Students of an introductory college-level Calculus sequence may eventually take a “Calculus II” course. This course is a main stay in the Engineering disciplines as well as a major in Mathematics. One of the main topics covered in this course is techniques of integration — u-substitution, integration by parts, trigonometric substitutions and trigonometric integrals, integration by partial fractions, etc. Of the three-part sequence, some Calculus students find the integration portion of Calculus II to be the hardest. Some specific complaints / shortcomings of students are generally as follows:

• Complaint: “I really liked math until we had to solve integrals. There’s no set procedure for solving them. You just have to guess. This isn’t what math is supposed to be like!”
• Complaint: “There are too many antiderivatives and trig identities to memorize.”
• Shortcoming : A student can muddle their way into Calculus II with substandard Precalculus and Trigonometry mechanics. However, techniques of integration makes a heavy demand on foundational mechanics. Conceptually, the ideas are simple and that’s not what a student ends up wrestling with. Thus, the student with weak technical ability often does poorly here.
• Shortcoming : A student could have reasonably strong basic skills, but could have muddled his/her way through Calculus I — a course focused primarily on limits and derivatives. Calculus II builds on these concepts and the student who hasn’t quite become proficient at differentiation (chain rule, product rule, differentiation of trig and inverse trig functions, etc.) is at an almost irrecoverable disadvantage — symbolic integration requires the ability to conjure antiderivatives.

I can empathize with both type of student complaints. The first is a result of the fact that many students are conditioned to work through mechanics in a mechanical way. Do this. Then that. Move that. And now you have the answer. It’s easy for this to happen, because the inexperienced or lazy teacher may not understand how to explain the thought process behind the mechanics — there is always a thought process. And the more a student thinks while going through the motions the deeper the understanding.

The second complaint is also a result of how students are trained to work with mathematics — memorize the quadratic formula, memorize trig identities, memorize that \(x^{2} + y^{2} = r^{2}\) is the formula for a circle with radius \(r\), etc. It is true that I and probably most mathematicians have these things memorized — but there is a material difference in how this material is memorized. I don’t have formulas memorized in the same way that I have the alphabet memorized. The alphabet is the alphabet and any other permutation of the letters is a reasonable order — one can argue that a true understanding of the order of the letters could come from understanding the history of the alphabet beyond the detail of “the word alphabet is from the Greek ‘alpha’ and ‘beta'”. In any case, practically all of us have have just gotten used to the order of the letters that make up the English alphabet. The quadratic formula, for example, is not to be memorized in this way. There is a comfort level that one can achieve with repeated use of it. However, there is a derivation! Students are too often asked to accept that the result is simply ‘magic’ or if the derivation is shown, it’s shown once and never re-emphasized. Thus, students are ‘pacified’ into accepting that the result isn’t magic and are then coerced into having to memorize the end result.

While there is, undoubtedly, a short term benefit to memorizing the end result, the cost comes in the long run. The short term benefits of memorization also tend to create chronic and curiously uniform weaknesses in math ability. These weaknesses become pronounced during the integration portion of Calculus II. As mentioned earlier, Calculus II is another course in a series of required courses for many Engineering disciplines and of course further pursuit of Mathematics. What is taught in Calculus II is used heavily in many other courses and a weak understanding of integration techniques (amongst other things) effectively means that further pursuit in STEM-related fields becomes progressively more difficult, if not impossible.

So without further ado, let me show you some typical \(u\)-substitution problems that highlight specific weaknesses. Remember, the purpose of these problems isn’t to badger a student into giving up on Mathematics, but rather these problems present an opportunity to tighten some of the mechanics shown in previous courses.

Factoring

When students are introduced to integration by \(u\)-substitution, they get into the habit of choosing \(u\) to be the highest degree power function (or polynomial depending on the problem). That is, without fail, a student who has become used to \(u\)-substitution at a superficial level will choose \(u = x^{3} + 1\) if the problem were written as
$$\int x^{2}\sqrt{x^{3} + 1}\ dx$$
since \(du = 3x^{2}\ dx\) which conveniently handles the \(x^{2}\ dx\) in the integrand.

But now, consider the following integral:
$$\int x^{3}\sqrt{x^{2} + 1}\ dx$$
A natural first choice is to choose \(u = x^{3}\). This is actually ok, in terms of a learning process. The student will quickly be mired in the mud with \(du = 3x^{2}\ dx\) as there is nothing further that can be done (admittedly, the adventurous student will try to somehow replace \(\sqrt{x^{2} + 1}\ dx\) with \(du\), but this is a different mechanical weakness that shows up). In any case, what’s a next possible choice? The student may then debate between \(u = \sqrt{x^{2} + 1}\), \(u = x^{2} + 1\), or \(u = x^{2}\). Of these, \(u = x^{2} + 1\) is the best choice, but many students give up at this point, since they are left with \(u = x^{2} + 1\) and \(du = 2x\ dx\) and there doesn’t seem to be a way to deal with \(x^{3}\ dx\). This highlights a weakness or perhaps an overconditioning into what factoring is:
$$x^{3} = x^{2}x$$

