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When integrating and differentiating, sometimes one direction is easy and the other is harder. A nice example is $\frac{d}{dx}\tan x=\sec^2x$, where differentiating is easy but integration (without knowledge of the other direction) is difficult. So to get round the difficult direction you might integrate it thus: $$\begin{align*} \int\sec^2xdx&=?\\ u&=\tan x\\ du&=\sec^2xdx\\ \int\sec^2xdx&=\int du,\\ &=u+c\\ &=\tan x+c \end{align*}$$ This is correct. However, it leaves a bitter taste in my mouth, and I would be hesitant to award any marks for this in an exam. The issue is with the "oracle": the student already knows the answer.

My question is twofold.

  1. Should the above solution get full marks?
  2. If not, how can I explain this to my students (and also to my tutors)?

I am really struggling with (2) - I am struggling to verbalise my issues with this approach*. I am asking (1) because the struggle with (2) makes me wonder if my initial assumption is incorrect (that is, I wonder if the answer to (1) is "yes").

Also, I could phrase the exam question to specifically ban this approach. However, that is beside the point - I want to better understand why this approach should be banned!

*I chose to study maths all those years ago because I thought I wouldn't have to write much...have opinions...communicate...that sort of thing...

user1729
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    If you think that a method for some type of problem is inappropriate than the onus is on you to set problems where the method fails. That way students have an incentive to learn other methods other than the professor's say-so. For what it's worth, I think that the answer to #1 here is a plain yes. – NiloCK Jun 23 '15 at 11:07
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    How else do you "integrate" without previous knowledge of derivatives of elementary functions? Riemann sums?? You might want to add what the preferred method for this particular integral is. – GPerez Jun 23 '15 at 11:08
  • @GPerez one method would be to view $\sec^2x$ as $\tan^2x+1$ then view $\tan^2x=\sin x\cdot \dfrac{\sin x}{\cos^2x}$, and use parts (+a substitution). – user1729 Jun 23 '15 at 11:19
  • (Sorry, I just re-read your comment - were you meaning adding this as a hint in the statement of the exam question, or saying how to do it here?) – user1729 Jun 23 '15 at 11:32
  • Also, I believe that memorising results is pointless - basic calculus courses give you the tools to determine many standard results (such as $\frac{d}{dx}\tan(x)=\sec^2x$) with only the knowledge of the derivatives of $\sin$ and $\cos$. So what is more useful is memorising methods. – user1729 Jun 23 '15 at 11:36
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    I see (maybe add that to the post!). But, my view differs. You seem to prefer that solutions be obtained by a series of steps, or something resembling a deductive process. I agree, but in this case I think any proposed "deductions" are contrived, especially since the average student will know the derivative of $\tan$ at this point. Furthermore, I think this type of answer reflects an exact understanding of what an "indefinite integral" is, namely the solution to $f'(x) = \mathrm{sec}^2x$, whereas arriving via a series a steps may even serve to hide the fact that the student has no idea what – GPerez Jun 23 '15 at 11:37
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    (...) the meaning of their solution is. Therefore I would award full marks, and I would even accept $$\int \mathrm{sec}^2 , dx = \tan x + k, \text{ because } \frac d{dx}\tan x = \mathrm{sec}^2 x$$ – GPerez Jun 23 '15 at 11:41
  • @GPerez Hmm - I am not sure I agree with your definition of integration! I still have memories of my lecturer pointing at the fundamental theorem of calculus and yelling at us, saying that we should all be running out of the lecture theater screaming at the propostity of this result. – user1729 Jun 23 '15 at 11:42
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    Actually I think we both agree on what integration proper is. But this isn't what we have here. Indeed the name of "indefinite integral" is quite unfortunate for what is the calculating of antiderivatives. FTC allows us to relate the two, but I personally would view them as different "operations". – GPerez Jun 23 '15 at 11:52
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    I cannot see why you'd want anything more than just the answer for this one. Please share what you think the best solution steps would be. – Sue VanHattum Jun 23 '15 at 14:11
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    Think of $\frac{d}{dx} \tan x = \sec^2 x$ as a given theorem. We want students to use theorems as opposed to solving everything from base principals. Clearly setting the solution path is important if we are not to frustrate those students who (properly) wish to use theorems. I find nothing more infuriating that for a good student to lose points after using a theorem in good faith to solve a problem. We want to promote students finding the most efficient solution in as much as feasible. With knowledge of trig (proper knowledge) the first solution is totally feasible. – James S. Cook Jun 23 '15 at 14:17
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    I think up $\int x dx = \frac{x^2}{2} + C$ by reasoning around differentiating the latter expression (though I can, if necessary, explicitly imagine its geometric meaning!). I believe that as long as you have covered the derivative of a function (in your case, $\tan(x)$; in my case, polynomials) in class, the approach you outline should receive full marks. Otherwise, one should just require the differentiation be written out (in your case, say, the quotient rule applied to $\frac{\sin(x)}{\cos(x)}$; in my case, perhaps even the definition of derivatives). – Benjamin Dickman Jun 23 '15 at 16:44
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    can you update the question showing us your "correct" solution? I may need to reevaluate the nature of truth and reality based on your response. – JTP - Apologise to Monica Jun 24 '15 at 12:16
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    Can someone elaborate what this post is getting at? It's not clear what "integrating by differentiating" even means. As far as I can glean, the issue is that the student hasn't memorized the table of obscure trig integrals, and they use explicit reasoning to work out the solution, rather than regurgitating it as a 'math fact'. – Tac-Tics Jun 25 '15 at 18:05
  • @Tac-Tics No, other way round: the student has memorized the table of obscure trig identities (they know that $\frac{d}{dx}\tan x=\sec^2x$, and hence conclude that $\int\sec^2xdx=\tan x+c$). – user1729 Jun 29 '15 at 11:37
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    It seems like knowing the answer should be good enough. If the question wanted a proof of the identity, it should be explicit about it. – Tac-Tics Jun 29 '15 at 21:35

