Rewriting a product of trigonometric ratios into terms you can already integrate

0:00Here is an integrand made of two ordinary trigonometric ratios multiplied together. The sine of twice the input, times the cosine of three times it. Every piece of that is familiar. The integral is not. Try substitution, which is the tool you reach for now. Substitute for twice the input -- the only inside piece whose own rate is a plain number. What comes back is a sine of the new letter times a cosine of one and a half times it.

0:35The mismatch has not gone anywhere. It has only been renamed. I put that leftover to a list of eleven standard integrals, allowing any of a hundred and twenty-six constant multiples in front, and allowing every row to be read at any of twenty-one straight lines in place of its input. It matches nothing. Substituting for the other argument does no better. Two arguments that disagree is a shape substitution cannot touch, because substitution changes what the input is called and never how many different things are being fed.

1:17So something else has to happen first, and here is the whole of it in one sentence. When the integrand carries trigonometric functions, use an identity to reach a form you already know how to integrate. That sentence has to carry an enormous amount of work, and it is easy to read it the wrong way round. The wrong way is: this is about identities, so learn your identities. The right way is: the identity is never the destination.

1:49The destination is always the same place -- a row of the standard list you already own. The identity's only job is to manufacture a match to a row that is already there. Hold on to that and everything that follows is one idea applied repeatedly, instead of a memory test with no organising principle. Three families of identity do almost all of the manufacturing, and each one attacks a different obstruction.

2:24The first drops a power. A ratio squared is not on the list; a ratio read at twice the input is. The second splits mismatched arguments. A product of two ratios with different arguments becomes a sum of single ratios, each with one argument of its own. The third peels a factor off. An odd power surrenders one factor, and the even power that is left converts into the other ratio, which makes a substitution available that was not available before.

3:00Those three names -- for dropping the power, for splitting the arguments, for peeling a factor -- are my scaffolding for you, not vocabulary anyone will examine you on. What is worth carrying is the pairing. Power too high: drop it. Arguments disagree: split them. Odd power with a substitution just out of reach: peel one off. Take the square of the cosine. Not on the list. The list has a row for the cosine, and squaring it leaves that row behind.

3:37The identity that helps expresses the cosine of twice an input through the square of the cosine. Rearrange it so the square is alone on one side. The square of the cosine is half of one plus the cosine of twice the input. I checked that at five separate inputs and it holds at every one. Now look at what the right-hand side is. A constant, and a cosine read at twice the input.

4:07Both of those are rows of the list. The half in front is a constant multiple, and the doubling is a straight line inside, which the previous idea taught you to carry. Integrate the two pieces separately: half the input, plus a quarter of the sine of twice it. Differentiate that back and you land on the square of the cosine again. It passes. Notice how little of that was about the identity. The identity was one line. The rest was a lookup.

4:42Back to the product that stalled. A sine of twice the input times a cosine of three times it. The identity for this shape turns a product of a sine and a cosine into half the sum of two sines: one read at the sum of the two arguments, one read at their difference. The sum of two and three is five. The difference of two and three is minus one.

5:08So the integrand is half of the sine of five times the input, plus the sine of minus the input. And now the two pieces have one argument each. The mismatch is gone -- not renamed, gone. Both pieces are the sine row read at a straight line, so both are on the list. Integrating gives minus a tenth of the cosine of five times the input, plus half the cosine of the input.

5:37That answer passes the same check: differentiate it and the original product comes back. But something happened in the middle of that which deserves to be said out loud, because most workings do it silently. The identity handed us a plus. Half the sine of five times the input, plus the sine of minus the input. The very next line of almost any working shows a minus. If you are reading carefully, this is exactly where you stop and suspect an error.

6:13There is no error. The sine is an odd function: at a negated input it returns the negative of what it returns at the input. So the sine of minus the input is minus the sine of the input, and the plus in front of it turns into a minus. I checked that at all five inputs, and I checked the cosine the same way. The cosine does it at none of them -- it is even, not odd.

6:42So the step is about the sine specifically, and not some general rule about negatives. And it is not cosmetic. I built the answer a reader would get if they kept the plus, and put it to the same test. It fails. The step is doing real work, and it takes one line to say. That product identity is usually quoted rather than shown, often with a bracketed nudge asking you why it is true.

7:12It is a question with an answer, and the answer takes fifteen seconds. Write the sine of a sum: sine of the first times cosine of the second, plus cosine of the first times sine of the second. Now write the sine of the difference: the same two products, with a minus between them. Add those two lines together. The second product cancels against itself, and you are left with twice the first product.

7:43Divide by two and you have exactly the identity we used. Every one of these families comes out of the sum and difference formulas the same way. They are not separate facts to be stored; they are one fact, added or subtracted. Now a harder one. The cube of the sine. There is an identity for the sine of three times an input, written through the sine and its cube. Rearrange it to isolate the cube.

8:15The cube of the sine is a quarter of three sines, less the sine of three times the input. Same test at five inputs, and it holds at every one. Both pieces are the sine row again -- one plain, one at a straight line -- so both are on the list. Integrate: minus three quarters of the cosine, plus one twelfth of the cosine of three times the input.

8:42That is one route, and it is the route the identity gives you. There is a second route to the same integral, and it uses no identity of that family at all. The cube of the sine is an odd power, so peel one sine off. What is left is the square of the sine, times one sine. Convert the square: the square of the sine is one less the square of the cosine.

