Integrating a product, and how the choice of first function decides whether it helps

0:00Here is an integral that looks harmless and is not. The variable, multiplied by the cosine of the variable. Go down any list of standard results and nothing on it has that shape. There is no row for a product. Try a substitution. A substitution wants the integrand to be one function of something, times the rate of change of that something. Here the two factors have nothing to do with each other. The variable is not the rate of change of the cosine, and the cosine is not the rate of change of the variable.

0:38So there is a gap, and the technique that fills it is unlike every other technique in this part of the subject. It does not evaluate anything. It takes one integral away and hands you a different one. Whether that is progress is a separate question, and it is the only question that matters. The technique comes from a rule you already have, and it takes one line to get it.

1:07The rate of change of a product is the first times the rate of change of the second, plus the second times the rate of change of the first. Before going on, I measured that. Six products were built -- a power against a sine, a power against an exponential, a sine against a cosine, a squared power against a cosine, an exponential against a sine, a power against a cosine -- and each one's rate of change was measured at five inputs, from both sides, and compared with the two-term expression.

1:44All six agree. The control is what a reader who has not met the rule writes instead: the rate of the first times the rate of the second. None of the six agrees with that. All six disagree. Now integrate both sides of the rule. Integrating a rate of change gives the thing itself, so the left side becomes simply the product. The product equals the integral of the first times the rate of the second, plus the integral of the second times the rate of the first.

2:20Move one of those integrals across. The integral of the first times the rate of the second equals the product, less the integral of the second times the rate of the first. That is the whole technique. It is the product rule, integrated once and rearranged, and it converts one integral into a different one. To use it you have to hand it two factors in two different roles, and the roles are not symmetric.

2:52One factor is going to be differentiated. Call it the first function. The other is going to be integrated. Call it the second function. Read what each role asks and the choice starts to make itself. The role that gets integrated asks: can I integrate you at all? Not easily -- at all. If I cannot, the formula cannot even be written down, because the formula needs that antiderivative in two places.

3:23The role that gets differentiated asks something gentler: does differentiating you make you smaller? A power differentiated is a lower power. Differentiate it enough times and it becomes a constant and then nothing. A sine differentiated is a cosine. Same size. Differentiate it forever and it never gets anywhere. So one role is about what you can do, and the other is about where you are going. Hold both. Back to the variable times the cosine, and take it both ways round, because the failure teaches more than the success.

4:05First the sensible way. Put the power in the differentiating role and the cosine in the integrating role. The cosine integrates to the sine. So the answer begins as the variable times the sine, and then subtracts the integral of the sine times one. That leftover integral is just the sine, which integrates to minus the cosine. Two lines and it is over. The variable times the sine, plus the cosine.

4:37Now the other way round. Put the cosine in the differentiating role and the power in the integrating role. The power integrates to half the square of the variable. So the answer begins as half the square times the cosine, and subtracts an integral in which the square is still sitting there. Look at what happened to the integral. It started with a power of one under the integral sign. It now has a power of two.

5:08The trade was legal. It was also backwards. You have swapped an integral you could not do for a harder one. That is one case. I wanted to know whether it is the pattern, so I wrote the trade down as a machine. The machine holds a state: a coefficient, a polynomial, and the name of one other factor. It applies the formula in whichever role it is told, and it counts how many applications the integral needs before nothing is left under the sign.

5:43It is never told which role is supposed to work. Every power from nought to seven, against each of the exponential, the sine and the cosine. Twenty-four integrals. With the power in the differentiating role, all twenty-four finish. None is still open at a cap of ten applications. And every one of them finishes in exactly as many applications as the power's own degree. A power of three takes three. A power of seven takes seven.

6:17The same twenty-four with the other factor first: none finishes. All twenty-four are still open at the cap, and by then the degree under the integral sign has climbed to seventeen. So the wrong choice does not merely fail to help. It moves away, one degree per application, and it does that every single time. But watch the word I did not use. I did not say the wrong choice is incorrect.

6:49Here is what happens if you take the cosine first once, notice, and then take the power for the rest of the way. The run still finishes. It takes three applications where one would have done. And the answer it assembles passes the same test as the short answer. Across six inputs the two answers differ by exactly one value, and that value is nought. Not equivalent answers. The same answer.

7:19Take the cosine first twice and it finishes in five. I ran the whole grid that way -- one deliberately wrong step, then the power -- across the first four powers against all three other factors. Twelve runs. Every one of them finishes, and every one takes exactly two applications more than it needed to: the wrong step, and the extra degree the wrong step created. So the trade is not a trap you fall into and cannot climb out of. It is arithmetic you have to do twice.

7:55That is worth knowing under exam pressure. A wrong first move costs you time, not the question. Hold on to the phrase 'because it is a power', though. In a few minutes it will stop being true. The usual advice at this point is a rule of thumb, and it is usually taught as a preference. Powers first, unless the other factor is a logarithm or an inverse ratio, in which case that one goes first instead.

