Two shapes worth spotting: the exponential pair, and the three surd forms

0:00Everything else in this part of the subject is a technique. Something you choose, and then do. The two things in this video are not techniques. They are shapes. Each one is derived once, and then handed over to be recognised on sight. After the derivation there is nothing left to choose. That changes what you practise. You are not learning to pick a method. You are learning to see that a bracket contains a function sitting beside its own rate of change, or that a root is hiding one of three fixed patterns once you complete the square underneath it.

0:40So the work moves from doing to spotting. And spotting can be taught, because the thing you are spotting has an exact description. Two warnings come with them, and I will hold both to the end. One of the three root results is proved where you meet it and the other two are only asserted. And the first result is written down two different ways, and one of the two evaluates nothing at all.

1:10Start with the first shape. An exponential, multiplying a bracket. Inside the bracket, a function, plus that same function's rate of change. Not any two things added together. A function, and its own derivative, side by side. When you see that, the answer is the exponential times the function. Full stop. No integral sign left on the right. Nothing further to do. Just the exponential, times the function that was sitting inside the bracket.

1:45Here is the simplest instance. An exponential times the bracket: inverse tangent of the variable, plus one over one plus the square of the variable. The second term is the derivative of the first. So the answer is the exponential times the inverse tangent, and you can write it down without doing any work at all. That is the whole point of a shape. The work happened once, when it was derived. What is left is recognition.

2:19So let me do the work once, because it is two lines and it explains why the answer has no integral sign in it. Split the integral in two: the exponential times the function, plus the exponential times the derivative. Take the first piece by parts. Put the function in the differentiating role and the exponential in the integrating role. The exponential integrates to itself. So that piece becomes the exponential times the function, minus the integral of the exponential times the function's derivative.

2:58Now look at what you have. The exponential times the function, minus something -- plus the second piece you split off at the start. And the something you are subtracting is exactly the second piece. They cancel. Completely. Nothing is left under an integral sign, and the answer is what is standing outside it. That is why the result is a closed form rather than a step. The by-parts leftover annihilates the term you were going to have to do anyway.

3:34Before I go on, I want to be honest about a trap in checking a claim like that. It would be easy to test this by building the integrand out of the answer. Take the function, differentiate it, add the two, multiply by the exponential. Then of course the answer works. You assumed it. So the derivative that sits inside the bracket is never written down here. It is measured, off the function itself, by the same difference quotient that judges every answer in this series.

4:09Ten functions were run that way: a constant, the variable, its square, the sine, the cosine, the inverse tangent, the logarithm, a reciprocal, the exponential and the inverse sine. All ten pass. None fails. None is unreadable. Now the control, and this is the part that makes it a measurement. Weight that derivative wrongly -- double it -- and only one of the ten still passes. One. And that one is the constant, whose rate of change is nought, so doubling nothing leaves nothing. The family holds exactly one such function.

4:50Turn every answer's sign over and none of the ten passes. So the second term in the bracket has to be the first one's derivative. Not something like it. It. Which brings me straight to the mistake this shape invites. An exponential times a sine. It is an exponential times something. Does the shape apply? No. And you can see why in one line: the bracket would have to hold the sine plus the cosine, and it holds only the sine.

5:23Measured: the exponential times the sine fails the test against the exponential times the sine. It passes against the exponential times the sine PLUS the cosine. One term short, and it belongs to a different case entirely -- the one where you apply the formula twice and solve for the integral on both sides. That is worth saying out loud, because in a set of exercises these two sit next to each other. An exponential times a sum, and an exponential times a sine, one line apart, and they are not the same problem.

6:05The test is never whether an exponential is present. It is whether the bracket beside it holds a function and that function's own derivative. Now the hard case, and it is the only genuinely hard thing in this half. Sometimes the split is there and you cannot see it, because the integrand has been written as a single fraction. An exponential, times the square of the variable plus one, over the square of the variable plus one -- written as a squared bracket underneath.

6:41There is no visible sum. There is one fraction. The move is to add and subtract inside the numerator until the fraction breaks into two pieces, and then to check that one piece is the derivative of the other. The function turns out to be the variable less one, over the variable plus one. Its derivative is two over the square of the variable plus one. Add those two together and you get the fraction you started with.