9:10I checked that rewrite too, at all five inputs. And now look at what is standing there. A function of the cosine, multiplied by a sine. The rate of the cosine is minus the sine. The derivative you need is already in the room. That is the substitution rule from the previous idea, and the search I built for it finds the cosine immediately. Substitute, integrate a simple polynomial, and come back: minus the cosine, plus a third of its cube.

9:46So we have two answers to one integral, and they look nothing like each other. One is built from cosines of the input and of three times the input. The other is a cosine and a cube. The usual way of resolving this is to say the two are equivalent, and leave the showing to you. Let us actually do the showing, because what comes out is stronger than equivalent.

10:12Expand the cosine of three times the input: it is four cubes of the cosine, less three cosines. I checked that identity too. Substitute that into the first answer. One twelfth of it becomes a third of the cube, less a quarter of the cosine. Add that to the minus three quarters of a cosine already sitting there. Minus three quarters less a quarter is minus one. Minus the cosine, plus a third of its cube. That is the second answer, letter for letter.

10:48Now hold on, because you have been trained to expect something else. Two answers to one integral are supposed to differ by a constant. Here the constant is nought. I measured this across six inputs. Their difference takes exactly one value, as it must -- and that value is zero at all six of them. To show that is a real measurement and not a machine that always says yes, I moved one answer by three quarters and ran it again.

11:21The moved answer still passes as an antiderivative, its difference from the other still takes exactly one value -- and now it agrees at none of the six. So this really is the rarer case. Not equivalent up to a constant. Equal. You now have two routes for an odd power, and it is tempting to argue about which is better. The more useful question is which is available, and that is decided by arithmetic rather than taste.

11:54Peeling works by leaving behind an even power of the sine, which converts wholesale into the cosine. That needs the remaining power to be even. I tried the first six powers of the sine. The peel exists for three of them and for no even one. Where it exists, the rewrite agrees with the power at every input, and the substitution search then finds the cosine at once. Where it does not -- every even power -- there is no factor to peel, and the identity route is not the shorter option. It is the only option.

12:33So: odd power, either route, and substitution is usually shorter. Even power, drop the power with an identity. The parity chooses for you. Some integrands carry three ratios rather than two, and there is a counting trap here. Take a sine, times a cosine of twice the input, times a sine of three times it. Three factors. It is very natural to reach for the identity three times. You need it twice. Turn the two sines into a difference of cosines, and then turn what that leaves into a sum.

13:13The count of applications is one less than the count of factors, because each application consumes two things and returns terms with one argument each. What comes out is four terms: a constant, and cosines read at twice, four times and six times the input. I checked the whole rewrite against the original at five inputs, and then put all four terms to the list. All four are rows. Nothing is left over.

13:47It would be easy to leave here believing that every trigonometric integral needs an identity. It does not. Here is a sum of two cubes over a product of two squares -- cubes of the sine and the cosine on top, their squares underneath. It matches no row of the list as it stands. I checked. But split it term by term and the first piece is a secant times a tangent, and the second is a cosecant times a cotangent.

14:19I verified that split at every input rather than trusting the algebra. Both of those are rows. The answer is the secant less the cosecant, and it passes the check. Not one identity was used. Only a split. Which is exactly why this belongs here. The instinct being trained is not identity-hunting. It is looking at an integrand and asking what would have to be true for it to be on the list.

14:51Two problems of this kind come with four answers offered and one to choose. They are settled at the same door as everything else. The first: a difference of two squares over their product. Divide through and it separates into the square of the secant less the square of the cosecant -- I measured that at every input. Both are rows. The answer is the tangent plus the cotangent. I put all four options to the test and exactly one passes: that one.

15:26The second: an exponential times one plus the input, over the square of a cosine of the input times the exponential. The numerator there is the rate of the product sitting inside the cosine. That is the shape from the previous idea -- the derivative is already there. So the answer is the tangent of that whole product. Again I ran all four options, and again exactly one passes. Notice you never had to guess. Four options is not a reason to guess; it is four things you can differentiate.

16:06One problem in this family cannot honestly be answered as it is set, and it is worth being told why. It asks for the integral of an inverse sine taken of a cosine. Every solver reaches for the same simplification: rewrite the cosine as a sine of the complementary angle, then cancel the inverse sine against the sine, leaving a quarter turn less the input. That cancellation is only legitimate while the complementary angle lies inside the inverse sine's own output interval.

16:45I tested it. At seven inputs inside a half turn of nought, the inverse sine of the cosine is exactly a quarter turn less the input. At eight inputs outside that stretch, it is that at none of them. And every one of those fifteen readings was actually taken. The zero is a measurement, not a machine refusing to answer. So the simplification is true on a stretch and false off it, and the problem as posed says nothing about which stretch it means.

17:19The honest thing is to state the restriction yourself, do the integral on it, and say what you assumed. Step back and count what actually happened across all of that. Three worked integrands were put to the standard list before any identity touched them -- a squared cosine, a mismatched product, a cubed sine. None of the three matched anything, even allowing any constant in front and any straight line inside.

17:50Then each was rewritten, and every single piece that came out landed on that same list, with nothing left over. That pattern is the idea. The identity moved the integrand; the list did the integrating. So when a trigonometric integral stops you, do not ask which identity this one wants. Ask what the list has that is close, and what would have to change for the integrand to become it.

18:19Power too high, drop it. Arguments disagreeing, split them. Odd power with a derivative nearly present, peel one off. And sometimes, nothing at all -- just a split. You are not learning identities. You are learning to manufacture a match.