Changing the variable until what is left is a standard form

0:00You now own a list of standard integrals and two tools for breaking a problem into pieces that are on it. Here is an integrand that defeats both. The sine of three times the input. It is on no row of the list, and no amount of splitting or rescaling will put it there. The list has a row for the sine. It does not have a row for the sine of something else.

0:26What you need is not another row. It is a way to change what the question is written in. That is the whole of this idea, and it has one sentence of advice attached to it about how to choose. The advice is good, and it is going to fail us four times in the next twenty minutes. Each failure is more instructive than the advice. Start with what licenses the swap, because it is not a convention.

0:56If you build a rule by feeding one rule into another, its rate at any input is the outer rule's rate, taken at the inner rule's value, times the inner rule's own rate. Read from left to right that is how you differentiate a composite. Read from right to left it says something else. If an integrand is already of that shape, you can integrate it by integrating the outer piece alone.

1:25Nothing in this video is told what the derivative of anything is, so that statement was measured instead. Six composites were built: a sine, a cosine, a logarithm, a square root, a tangent and an exponential, each taken of a different inner rule. For each, three separate measurements. The rate of the whole thing. The rate of the inner rule. And the outer rule's rate, read at the place the inner rule sends the input to.

1:57At every one of five inputs, in all six, the first was the product of the other two. That product is the entire method. Everything after this is about recognising it. So here is the advice, in one sentence. Substitute for something whose own derivative is already sitting among the factors of the integrand. That is a testable instruction, and testing it is more useful than repeating it. Write the integrand not as one lump but as the factors it is actually written as, because the advice is a claim about factors.

2:38Then list the candidates: the sub-expressions the integrand itself contains. An argument, or a ratio appearing inside it. The bare input is never a candidate. Every integrand contains it, and it would make the search say yes for nothing. Now ask each candidate two questions. First: does its rate settle to the same number at every input? That is the easy case, because a number can always be pulled out in front.

3:11Second, for the rest: is its rate a constant times some combination of the factors already standing there? Every non-empty combination is tried, against a grid of a hundred and twenty-six constants. A candidate that answers neither is a candidate the advice does not reach. Eleven worked integrands were run through this, together with the eleven answers that go with them. First the answers, at the same door every claim in this series goes through. Measure the answer's rate at five inputs and compare it with the integrand's own value there.

3:54All eleven pass. Nothing below is about an answer being wrong. Now the search. Two of the eleven are answered by a number: the candidate's rate is the same constant everywhere, so it comes out in front and the rest is a lookup. Six of the eleven are answered by a factor already standing there. The candidate's rate is a stated constant times something already standing in the integrand. Six plus two is eight. Three are answered by neither.

4:28Three out of eleven is not a rounding error. It is nearly a third of the worked material. And those three are exactly the three that get rewritten before anything is substituted at all. Before the interesting failures, one substitution carried all the way through, because the step people drop is the step that costs them. The sine of three times the input. The candidate is three times the input, and its rate is three at every input, so it is the easy case.

5:05Change the variable. But the piece after the integrand has to change too: a step in the new letter is three steps in the old one. Forget that and you get minus the cosine of three times the input. Remember it and you get the same thing divided by three. Which of those is right is not a matter of taste, and the difference is not something a constant of integration can absorb.

5:34Here is that, counted. Take the correct answer and add each of forty-one different constants to it. Put all forty-one through the rate door. All forty-one pass. That is what a family looks like. Now take the version with the divisor dropped, and try to repair it with the same forty-one constants. None of them pass. Not one. An additive constant is free. A multiplicative one is a different function, and the door knows.

6:08Sometimes one change of variable leaves another to be made. The fourth power of the tangent of a square root, times the square of the secant of that root, all over the root. Run the search. It finds the root: the root's rate is a half times the factor one over the root. Substitute, and you are left with the fourth power of a tangent times the square of a secant. Which needs another substitution.