8:27Two sentences, and the second overrules the first, and nobody says why. Here is why, and it is not a preference at all. Remember what the integrating role demands. It demands an antiderivative, in hand, before you can write the formula down. So I kept an explicit list of what can be integrated -- a constant, a power, an exponential, a sine, a cosine -- and, deliberately, what cannot: a logarithm, an inverse tangent, an inverse sine. Those three are exactly the functions whose integrals are what we are trying to build.

9:08Then twenty items were asked how many of their two roles that list permits. Eleven admit both roles. Seven admit exactly one. And two admit neither. Every one of those seven has a logarithm or an inverse ratio in it. So the second tier is not an exception bolted onto the first. It is what is left when you delete the roles that cannot be taken. Those functions go first because they are the ones nobody can integrate, so they had better be the ones that get differentiated.

9:45The first tier is a choice, decided by which trade is favourable. The second tier is not a choice at all. And the two items that admitted neither role? They are the interesting ones. They are lone functions. A logarithm, on its own. An inverse tangent, on its own. There is no product, so there is nothing to put in two roles. Supply one. Write the logarithm as the logarithm times one.

10:17That is not a trick. One is a perfectly good function; it integrates to the variable. Now the roles are forced, because the logarithm cannot be integrated: the logarithm differentiates, the one integrates. The formula gives the variable times the logarithm, less the integral of the variable times the rate of change of the logarithm. That rate is one over the variable. Variable over variable is one. The leftover integral is the integral of one.

10:54The variable times the logarithm, less the variable. Both of those lone functions were put through the same test as everything else in this video, and both answers pass. Supply the constant factor to the two stuck items and every one of the twenty admits at least one role. Nine forced, eleven open, none stuck. So the technique is not for products. It is for anything you can honestly write as one.

11:27A question that troubles careful students, and deserves a straight answer. The second factor gets integrated. Every integral carries an arbitrary constant. So should you carry one here? The usual answer is that it is unnecessary. I would rather show it than say it. So the whole answer for the power against the cosine was built again, with an arbitrary constant added to the antiderivative of the second factor and carried all the way through, at five different constants.

12:03It appears twice. Once in the boundary term, multiplied by the first function. Once inside the leftover integral, multiplied by the rate of change of the first function -- and integrating that gives the same product back. One is added and one is subtracted, because of the minus sign in the formula. At every one of the five constants the difference from the plain answer takes exactly one value across six inputs, and that value is nought.

12:37The control is what happens if you turn the formula's minus sign into a plus. Then the two terms add instead of annihilating, and the difference takes six distinct values, and is nought at none of them. So the cancellation is the minus sign's doing. Carrying the constant is not wrong. It is work that undoes itself. Here is a harder one, and it needs a technique inside a technique.

13:08An inverse sine, multiplied by the variable over the root of one less the square of the variable. The inverse sine cannot be integrated, so the roles are decided for you. The inverse sine differentiates. The other factor integrates. But you cannot write the formula down yet, because that other factor is not a standard result either. You have to integrate it first, by substitution: the thing under the root has a rate of change of minus twice the variable, and the variable is sitting right there on top.

13:46That piece integrates to minus the root of one less the square. Only now can the formula be written. The answer begins as minus the root times the inverse sine, and subtracts an integral of minus the root times the rate of change of the inverse sine. And the rate of change of the inverse sine is one over that very root. The root cancels. What is left is the integral of one.

14:16The variable, less the root of one less the square, times the inverse sine. That answer went through the same test and passes. So did the little substitution underneath it, on its own. Notice the shape of the work. The substitution was used to prepare a factor. It did not do the integral. Now the case students abandon one line before the end. An exponential, multiplied by a sine. Neither factor gets smaller when you differentiate it. The exponential comes back as itself. The sine comes back as a cosine. There is no power to run down.

15:00Apply the formula anyway. You get a boundary term and a new integral -- the exponential against a cosine. Apply it again, keeping the same role for the same kind of factor. You get another boundary term, and the integral you get back is the exponential against the sine. Which is where you started. That looks like total failure. It is the opposite. Let the unknown integral be a letter. What the two applications produced is an equation: the unknown equals two known terms, minus the unknown.

15:38The machine reports the multiple it came back as: minus one, after two applications. Minus one is not one, so there is an equation to solve. Add the unknown to both sides and divide by two. Half the exponential, times the sine less the cosine. Two things about that, and both are measured. First: it works whichever role you start with. Put the exponential in the differentiating role both times instead, and the state also comes back as minus one times the original, in two applications. The two answers both pass the door, and across six inputs their difference takes exactly one value, which is nought. Not equivalent answers. The same answer.