Even when shown the above step, students are befuddled. “Great, now what?” is a common reaction. Well, now, it’s a matter of working those Algebra skills and training the eyes to see. First, \(u-1 = x^{2}\) and \(\frac{du}{2} = x\ dx\) which means that \(x^{2}x\ dx = \frac{u-1}{2}\ du\) and this gives $$\int x^{3}\sqrt{x^{2} + 1}\ dx = \int \frac{(u-1)}{2}\sqrt{u}\ du$$

Once again, even shown this step, students stare not knowing what to do next. This is a result of the fact that students are conditioned to think of factoring polynomials or to think of factoring \(x^{k}\) where \(k\) is an integer. Thus, when faced with \(\frac{u-1}{2}\sqrt{u}\) students don’t consider the option that
$$\int \frac{(u-1)}{2}\sqrt{u}\ du = \int \frac{1}{2}u^{\frac{3}{2}} – \frac{1}{2}u^{\frac{1}{2}}\ du$$
which is easy to integrate directly.

Thus, the purpose of this problem or problems of this type isn’t entirely about \(u\)-substitution. It’s about broadening those overconditioned skills, strengthening those weakly developed Algebra skills, or just expanding the field of vision.

Adding Zero

Consider
$$\int \frac{dx}{e^{x} + 1}$$
as a \(u\)-substitution problem. Eek!! There’s nothing to do here!! And students are completely lost. A choice of \(u = e^{x} + 1\) doesn’t immediately work since \(du = e^{x}\ dx\) and there is no \(e^{x}dx\) term in the integrand. So? Immediately give up. This problem is trickier than the previous one and requires a student to first see that \(dx = 1\cdot\ dx\). Overconditioning to just seeing \(x\) with “no number in front” makes students forget that \(x = 1\cdot\ x\). Again, students are shown at some point that \(x = 1\cdot\ x\), but much like the derivation of the quadratic formula, there is no further emphasis of the hidden, but leading one as the coefficient of \(x\).

Again, students will stare at this hint as useless. And in some sense it is a useless hint. The next thing a student has to do is to consider more options. Adding zero or multiplying by one is a useful technique for rewriting problems. Here, we want to add zero. Notice first that
$$\int \frac{dx}{e^{x} + 1} = \int \frac{1}{e^{x} + 1}\ dx$$

Thus,
$$\int \frac{1}{e^{x} + 1}\ dx = \int \frac{e^{x} + 1 – e^{x}}{e^{x} + 1}\ dx$$
which yields
$$\int \frac{e^{x} + 1 – e^{x}}{e^{x} + 1}\ dx = \int 1 – \frac{e^{x}}{e^{x} + 1}\ dx$$
and is straightforward to integrate.

Adding and subtracting by \(e^{x}\) is a difficult step to intuit, especially if this idea has never been introduced or has been minimally emphasized. The technique of adding zero is sometimes called “pivoting” and is often exploited in proofs using the “triangle inequality”.

Multiplying By One

“Anything multiplied by one is itself” is a pithy arithmetic factoid that students are taught when they first encounter multiplication. By the time they get to Algebra, the notion of “multiplying by one” remains largely unexplored in a non-arithmetic way. Strangely, students get plenty of exposure to “multiplying by one”, but it is called all sorts of different things: cross multiply, multiply/divide both sides by <number>, make a common denominator, etc. This additional vocabulary, while apt for the specific process in mind, masks the “multiply by one” technique.

Consider
$$\int \frac{2}{e^{-x} + e^{x}}\ dx$$

This is another problem that defeats students almost immediately. The presence of two exponentials both in the denominator with nothing “nice” in the numerator to help out creates a sense of hopelessness. One can actually make this problem “worse” by writing it as
$$\int \frac{1}{\cosh(x)}\ dx$$
where \(\cosh(x)\) is hyperbolic cosine. Regardless, the exponentials are still there. The student would have had a relatively easier time with
$$\int \frac{e^{x} – e^{-x}}{e^{-x} + e^{x}}\ dx$$ than the original problem posed. Why? This last problem is a direct, one-step \(u\)-substitution. the original problem needs to be massaged a bit. And there are several ways of massaging. One way, is to multiply the integrand by one via \(\frac{e^{x}}{e^{x}}\) resulting in
$$\int \frac{2e^{x}}{1 + e^{2x}}\ dx$$

Now, some students can finish the problem from here. Other students will try \(u = e^{2x}\) and find themselves in a deeper hole since \(du = 2e^{2x}\) which doesn’t help entirely with the numerator.

What’s the new issue that’s made itself clear? Students forget, misunderstand, don’t know, etc. that \(e^{2x}\) is the same as \(e^{x}e^{x} = (e^{x})^{2}\). Without recognizing this, solving this integral is hopeless. This lack of understanding is often due to an overconditioning on working with exponents and exponentials. There is a heavy focus in an Algebra course to make sure a student understands that \(x^{2}x^{3} = x^{5}\) and that \(3^{x}2^{x} = 6^{x}\). However, \(e\) is often introduced in Algebra II or Precalculus and understanding how to work with \(e^{kx}\) is typically limited to working on problems of compounded growth (bacteria or continuously compounded interest).

