Six formulas for quadratic denominators, and completing the square to reach them

0:00Here is an integrand that looks like nothing much. One, over a quadratic. A square term, a term in the input, a constant. You own a list of standard integrals. Nothing on it has a quadratic underneath. Substitution does not reach it either -- there is no inside piece whose rate is sitting in the numerator, because the numerator is one. So a new set of results gets handed over. Six of them, and then four more that build on the six.

0:34Ten things to remember, apparently. It is not ten things. It is three things to remember and one thing to do, and the whole of this idea is getting that sorting right. Start with the six, because they are not six. Look only at what is underneath, and there are three denominators. The input's square less a constant's square. The constant's square less the input's. And the two added. A difference, the same difference the other way round, and a sum.

1:08Each of those three appears twice: once bare, and once under a square root. Three times two is six. That is the whole structure, and it is worth more to you than the order they usually come in. I checked all six as mathematics rather than copying them down. Each one is two rules that know nothing about each other -- the integrand, and the answer -- and the answer's rate is measured and compared with the integrand's value.

1:39Five inputs inside the stretch each row actually has, at four different values of the constant. Twenty-four checks, twenty-four passes. And the same twenty-four answers with the sign turned over all fail, which is what makes the passes worth anything at all. Now the second piece of sorting, and it is the one that will save you time in an exam. Four of the six answers are logarithms. Two are inverse functions -- an inverse tangent and an inverse sine.

2:15You could learn that off the notation. I would rather measure it. Take each answer and run it to the edge of the stretch it lives on, or, if it has no edge, out to a far input. Four of them run away: they pass four in size and keep going. Two of them stay put, never reaching two. The two that stay are exactly the two the notation calls inverse functions. A logarithm has nowhere to stop; an inverse function does.

2:47So here is the sorting worth carrying. A sum of two squares gives an inverse tangent. The difference the other way round, under a root, gives an inverse sine. Everything else gives a logarithm. And two of the six are the same result written twice. The first row and the second row have the same denominator with its sign turned over -- and their answers are related the same way.

3:17The second answer is minus the first. I read both at six inputs, some inside the constant and some beyond it, and all six were actually read. Their difference takes exactly one value across those six, which is what two answers to one integral must do -- and here that value is nought. They are not different functions. They are one function written twice, because the two are used on opposite sides of the constant and it is convenient to have the logarithm's argument come out positive either way.

3:55So the count of things to learn drops again. Three denominators, and two of the six answers are one answer. Where do the six come from? Two of them are proved one way and four another, and the difference is not a matter of taste. A difference of two squares has real zeros. I went looking with a halving search and found one within a thousandth of the constant, both ways round, at all four constants.

4:27Having zeros means the fraction can be split into two simpler ones -- and I checked that split as an identity at five inputs rather than quoting it. Once it is split, each piece is a reciprocal of something linear, and you already know that integral. A sum of two squares has no real zeros. The same halving search across two thousand units of input returns nothing at all, at every constant tried, and the denominator never falls below the constant's square.

5:02Nothing to split it into. So that proof has to do something else entirely, and so do the three under the root. What they do is substitute, and each of the four picks a trigonometric ratio to substitute for. The sum of squares, bare: put the input equal to the constant times the tangent. The difference, under a root: the secant. The difference the other way round, under a root: the sine. The sum, under a root: the tangent again.

5:36I did not take those on trust. Each one is carried out as arithmetic. Read the integrand at the substituted input, multiply by that input's own rate, and compare the product with what the proof says is left over. All four hold, at five angles and four constants. Sixteen checks, sixteen passes. And hand the sum of squares the sine substitution instead of the tangent, and it agrees at none of the four. The choice of ratio is doing work.

6:12Now look at what the four substitutions actually leave behind, because this is the part nobody says out loud. Two of them leave a constant. Integrate a constant and you are done. The other two leave the secant. The secant is not on the standard list. Its integral is not something you can look up in the rows you started with. It had to be derived, and you derived it earlier, when you were learning substitution.

6:43I checked it here too: the logarithm of the modulus of the secant plus the tangent has the secant for its rate, at all five angles. And I put the secant beside four ordinary rows -- the sine, the cosine, the tangent and the input itself -- and it is none of them. So these six results are not a fresh start. Two of the four substitution proofs stand directly on work you did before this, and if you skipped that, you cannot follow these.

7:17This is a second storey on the same building. That is the three things to remember. Here is the one thing to do. Nobody hands you a difference of two squares. They hand you a general quadratic, with a middle term. Complete the square. Pull the leading coefficient out, write what is left as a shifted square plus a leftover, and the leftover is a number. That number's sign is the entire decision.

7:48Leftover positive, leading coefficient positive: you have a sum of two squares. Leftover negative: a difference. Leading coefficient negative with a negative leftover: the difference the other way round. Three signs, three shapes, and then a lookup. That is not an approximation and it is not a rule of thumb. It is arithmetic on the coefficients, and I ran it. Forty-five quadratics. Five leading coefficients, three middle terms, three constants.