16:31This is the one case in the whole topic where the choice genuinely does not matter, and it does not matter for a reason: the loop is going to close either way. Second, and this is the trap: alternate the roles. Take the sine to differentiate on the first application and the exponential to differentiate on the second. That also comes back after two applications. But it comes back as PLUS one times the original.

17:02The unknown equals two terms plus the unknown. Cancel and you are told that nought equals nought, which is true and useless. The machine returns no answer there. It refuses, rather than dividing by nought. You did not make an error. You undid your first application with your second. And one more warning. Two applications do not always mean a loop. A squared power against an exponential also takes two applications, and what it reaches is not a multiple of anything -- it is finished, with three boundary terms and nothing left under an integral sign.

17:46Now the honest limit, and it is sharper than it is usually stated. Everything the machine counted earlier was really counting one thing: how many steps the exponent on the variable needs to reach nought. A power of three needs three steps. Three, two, one, nought. Then there is no power left and the integral is a standard one. So let the exponent be any fraction at all, and ask the same question.

18:16Every exponent from nought to three and a half, in halves, against a sine and against a cosine. Sixteen runs. Eight finish. Eight do not. And every exponent that finished is a whole number. The halves go one half, minus one half, minus three halves, minus five halves. They step past nought and never land on it. The power does not disappear, it moves into a root underneath, and the run is endless.

18:48The control matters here. Give the same machine permission to step in halves and all sixteen finish, and the exponents that finished are no longer all whole numbers. So the eight is a reading of the whole numbers, not a count of everything I offered it. The variable times a sine: exponent one, finishes. The root of the variable times a sine: exponent a half, never finishes. One radical apart, and on opposite sides of what this technique can reach.

19:24I want to be careful about what has just been shown, because this is where confident statements get made that are bigger than the evidence. What the machine measured is that the trade does not close on the root. That is a statement about this technique. It is not a statement that no answer exists. My machine cannot decide that, and it does not claim to. What it can do is refuse the answer you would write by analogy.

19:54The variable times a sine gives the sine, less the variable times the cosine. Clean, and it passes. So write the analogous thing for the root: the sine, less the root of the variable times the cosine. Measure its rate of change and compare it with the integrand it is supposed to belong to. It fails. That is the useful lesson. The pattern that worked for the power is not a pattern about the shape of the answer. It was a consequence of the exponent reaching nought, and once it does not, nothing carries over.

20:36One more thing, and it is the kind of detail that quietly costs marks. Suppose you meet the inverse sine of twice the variable over one plus its square. Almost everyone rewrites that as twice the inverse tangent of the variable, and then integrates by parts with a factor of one supplied. The rewriting makes it easy. The rewriting is not an identity. I put the two expressions side by side at nineteen inputs, from minus two and a half to two and a half.

21:11They agree at seven of them and differ at twelve. And the seven are exactly the inputs whose modulus is under one. The rate of change says it more sharply still. Inside that stretch, the inverse sine's measured rate is two over one plus the square -- which is twice the rate of the inverse tangent, as expected. Outside it, that is simply wrong, and the same expression with its sign turned over is right.

21:42So the rewriting carries a restriction. It holds while the modulus of the variable stays under one, and outside that it changes sign on you. If the question does not state the restriction, state it yourself. It is the difference between an answer and half an answer. Let me put the whole decision on one board, because in practice you make it in about four seconds. You have a product. Ask two questions in order.

22:14First: is one of the factors something you cannot integrate -- a logarithm, an inverse sine, an inverse tangent? If yes, that one differentiates. The decision is made, and it was never a choice. That covered seven of the twenty items I ran. Second, if both factors can be integrated: is one of them a power? Then the power differentiates, because the power is the only thing here that gets smaller when you differentiate it.

22:47That is the other tier, and the twenty-four counted runs are its evidence. There is one leftover case, and you have already met it: an exponential against a sine or cosine, where neither factor shrinks. There the choice does not matter and the loop closes either way. And if there is no product at all, supply one. Multiply by one and go back to the first question. Four seconds. Two questions, and a rule for the case where both answers are the same.

23:23What is worth taking out of here. This technique evaluates nothing. It trades one integral for another, and the trade is favourable or it is not. Favourable has a meaning you can count: the exponent on the power comes down by one each application, and the run ends when it reaches nought. Twenty-four integrals finish that way and take exactly the power's own degree of applications. Not one of the twenty-four finishes with the roles the other way round.

23:57The rule of thumb's second tier is not an exception. A factor you cannot integrate cannot take the integrating role, so the choice was never open. A constant on the second factor's antiderivative cancels itself, and the minus sign in the formula is why. When the original comes back, read the multiple. Minus one gives you an equation. Plus one means you undid your own work. And when the power is a root rather than a whole number, the exponent steps past nought instead of landing on it, and the trade never closes.

24:37The formula is one line. Everything else in this video was the question of whether applying it helps -- and that question has an answer you can count, not one you have to remember.