Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Products of powers of sines and cosines. The integral of sine X to the power M times cosine X to the power NDX where M&N are greater than or equal to 0 and they're integers. We're going to have a case one if M is odd. If M is odd, we can rewrite M AS2K plus one where K is some integer. So then we'd have the integral of sine X to the 2K plus one times cosine X to the N power DX. We're going to pull out one single sine X and then we're going to have an even power of sine X to 2K. We're going to then take those and switch those into one minus cosine squared X using the Pythagorean identity. So we're pulling out a single sine X and then we're going to change that sine squared X to 1 minus cosine squared X to the K power. So then if we let U equal cosine X, our DU would be the negative sine X. So we're going to be able to pull the sine X and the DX out and replace it with negative DU. And we'd get 1 -, U ^2 to the K * U to the N, the integral with the negative in the du replacing that sine XDX. Then we should be able to evaluate that. We may have to do some multiplication something, but U to any power we can evaluate. So the next one would be if M is even and N is odd, it's going to be the same concept. We're going to let N equal 2K plus one. So we're going to have the integral of sine X to the M power, cosine X to the 2K plus one power DX. We're going to pull out a single cosine X. We're going to have sine MX, cosine squared X to the KDX. We're going to switch these cosine squared X's to sine squared X's by using the Pythagorean identity that says 1 minus sine squared X is cosine squared X. So then if we let U equals sine, XDU is cosine XDX. So here's my cosine XDX I'm going to take out, replace it with DUI, get U to the M times the quantity 1 -, U ^2 to the KDU. Now, if M&N are both even, this one gets to be a little tricky. We're going to use our power reducing formula that says we're going to substitute sine squared X equaling 1 minus cosine 2X over 2 and cosine squared X = 1 plus cosine 2X over 2. Then we're going to simplify and attempt to use case one or case 2. If not possible, we're going to do case 3 again. If not possible, the next time through we're going to do case 3 again. So cosine squared X1 plus cosine the quantity 2X over 2. If we thought about taking the square root, we could have it cosine X equal positive negative sqrt 1 plus cosine 2X over 2. Now variables are just variables, so we could let 2X equal Y so that X would equal y / 2. So another way of writing this if we wanted was cosine y / 2 equal positive negative square root 1 plus cosine y / 2 or sine y / 2 equal positive negative sqrt 1 minus cosine y / 2. Now you may not remember how those formulas came about, so let's just do a quick refresher. If we had cosine of X + X, that's really just cosine of 2X, and our formula would say cosine X, cosine X minus sine X, sine X. So we'd have cosine 2X equaling cosine squared X. If instead of the sine squared, we replaced it with 1 minus cosine squared X by Pythagorean identity, we'd have cosine 2X equaling 2, cosine squared X -, 1. So then if we added one to the other side, we'd get one plus cosine 2X, and then if we divided by two, we'd get that cosine squared X by itself, which is what we referred to back here. Cosine squared X equal 1 plus cosine 2X over 2. That's how it came about. We could do the same thing, but instead of taking out the sine squared X, we'd take out the cosine squared X. That's how we'd get the other equation. If we look at some examples, zero to π of sine X / 2 to the fifth power DX. So we're going to pull out one of the sines. This is going to leave a sine to the 4th X / 2, which we could think of as sine squared X / 2 ^2. Powers to powers we multiply. So we're going to take the sine squared inside and we're going to change it to 1 minus cosine squared X / 2 ^2. We're going to foil this out. So we're going to get 1 - 2 cosine X / 2 ^2 plus cosine X / 2 to the fourth times the sine X / 2 that we pulled out a minute ago, the integral zero to π DX. So now if we thought about letting U equal cosine X / 2, then DU would be negative 1/2 sine X / 2 DX or -2 DU equals the sine X / 2 DX. So here's my sine X / 2 and here's my DX. So I'm going to take that out and put a -2 DU. My negative is hiding up there -2 DU 1 minus two U ^2 + U to the 4th. If we change our bounds, cosine of 0 is 1. So that's our new lower bound. Cosine of π / 2 is 0. So that's our new upper bound. Now we're just going to integrate. So we have -2 the integral of one is U. We're going to use the power rule. So we're going to add 1 to the exponent and divide by the new exponent. We're going to add 1 to the exponent, divide by the new exponent. We're going to put in our upper bounds minus our lower bounds, and we're going to see that that's going to give us 16 fifteenths when we simplify it. Let's look at another example. If we thought about zero to π halves of sine squared 2 Theta, cosine cubed 2 Theta D Theta. So this time the term on the sine is not odd, but the sine, the quantity on the cosine is odd. So we're going to pull out a single cosine. We're going to switch the rest of those cosines to sines using Pythagorean identity. So we have one cosine 2 Theta out here. Here's the sine squared 2 Theta we haven't touched, and then the other cosine squared 2 Theta is going to turn into one minus sine squared 2 Theta. If I let U equals sine 2 Theta, DU is going to be two cosine 2 Theta D Theta. So 1/2 DU is going to equal cosine 2 Theta D Theta. So now we're going to put in 1/2 DU for the cosine 2 Theta D Theta part. And then we're going to have U ^2 times the quantity 1 -, U ^2. Now when we change our bounds, we'd get U of π halves being zero and U of 0 equaling 0. So actually changing our bounds right here we could be done because we know if the bounds are the same, it's going to equal 0. So let's just ignore the bounds for a second so we can actually see what's going on. If we distribute it out and use the power rule. Add 1 to the exponent, divide by the new exponent, add 