7:13Now, 'add and subtract until it works' is exactly the kind of step that gets asserted rather than shown. So here it was not spotted. It was searched for. A grid of one hundred and forty-seven rational candidates was fixed in advance. Five integrands were written down with their functions thrown away, and the grid was asked to find each one back. All five are found, each exactly once, and the candidate found is in every case the function that was thrown away.

7:48The match is not checked at a convenient input either. It is checked as an identity of exact fractions: multiplied out and compared coefficient by coefficient, where one coefficient out of place is a refusal. And the control is an integrand that no function beside its own derivative can produce -- a plain reciprocal square. The same grid returns nothing at all for it. The search can fail, which is what makes finding something mean anything.

8:24Second shape, and now we are under a square root. Three results, and they are usually handed to you as a numbered list of three. The first: the root of the square of the variable, less the square of a constant. The second: the same, but with a plus. The root of the square of the variable, plus the square of a constant. The third: the other way round. The root of the square of the constant, less the square of the variable.

8:55Each has an answer, and each answer starts the same way: half the variable, times the root itself. The first two then carry a logarithm -- the logarithm of the modulus of the variable plus the root. The third carries something completely different: an inverse sine. That is the map. One shared first term, and then a fork: two logarithms and one inverse sine. Look at the first two side by side and you will see the problem.

9:29They are the same formula. The sign under the root changes, and the sign before the logarithm changes with it. Nothing else moves. Which is why they are the pair everybody merges. You remember the shape and lose the signs. So instead of writing them as two formulas, I wrote them as one family with two signs in it, set independently, and ran all four combinations at three different constants. Twelve answers.

10:01Exactly six pass and six fail. And every one that passes has the two signs agreeing. A plus under the root wants a plus before the logarithm. A minus wants a minus. Read that the other way and it is a warning. Get one of the two signs right and the other wrong, and you do not have the other formula. You have neither. Half the combinations are simply not integrals of anything you were looking at.

10:32With every sign turned over, none of the twelve passes. The third result was run separately: at each of three constants it passes, and with the sign on its inverse sine coefficient turned over, none of the three does. Now, where does the first one come from? By parts, with a factor of one supplied as the second function -- the same trick as integrating a lone logarithm. The root differentiates. The one integrates, to the variable.

11:07That gives the variable times the root, minus an integral of the square of the variable over the root. And that leftover integral looks worse than what you started with. It is not. The square of the variable over the root is the root itself, with the constant's square over the root taken away. That is the step everything turns on. Substitute it and the leftover integral splits into two: the original integral again, and the constant's square times an integral of one over the root.

11:43So the original integral has come back on the right, with a minus sign in front of it. Move it across. Two of them on the left, one on the right, divide by two, and there is the first result. The same shape of argument as the self-closing case from before: when the thing you are trying to find reappears, you are not stuck, you are one line from an answer.

12:11That middle step deserves a moment, because it is doing all the work and it is easy to nod through. The claim is that the square of the variable over the root equals the root itself, minus the constant's square over the root. It is true for a reason you can see in one line: the root squared IS the square of the variable minus the constant's square, so dividing the root by itself puts it back.

12:40But 'you can see it in one line' is not a measurement, so it was measured. At three constants and both signs -- six readings -- the difference between the two sides takes exactly one value, and that value is nought. The control: turn the sign of the second piece over, so the two pieces add instead of subtracting, and the difference takes five distinct values every time and is nought at none.

13:09A second control adds five to one side, where the difference is still a single value -- and it is five, not nought. So 'one value' and 'that value is nought' are two separate readings here, not one. And the other leftover integral, the one over the root, is checked too: the logarithm of the variable plus the root passes against it, at all six. Now the first warning I promised, and it is about honesty rather than mathematics.

13:43Wherever you meet these three, the first is derived in full. The other two arrive complete, preceded by one clause saying they go the same way. That clause is doing a lot of work, and a video that runs one derivation and then implies three would be overclaiming. So I ran the other two by the same route. The second is the first with one sign changed, and it is already inside those six readings.

14:13The third needs its own version of the middle step, and it comes out differently: the square of the variable over the root is the root back with the OTHER sign, plus the constant's square over the root. At three constants that difference takes one value, nought, every time. And its leftover integral is the inverse sine of the variable over the constant, which passes against one over the root at all three.