6:38But the search found something else too, in the same pass. The tangent of the root. Its rate is a half times two of those factors at once: the square of the secant, and one over the root. That candidate does the whole thing in one move. Two candidates, one integrand, both correct. The advice does not rank them, and neither does the search. It reports both and lets you pick.

7:06The answer, two fifths of the fifth power of the tangent of the root, passes the door either way. Two familiar integrals come out of this immediately, and they are worth watching because the move is the same both times. The tangent. Written as a sine over a cosine, its factors are the sine, and one over the cosine. The search finds the cosine. Its rate is minus one times the sine standing beside it.

7:39So substitute for the denominator. What is left is one over the new letter, and that is on the list. The answer is a logarithm. The cotangent goes the same way with the roles swapped. The search finds the sine, whose rate is plus one times the cosine standing beside it. Two integrals, no cleverness, and the advice earned both of them. Hold on to the pattern, because it is exactly what will be missing in three minutes.

8:13The tangent's answer is written two ways, and students routinely believe they have found a contradiction. Minus the logarithm of the modulus of the cosine. And the logarithm of the modulus of the secant. In this video those are not two names for one thing. They are two separate constructions. The logarithm here is found by halving, and the secant and the cosine are separate measurements. Neither answer is built out of the other.

8:46Read both at five inputs. They agree at all five. Not to within a constant. Exactly. That is worth saying carefully, because the previous idea in this series was about answers that differ by a constant and still name one family. This is not that. These two are the same function. The control is the check that makes the agreement mean something. Move one of them by half a unit and they differ at all five inputs.

9:18So the door can tell agreement from near-agreement, and it says these two agree. Now the secant. The integrand is one factor: the secant itself. The candidates are the sub-expressions it contains. The secant. The cosine. The sine. The tangent. Four candidates. One factor. A hundred and twenty-six constants to scale by. The search returns nothing. Not nothing found yet. Nothing, over every combination it was given. None of those four candidates has a rate that is any constant multiple of the secant, and none of them has a rate that is a number.

10:02So the advice, applied honestly and exhaustively, says: I have no suggestion here. The cosecant returns the same nothing for the same reason. And there is a third. An odd power of the sine times an even power of the cosine. Two candidates, three combinations, nothing. Three integrands where the advice is silent. And all three are standard, and all three have to be done. So what do you do when the derivative you want is not there?

10:38You put it there. Multiply the secant above and below by the sum of the secant and the tangent. Multiplying by one changes nothing. The numerator becomes the square of the secant plus the secant times the tangent. The denominator becomes the sum of the secant and the tangent. Before running the search again, check that the rewrite is the same function. It was put beside the original at every input, and it agrees at all of them.

11:12Now run the search on the rewrite. It finds exactly one candidate: the sum of the secant and the tangent. Its rate is one times the numerator, exactly. That is not luck. Look at what was chosen to multiply by. The sum was picked because its derivative is what the numerator would become. The target came first, and the choice followed from it. Said as a trick to memorise, this move is arbitrary. Said as a target chosen first, it is the only thing to try.

11:50The cosecant is the same move, and the door is stricter about it than most people are. Multiply above and below by the sum of the cosecant and the cotangent. The rewrite agrees with the cosecant at every input tried. The search finds the sum of the cosecant and the cotangent. But it reports the constant as minus one, not one. That minus sign is not decoration. The rate of that sum is the negative of the numerator, and if you carry it as a plus your answer has the wrong sign.

12:28The third silent case is filled a different way and it is worth the contrast. An odd power of the sine times an even power of the cosine. Peel one sine off the odd power. Convert what is left of it using the identity that relates the two squares. The rewrite agrees with the original at every input, and now the search finds the cosine, whose rate is minus one times the peeled-off sine.

12:59Three silent integrands. Three rewrites. Each one checked as a function before it was checked as a search, and each one found exactly one candidate afterwards. There is a fourth case, and it is the sharpest one, because here the advice is not silent. The reciprocal of one plus the tangent. Written as a ratio of cosines, its factors are the cosine, and one over the sum of the cosine and the sine.