Letting \(u = e^{x}\) gives \(du = e^{x}\ dx\) and yields
$$\int \frac{2}{1 + u^{2}}\ du$$

and now the coup de grâce in the student’s spiral into math misery … they haven’t had enough practice with their inverse trig derivatives since \(\int \frac{2}{1 + u^{2}}\ du = 2\arctan(u) + C\). One can solve this without simply recognizing that \(\frac{1}{1 + u^{2}}\) is the derivative of \(\arctan(u)\). To do so, would require making the trigonometric substitution \(u = \tan(\theta)\). But in a typical Calculus II course, trigonometric substitutions are handled separately from the basic \(u\)-substitution. Regardless, either way requires some command of trigonometry.

Doing It Twice

Students can find themselves frustrated when they have made a correct choice for \(u\) but the resultant integral isn’t directly solved. Consider

$$\int \frac{\ln(\ln(x))}{x\ln(x)}\ dx$$

Students may not immediately recognize that \(\frac{d}{dx}\ln(\ln(x)) = \frac{1}{x\ln(x)}\). Instead, a natural first choice for students is \(u = \ln(x)\) with \(du = \frac{1}{x}\ dx\) which gives
$$\int \frac{\ln(u)}{u}\ du$$

(There are other mistakes a student can make at this step — e.g., confusing how the substitution should work given the multiple instances of \(\ln\), but that’s a different issue.) Once a student has arrived at the above step, they will look perplexed and may ask, “Do I have to do another \(u\)-substitution?”. If a student, does indeed, ask this question, that’s great! If not, and they look confused, it is most likely because they are not used to a multi-step problem. In previous math classes and even in Calculus II, problems are typically set up so that the student does one step that’s associated with the new technique and the remainder of the steps are “clean up”. In the problem above, the technique revolves around “undoing the chain rule” via \(u\)-substitution and the first choice of \(u = \ln(x)\) seems to be a correct choice. However, this didn’t result in the problem entering its “clean up” phase. As a result, uncertainty and unfamiliarity sets in.

The next step is \(v = \ln(u)\) which gives \(dv = \frac{1}{u}\ du\) and yields
$$\int v\ dv$$ which becomes $$\frac{v^{2}}{2} + C$$ and undoing all the substitutions gives $$\frac{[\ln(\ln(x))]^{2}}{2} + C$$

The confusion sets in at two different spots. The first, is the idea that there is a second substitution. The second is “all the letters”. Students will want to erroneously use \(u = \ln(u)\) in the second substitution. Conceptually, this is fine. Syntactically, this is wrong. The teacher should step in at this point and promote good syntax. The use of another letter is essential since it clearly separates what’s what and helps to show that if had been seen earlier, \(u = \ln(\ln(x))\) would have been the “one step” choice for \(u\)-substitution.

Final Thoughts

The teacher would be well-advised to encourage more problems of the sort provided here. They help break students out of the mentality of one-step problems, promote the synthesis of other techniques, and allow the student an opportunity to think. While one-step problems have their place in the learning process, the teacher oughtn’t stop there. As the coursework becomes more advanced and as problems become more complicated, it will become essential for the student to be able to break down problems in a systematic way. This is what mathematical thinking provides.

The problems given above aren’t intended to stymie students to the point of anger. The purpose of these problems is to tighten those skills that haven’t been solidified, introduce techniques beyond just the Calculus techniques that should have been learned in previous courses, and to help demonstrate that a fusion of many techniques allows for a more rich understanding into the structure of a problem.

While many of these problems are contrived — i.e., “When will I ever have to do this?” — their purpose is to help develop mechanics. Those mechanics are what students will use, if they know them. For example, $$\int \frac{1}{1 + x^{2}}\ dx$$ is related to the Cauchy distribution. The Cauchy distribution has “practical” use in physics and in mathematical finance. The Cauchy distribution is also a standard example of a distribution that has no mean (not to be confused with having a mean of zero, but rather that the mean does not exist at all!).

There is another weakness that I haven’t discussed here: fractions. I have seen too many students come into the Calculus sequence with nary a clue as to how to work with fractions. Detecting a student’s weakness with fractions doesn’t require integration techniques and that’s why I have not made a larger point of it.

Finally, Calculus II is unforgiving in this way. Those small, mechanical weaknesses in student ability is the actual stumbling block. The concepts are easy by comparison. A teacher of Calculus II should make every reasonable attempt to highlight and exploit these weaknesses in their students so that their pursuit of a STEM-related career does not end here.

Having earned a BS in Electrical Engineering and an MS and PhD in Mathematics, working in a number of fields outside of education, I can safely say that the student will have the opportunity to use what is learned in the Calculus sequence if they pursue a STEM track.