, 2013 at 3:17 am

Thanks for the tips and tricks! Very helpful!

dedusuiu Says:
February 23, 2018 at 8:22 pm

here sure

dedusuiu Says:
February 23, 2018 at 8:27 pm

also we see as say prof dr mircea orasanu and prof horia orasanu as followed
MULTIPLE INTEGRAL INTEGRATION AND CYLINDRICAL COORDINATES
ABSTRACT
In fact, one could make axiom systems very much like natural deduction systems by adding axioms reflecting typical inferences for natural deduction systems. However, since axioms systems don’t use subproofs, they can’t capture such basic inference patterns as Conditional Proof, Indirect Proof, or Proof by Cases.

moceacu Says:
March 2, 2019 at 8:18 am

for the subject so called Calculus Organizing of integration there are important many aspects and ideas that contributed in main domain these are observed by prof dr mircea orasanu and prof drd horia orasanu and followed specially for integral relations associated of CONSTRAINTS OPTIMIZATIONS that are also ideas of prof dr Constantin Udriste

prof dr mircea orasanu Says:
March 18, 2019 at 8:25 am

these are fundamental integrals for study of important domains in ANALYSIS and complex analysis as observed prof dr mircea orasanu and prof drd horia orasanu and thus followed to study important situations as a position vs. time graph, the slope of the tangent line corresponds to velocity and the steepness corresponds to speed. A velocity vs. time graph is a graphical mapping of all slopes on a position vs. time graph. The SI unit for velocity is the meter per second, m/s.
• Average velocity is displacement divided by time, while average speed is distance divided by time. Speed is always positive or zero, but average velocity can be positive, negative, or zero. Average velocity only depends on initial and final positions, along with the time interval between them; it is independent of the motion in between. In the example above, if the motion lasted for 10 s, the average velocity would be (+1 ft) / (10 s) = +0.1 ft/s, but the average speed would be (15 ft) / (10 s) = 1.5 ft/s. Note: average velocity is not simply the average of the initial and final velocities. This would only be the case if acceleration is a constant.
26. Come up with a nonnumerical scenario that clearly shows that average velocity is not this simple average.
• Acceleration is the rate at which velocity is changing w/ resp. to time. That is, a(t) = v’(t) = x’’(t). Like velocity, acceleration can be positive, negative, or zero. We can say, “Acceleration is to velocity as velocity is to position.” In common language, deceleration refers to slowing down, but a negative acceleration could correspond to speeding up. It all de pas observed prof dr mircea orasanu and prof horia orasanu here appear as followed

Kingsley Kayira Says:
November 13, 2019 at 12:12 pm

so how can u know the integration technique required to use coz some terms are so confusing

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