8:26Each one has its square completed by arithmetic on the coefficients alone. One of the forty-five turns out to be a perfect square with no leftover at all, so it is set aside rather than routed. The other forty-four get sorted by the two signs into the three shapes -- and into a fourth outcome, no real stretch at all, which I will come back to. Then every one of the forty-four answers that the sorting produces goes to the same door as everything else: measure its rate, compare with the integrand, at three inputs each.

9:05Forty-four pass. None fails. None goes unread. That is what it means to say the sign decides. Not that it usually decides, or that it decides in the cases anyone bothers to write down. Forty-four out of forty-four. Now run the same forty-four under a square root, and something changes. Six of them have no answer at all. Those are the ones where the leading coefficient is negative and the leftover is positive -- which means the quadratic is negative at every input there is.

9:44A square root of a negative number is not a number, so there is no stretch for the integral to live on. Not a hard integral. No integral. And here is the thing worth noticing: those six were identified by the sorting, from the two signs, before anything was read. The other thirty-eight all pass. So the same two signs that choose the shape also tell you, before you start, whether there is anything to find.

10:17One detail causes more lost marks here than anything else, so let us measure what it costs. The leading coefficient. Two in front of the square, or three, or five. It has to come out of the integral before you complete the square, not after. Out of a bare quadratic it comes out as a reciprocal; out of one under a root, as a reciprocal square root. Complete the square without pulling it out first, and the shift is wrong, the leftover is wrong, and the factor in front is missing.

10:52I built exactly that reader and put their answers to the door. Where the leading coefficient really is one, they get all eight right, bare and under a root. Which is the control -- the machinery is not simply refusing everything. Where it is anything else, they get none of the thirty-two right bare, and none of the twenty-six under a root. Not fewer marks. Zero. Everything so far had a one on top. What if the numerator is linear?

11:26There is one move, and it is worth understanding rather than remembering. The denominator has its own rate -- differentiate the quadratic and you get something linear. So write the numerator as some multiple of that rate, plus a leftover constant. Two unknowns. Find them by comparing coefficients: the terms in the input on both sides, and the constant terms on both sides. Two equations, two unknowns, done. Now the integral has split into two. The first piece is a rate over the thing it is the rate of, which is a logarithm. The second piece is a constant over a quadratic, which is one of the six.

12:13A logarithm and an inverse function, side by side, from one integrand. I ran that over thirty-three numerator-and-quadratic pairs. The full answer passes every single time. Thirty-three from thirty-three. Then I asked the question a hurried student's working answers by accident: what if you only do the first piece? The logarithm alone answers two of the thirty-three. And those two are exactly the two where the second unknown comes out nought -- where the numerator already was a multiple of the denominator's rate and there was nothing left to split off.

12:57The second piece alone answers none of them. Under a root the numbers are twenty-seven from twenty-seven, one, and none, with the same one explanation. So the split is not decoration. Both pieces carry part of the answer, except in the case where the split does not happen at all. One trap left, and it is a good one. Two of the six are the rooted difference of squares and the rooted sum. Their answers are written almost identically -- a logarithm of the input plus a root.

13:35The only difference is a sign under that root. It is tempting to treat them as one result. So I put each one's answer to the other one's integrand, at four constants. All eight fail. And their difference is not even a constant: across five inputs it takes five different values, so they are not two answers to one question in any sense at all. They look alike because a sign under a root is small on the page. They are not alike, and the proofs are not alike either -- one substitutes the secant and the other the tangent.

14:15Let us put it together, end to end, on seven integrands that cover every shape this idea has. A bare difference of the input's square and sixteen. Write sixteen as four squared, match the first row, and the answer is an eighth of a logarithm of a ratio. Twice the input less its square, under a root. Complete the square and it becomes one less a shifted square -- the inverse sine shape. The answer is the inverse sine of the input less one.

14:50A monic quadratic that completes to a shifted square plus four: an inverse tangent. A leading coefficient of three: the three comes out as a reciprocal, the square is completed inside, and a difference of squares is left. A leading coefficient of five, under a root: the five comes out as a reciprocal square root instead. And a linear numerator over each of a bare quadratic and a root -- a logarithm plus an inverse tangent in one, a root plus an inverse sine in the other.

15:28All seven pass the rate door. All seven with the sign turned over fail. Two problems of this kind offer you four answers and ask you to pick. You never have to pick. You differentiate. The first: a monic quadratic that completes to a shifted square plus one. Four options, and exactly one has the integrand for its rate. The second: a root, whose quadratic completes to a constant less a shifted square scaled by four. The four comes out and leaves an inverse sine of a ratio.

16:07Again four options, again exactly one passes. The near misses are the instructive part: a missing shift, a missing half, a sign turned over inside the inverse sine. Each of those is a real mistake somebody makes, and each of them fails the moment you differentiate it. So: what do you actually have to carry out of here? Three denominators. A difference of two squares, the same difference turned round, and a sum.

16:41Each of them bare and each under a root. Two of the six answers are one answer written twice, and only two of the six are inverse functions. And one move: complete the square, look at the sign of what is left, and pull the leading coefficient out before you do. A linear numerator adds one split and nothing else. That is not ten things. It is a small structure with a decision in the middle of it, and the decision is made by the sign of a number you have already worked out.

17:15The list is not the point. Knowing which row you are heading for before you start is the point.