8 Answers8

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Your student should get full marks.

In fact, I would say that even a more complicated example, like $$\int 2x\cos(x^2) dx = \sin(x^2) + C$$

should be awarded full points as long as the student justifies this by differentiating $\sin(x^2)$. In fact, this solution demonstrates deeper understanding of the meaning of these symbols than the variable substitution would (the student actually recognizes the chain rule in action).

In general, while differentiation of elementary functions is a mindless algorithm (after learning the basic ones), integration is an art, which fundamentally proceeds by exactly the kind of insight in the OP. Namely, at some point you have to recognize something which you know to be an antiderivative. All of the other techniques (variable sub, trig sub, partial fractions, etc) are just techniques for manipulating the integral to the point where you can clearly see it.

Steven Gubkin
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    +1 for the "integrating by happening to notice it's the derivative of this" method (that's what I call it to the students) and also the "all other techniques are for turning it into something you can just do". You know, there was an exam question here at my uni once with this: $\int \frac{\ln x \cos x - \frac{\sin x}{x}}{(\ln x)^2} dx$. I'm pretty sure every familiar technique for this would fail and the only way is to think "it looks sort of like what the quotient rule would produce". Nasty, but does ram home the point that sometimes, you just want to notice a derivative. – DavidButlerUofA Jun 23 '15 at 19:57
  • @DavidButlerUofA That is a great example of the power of "the method"! – Steven Gubkin Jun 23 '15 at 21:17
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    Actually, you can separate it into two fractions $\frac{\cos{x}}{\ln{x}} - \frac{\sin{x}}{x(\ln{x})^{2}}$ and integrate the second one by parts, with $u = \sin{x}, dv = \frac{dx}{x(\ln{x})^{2}}$. Your result is $-\frac{\sin{x}}{\ln{x}} + \int \frac{\cos{x}}{\ln{x}}dx$. Since this is subtracted from the first integral, the $\int \frac{\cos{x}}{\ln{x}}dx$ terms cancel and the $-\frac{\sin{x}}{\ln{x}}$ term becomes positive, giving your final answer of $\frac{\sin{x}}{\ln{x}} + C$. Not that I disagree overall; I still think this is a strong example. – MintChocolateIceCream Jun 24 '15 at 03:06
  • Wow. In the course this example was in, they hadn't learned integration by parts, so the lecturer really was going for "notice a derivative", but I do like that it can be done by substitution and parts. – DavidButlerUofA Jun 24 '15 at 03:38
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    facepalm Of course you can do it by substitution and by parts! The quotient rule is just the chain rule and the product rule, and the reverse of those are substitution and parts. – DavidButlerUofA Jun 24 '15 at 03:53
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I agree with the other post that you should give full credit unless there were clear directions saying what would and wouldn't be acceptable approaches.