14:44So they do go the same way. I am telling you that because I checked it, not because a sentence said so. In practice you will almost never meet those three cleanly. You will meet a general quadratic under a root. The first step is always the same: complete the square underneath. Take the coefficient of the variable, halve it, and that is your shift. Then whatever is left over after subtracting the square of that half is your constant.

15:18And now the destination is decided for you, by two signs. If the squared term is positive and the leftover is positive, you land on the second result. Positive squared term, negative leftover, and you land on the first. If the squared term is negative, you land on the third -- provided there is anything left to take a root of at all. That is arithmetic, so it was done as arithmetic. A hundred and sixty-two quadratics under a root: a leading term of either sign, against every shift and constant from minus four to four.

15:59Fifty-four go to the first result. Twenty-two to the second. Fifty-four to the third. And thirty-two have no root to take at all -- there is no real square root of what is underneath, anywhere. But a classification that nothing checks is just a filing system, so fifty of those cases were then put to the same test as everything else, with the answer their own classification names, shift and all.

16:31Thirty-six pass. None fails. None is unreadable. The other fourteen are the ones the classification had already refused. With every answer's sign turned over, none of the thirty-six passes. One more thing that run showed, and it is a caution about testing. Read the same fifty at one fixed set of inputs, chosen without regard to where each root actually exists, and seventeen of the thirty-six cannot be read at all.

17:04That is not a failure. It is the test declining to answer where there is nothing to answer about -- and it matters, because a test that quietly counted those as passes would be telling you nothing. The lesson for you at a desk is the same one. When you finish one of these, check your answer somewhere the root is real. Half the marks lost on this shape are lost by evaluating at a point that is not in the domain.

17:35Here is how the near-twin actually gets examined, and it is worth seeing once. A quadratic under a root whose completed form is a shifted square, less nine. That is the first result, with the constant three. Four answers are offered. All four have the same first term -- half the shifted variable, times the root. All four carry a logarithm of the same thing. They differ in two places only: the sign before the logarithm, and the number in front of it. Either a minus or a plus, and either four and a half or one and a half.

18:16Nothing else. No arithmetic to do. Just the near-twin, and the coefficient. Exactly one of the four passes, and it is the one with the minus sign and four and a half -- half the constant's square, since the constant is three. So the distinction that looked like pedantry when we compared the two formulas is the entire question here. Merge the twins and all four options look equally plausible.

18:47Last thing, and it is the second warning. The first result -- the exponential shape -- is written down in two different ways, and they are not the same statement. One is what I gave you. The integral of the exponential times the bracket equals the exponential times the function. A closed form. No integral sign on the right. The other keeps an integral sign and a differential on the right: the integral of the exponential times the bracket equals the integral of the exponential times the function.

19:25Look at what that second one actually says. It does not evaluate the integral. It just drops a term from inside it and hands you another integral to do. As a result, it is useless. As a line to revise from, it is worse than useless, because it looks like a formula. Here is the measurement. Differentiate both readings and they agree exactly where the function's own rate of change is nought.

19:56Across the family of ten, that is one function -- the constant -- and nine where they disagree. So the second version says the same thing as the first in exactly the case where there was nothing to say. Everywhere else it asserts something different, and something false. If you have both versions in front of you, the one to keep is the one with no integral sign on the right.

20:24What is worth carrying out of here. Neither of these is a technique. Each is a shape, derived once, then recognised. First shape: an exponential times a bracket holding a function and its own derivative. Answer: the exponential times the function, and nothing else. Ten functions tested; weight that derivative wrongly and nine of the ten stop working. If the split is not visible, manufacture it. Add and subtract inside the numerator until one piece is the derivative of the other -- and check it as an identity, not at one convenient input.

21:07An exponential times a sine is not this shape. It is one term short. Second shape: three roots. One shared first term, then two logarithms and an inverse sine. The two logarithmic ones are one formula with two signs that move together. Six of the twelve combinations are right and six are wrong, and getting one sign right and the other wrong leaves you with neither. For anything general, complete the square first. The sign of the squared term and the sign of what is left over pick the destination between them, and nothing else has to be decided.

21:50And when a result appears twice in two different forms, read the right-hand sides. If one of them still has an integral sign in it, it is not the result -- it is a step that has been mistaken for one.