13:33Run the search. It finds something. The sine, whose rate is one times the cosine standing beside it. So the advice speaks. Substitute for the sine. And it does not work. What is left over after you take the sine's rate out is one over the sum of the cosine and the sine, and that is not a rule of the sine at all. Here is why, measured rather than asserted.

14:01If the leftover were some rule of the sine, then anywhere two inputs have the same sine, the leftover would have to take the same value. Half a turn less than an input has the same sine as that input. That was checked at five inputs and it holds at all five. The square of the sine, which really is a rule of the sine, takes the same value at all five as well. So the test is capable of saying yes.

14:31The leftover takes a different value at all five. A rule of the sine cannot do that. The candidate the advice handed you is a dead end, and the same thing happens on three more integrands of the same shape. This is worse than silence. Silence sends you looking. A false lead sends you down a page of algebra that cannot close. The repair is the same idea as multiplying by one, in a different costume.

15:05You want the numerator to be the derivative of the denominator. The denominator is the sum of the cosine and the sine, so its rate is the cosine less the sine. The numerator you have is the cosine. So add a sine and subtract it again. Now split. Half of the sum over the sum, which is a half. Plus half of the difference over the sum, which is exactly the shape you wanted.

15:34The rewrite was put beside the original integrand at every input and agrees. And the search on the second piece finds the sum of the cosine and the sine, whose rate is one times the numerator standing there. Notice what happened to the advice. It was not extended. It was reversed. The advice looks at the integrand and asks what is there. This move decides what it wants to be there and rewrites until it is.

16:05That is the actual method, and it covers the easy cases too. Looking for a derivative already present is just the case where no rewriting is needed. One step gets skipped so often it deserves its own beat. The question was asked about the original letter, so the answer has to come back to it. Twice the input, times the sine of one more than its square. Substitute for one more than the square and the answer in the new letter is minus a cosine.

16:40Put the original letter back and you get minus the cosine of one more than the square. That passes the door. Leave it in the new letter and you get minus the cosine, read as a rule of the input. That fails the door. And it does not even agree with the right answer at any of the five inputs tried. An indefinite integral is a function, not a number. A function written in the wrong letter is a different function.

17:13Two of the items you will meet offer four answers and ask which one is right. They need no new idea at all. The first has an integrand that is a sum's own rate, divided by that sum. A tenth power and a power of ten, added. Do not choose by looking. Put all four options through the same rate door. Three fail. The logarithm of the sum passes. That is the answer, and it was measured, not recognised.

17:44The second is the reciprocal of a product of two squares. Rewrite it first, and the rewrite is checked: that integrand is measured to be the square of the secant plus the square of the cosecant, at every input. Which makes it two lookups added, and the answer is the tangent less the cotangent. Put all four options through the door. Exactly one passes, and it is that one. A four-option item is not a guessing game. It is four claims and one test.

18:20Last, the thing that turns a long list of practice into a short one. Twelve integrands were built to the four shapes this material keeps meeting, and put through both the door and the search. All twelve answers pass the door. The search answers three of them with a number. Those are the three with a straight line inside something: the rate is a constant, so it comes out in front.

18:47It answers the other nine with a factor already there. Nothing was left unclassified. But three of those nine are the false leads again. Each one is a reciprocal of one plus or minus a ratio, and in every one the advice names a candidate that does not finish. So the practice set is not a list of items. It is four shapes. A straight line inside: substitute for the line and carry the constant.

19:19A rate already among the factors: substitute for the thing whose rate it is. A whole composite standing there: substitute for the composite. And a reciprocal of one plus a ratio: do not trust what the advice finds. Rewrite until the derivative you want is in the numerator. Three of those four are the advice. The fourth is the method the advice is a special case of. Decide what derivative you want. Then rewrite until it is there.