I think it would be very, very hard to convey to students why using the integral of tan, which they happen to know, is off limits, while using other integrals they know is allowed, other than as a detailed and arbitrary decree to make the problem hard.

Students haven't gotten an intuition for why some methods are "clever", some are "good", and some are "unacceptable". They're struggling just to know the difference between valid and invalid. So, in general, questions which ask them to use a particular method have to very clear, and can't depend on them sharing the professor's understanding of what approaches are preferable.

Henry Towsner
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A bit too long for a legit comment... I think a very important question in the background here is that of whether we want to teach students that "correct" math involves primarily adherence to somewhat-arbitrary, even if clear, rules set down by the teacher, or whether there is an underlying reality, and best-practices, etc. Some of the "rule" nonsense can be dispelled by asking questions more forthrightly, rather than having them take place in a secret/tacit context, that is, say "evaluate this integral by reducing to simpler cases involving only sine, cosine, and log ..." ... although once the rule is spelled out, it kinda ruins the "examination drama" of the question. Indeed, asking questions more forthrightly also is a diagnostic for questions whose implicit rules "seem silly when you say them out loud".

This happened to remind me of my bafflement as a little kid when being told that $25\times 4=100$ was not "obvious". Had to show work!

It is true that there is a subtle, perhaps profound, difference between guessing and checking, and having a device to more-systematically determine things. The problem is, there is no reasonable, human-executable algorithm to express integrals in elementary terms. Many are not so-expressible, as Liouville already found. There are a few limited machinations to rearrange an integral until it is recognizable as a derivative of a familiar thing. This is not "systematic". Ever-more a mystery to me why we portray this to students as an essential intellectual enterprise... except, duh, it's easy to make textbooks that do this, and easy to teach such classes. But, seriously, let not try to make the students pretend to see something that simply isn't there.

paul garrett
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Whether you would give this solution full marks depends on which derivatives you expect the students to remember off by heart.

If you expect the students to know off by heart that $\frac{\text{d}}{\text{d}x} \tan x = (\sec x)^2$, then you must accept any answer that directly says $\int (\sec x)^2 \text{d}x = \tan x + C$. The whole point of an indefinite integral is to write down the functions that differentiate to give the integrand, and if you have spent time making them remember derivatives, then this is therefore the same as making them remember integrals.

On the other hand, if you always expect the students to derive the derivative of $\tan x$ by first writing it as $\frac{\sin x}{\cos x}$ and using the quotient rule, and they never get full marks for any question involving the derivative of $\tan x$ unless they do this, then you are justified in feeling uncomfortable with the working you see above. However, I agree with Steven Gubkin you would still have to accept working that includes a derivation of this derivative and then does the integration directly (unless you specifically asked for substitution of course).

So analyse what derivatives you ask them to just know, and these are precisely the integrals you are implicitly expecting them to just know.

PS: In general I would never not give marks if a student gave a clearly-described method that actually worked. For example, the integral $\int \sin x \cos x \text{d}x$ is usually done by rewriting using a trig identity, but if a student did it by substitution or parts and it worked, then I'd be fine with that.

DavidButlerUofA
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I think it's important here that you think about the purpose of an assessment. I tell my first year college students (often to their shock) that the purpose of a test question isn't to see if a student can find the correct answer. The purpose is to see if a student understands the concepts discussed in class. From that perspective, I think you're justified in not giving credit for answers using methods that didn't demonstrate the skills you were trying to evaluate.