1 to the exponent, divide by the new exponent, putting in the substitution back. So we get one half 1/3 sine 2 Theta cubed -1 fifth sine 2 Theta to the fifth. Then if we put in that Pi halves and 0, we can actually see that it is going to equal 0. So if you change your bounds as you go, sometimes you save yourself quite a bit of work by realizing that if the bounds are the same, it's always going to be 0. Looking at the next example, zero to π of square root 1 minus cosine 2 XDX. We don't know how to deal with the square root, but we do have some properties that say, well, if we had sine y / 2 equaling positive negative square root, we could multiply the square root to the other side and now we'd have this positive negative sqrt 1 minus cosine Y, which is what we have here. So we're going to have this equaling zero to π root 2. Because of this positive negative. We have to realize that we have to take into account what's happening. Well, if we think about zero to π and 0 to π, our sine value is always positive. If you didn't see that, you could always put in absolute values there and you'd be OK, because we know the inside of a square root is always positive. So we needed the positive case here. We can force that by using absolute values or realizing that the sine value from zero to π is going to be positive. So from there, we're just going to integrate. We're going to get negative root 2 cosine X from zero to π. So negative root 2 upper bound minus lower bound is going to give us 2 radical 2 when we multiply it all out. The next thing we're going to do is we're going to look at product to sum formulas. And this is just a quick refresher from trigonometry. If I have cosine of alpha minus beta XI could put those in brackets if we prefer. That's just the cosine of the first angle, cosine of the second. We switch the sine for cosine. So it's going to be plus sine of the first angle times sine of the second. Now I'm going to do the same thing, but instead of a minus, it's going to be a plus. And when I add these two equations together, you're going to see that these sines down here cancel. So I'm going to get cosine alpha minus beta X plus cosine alpha plus beta X equal to cosine alpha X cosine beta X. If I divide by two, this is one of the formulas that we're going to want to use in a few moments. There's going to actually be three of these. So the next one's going to come from subtracting those original 2 formulas instead of adding them. So I'm going to have this minus this. This time the cosines are going to cancel and we're going to get left to sine alpha X sine beta X. If I divide by two, this would be our second product to sum formula. And our third one comes about from doing the exact same thing, but instead of using cosines we would start out with our sines. So product of some identity is just a summary. Sine MX sine NX is 1/2 cosine M minus NX minus cosine M plus NX. Sine MX cosine NX equals 1/2 sine M minus NX plus sine M plus NX. Cosine MX cosine NX equals 1/2 cosine M minus NX plus cosine M plus NX. Looking at an example here we have a product of sine times cosine. So we're going to use our product formula that said sine of the two angles subtracted plus sine of the two angles added. So we'd get sine of negative X. And by our even and odd identities we know that that's really just negative sine X plus sine 5X. So now when we evaluate the integrals, the 1/2 cosine X -, 1/5 cosine 5X, putting in our upper bound minus our lower bound, we would get a final answer of negative 2/5. Here's 10 to π of eight sine to the fourth Y cosine squared YDY. This is our third case where they're both even. So we're going to switch them both into equivalent formulas using the double angle concept. So sine Y to the 4th power, we're going to think of sine squared y ^2. So we'd have 1 minus cosine two y / 2 quantity squared and the cosine squared Y is going to be 1 plus cosine two y / 2. Now this square with the two in the denominator and this other two in the denominator is going to cancel with that 8. So first thing is we can get rid of all those coefficients zero to π. I'm going to recommend that we realize that one minus cosine 2Y and one plus cosine 2Y is the difference of two squares and multiply that first. So we're going to leave one of those 1 minus cosine 2 YS alone and we're going to do the difference of two squares, which is going to give us 1 minus cosine squared 2Y. Now it really won't matter. It just makes the math a little easier because now we are going to have to multiply again. We're going to foil again and when we foil this time we get one minus cosine squared 2Y, minus cosine 2Y plus cosine cubed 2Y DY. So the one pretty easy, it's just going to be Y when we integrate it, the cosine squared 2Y we can't do. So we're going to use the double angle formula again and get one plus cosine twice the angle 2Y or four y / 2. Here the integral of cosine 2Y is just going to give us negative 1/2 sine 2Y cosine cubed. So we're going to pull out one of the cosines and we're going to switch the rest to sine. So 1 minus sine squared 2Y DY down here. At this end, if we thought about letting U equals sine 2Y DU would equal to cosine 2Y DY or 1/2 DU equal cosine 2Y DY. If we do our upper bounds and lower bound change, because we've just done AU sub, we realized that we have zero to 0. So this term is actually just going to be 0 and we can pretty much ignore it down here at the end. So if we stick in our upper bound minus our lower bound here we have π -, 0. If we integrate, we think about the 1/2 we'd get -1 half Y. If we think about integrating this negative cosine four y / 2, that's going to give me 1/8 of the minus I left out in front minus. If you distribute negative times positive negative or just leave it out in front 1/8 cosine 4Y0 to π. And then this last piece, remember was zero to 0. So we can pretty much ignore it. When I substitute in my upper bound minus my lower bound, I'm going to get 1/2 Pi plus sine of four Pi is 0 - 0 + 0. So we're going to end up with just π -, 1/2 Pi or Pi halves. Thank you and have a wonderful day.