However, the other side of that coin is that it's up to the instructor to write questions that can't be answered using other methods. Sometimes this is as simple as specifying a method. For example, if I want to see if students can use the elimination method to solve a system of linear equations, I'll start the question with, "Solve the following systems of equations using the elimination method . . ."

G. Allen
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    +1 for the purpose of a test question isn't to see if a student can find the correct answer and for it's up to the instructor to write questions that can't be answered using other methods, both of which are things I've written a lot about (in bits and pieces, here and there) online for nearly 20 years. By the way, if a student argues the first one by saying this isn't how they'll be evaluated in the workplace (rarely happened with me, but I see it online a lot), then say fine, you’ll do it that way, which means no partial credit. – Dave L Renfro Mar 28 '18 at 18:17
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To agree with most of the answers here, and offer a similar-in-spirit example from a connected field, dealing with (definite) integrals in mathematical statistics is all about re-arranging them to look like a "known" probability density function (PDF). In essence this is also "recognizing a derivative" (better, "recognizing that we can arrive at a derivative") since the PDF is the derivative of the distribution function which is an integral. Sometimes we have also to transform the interval of integration in order to match it with the "benchmark" interval for which the distribution is known... at which point the integral "magically disappears from view" since it just became equal to the cumulative distribution function integrated over the support, and this always equals unity... leaving you with all the stuff you had to attach to it during the transformations -that admittedly may still include an integral, but of those that we nowadays consider "closed-form" since they do not have an analytical solution but are totally mapped numerically like the cumulative distribution function of the standard normal distribution.

Sometimes we arrive not at an integrand that represents a PDF (i.e. the derivative), and so at the cumulative distribution function, but at the moment generating function, which is also a known associated function with the distribution.

To provide the simplest example, if we are given the integral

$$I =\int_{-\infty}^{\infty} \exp\{-0.5x^2-ax\}dx$$ what we will do in mathematical statistics is

"Ah, exponential with the variable squared" $\rightarrow$ normal distribution: add and subtract the normalizing factor, and separate the two terms

$$I = \sqrt{2\pi}\int_{-\infty}^{\infty}e^{-ax} \left(\frac {1}{\sqrt {2\pi}}\exp\{-0.5x^2\}\right)dx$$

The term in the parenthesis is the probbaility density function of the standard normal distribution , typically denoted $\phi()$, so we have

$$I = \sqrt{2\pi}\int_{-\infty}^{\infty}e^{-ax} \phi(x)dx$$

The integral is now the moment generating function of the standard normal (the support is the "benchmark" one), $MGF(s) = E(e^{sx})$ evaluated at $s=-a$ so

$$I = \sqrt{2\pi}\cdot \exp\{0.5a^2\}$$

It is crucial to develop the skill to recognize the "kernels" of the probability density functions (the terms that involve the variable) as well as other exhaustively worked out entities in the field like the moment generating function, or the cumulant generating function. And so, earning marks for it is only natural.

For a more complicated example, see this post over at Cross Validated, where variable transformation is required apart from accounting for normalizing factors (especially under eq. $(3)$ there).

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Alecos Papadopoulos
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I kind of had the same thought as the OP since I'm currently teaching a Cal 2 summer course. I don't think that solving $\int \sec^{2}{x}dx$ by using prior knowledge from Cal 1 is wrong necessarily, but it's interesting to me that textbooks don't provide an alternative method of reasoning the solution. In case anyone is interested, you could also solve it using integration by parts and letting $u=\sec^{3}{x}, dv = \cos{x}dx$. The rest of the steps are fairly easy if you're comfortable with trigonometric integrals. I'm not aware if there's another simpler method.

I know this isn't exactly what the OP asked for but I figured it was relevant and would give the other commenters an alternative solution to reference.

MintChocolateIceCream
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You should give full marks for an "easy" answer, if it is mathematically valid.

Perhaps the classic example was when future mathematician Carl Friedrich Gauss was a grade school boy, and the teacher asked the students to add the numbers between 1 and 100. Gauss solved the problem in about 30 seconds.

He noticed that you could add 100+1, 99+2, 98+3, etc., and that there were 50 such pairings adding up to 101. So 101 X 50 =5050.

Tom Au
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