a-use-the-reduction-formula-in-example-6-to-show-that-int_0pi-2-sin-n-x-d-xfracn-1n-int_0pi-2-sin--2

Question

(a) Use the reduction formula in Example 6 to show that
$$\int_{0}^{\pi / 2} \sin ^{n} x d x=\frac{n-1}{n} \int_{0}^{\pi / 2} \sin ^{n-2} x d x$$
where $n \geqslant 2$ is an integer.
(b) Use part (a) to evaluate $\int_{0}^{\pi / 2} \sin ^{3} x d x$ and $\int_{0}^{\pi / 2} \sin ^{5} x d x$
(c) Use part (a) to show that, for odd powers of sine,
$$\int_{0}^{\pi / 2} \sin ^{2 n+1} x d x=\frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot 2 n}{3 \cdot 5 \cdot 7 \cdot \cdots \cdot(2 n+1)}$$

Jorge Olivares · Accepted Answer

four part, eh? We recalled a reduction formula. That is the one that we have here. So if we want to compute dainty were from Cyril, they were too. Sign to the poor orphan effects T X. We just need to evaluate the right hand side of the reduction formula, meaning this part here. So we get minus one over end course and effects, saying to a power off in minus one x and we need to evaluate between civil on power too, plus 10 minus one over end 18 to grow from cereal. A word to him, upside to the power of Finn minus two affects Deeks. So if we have always the first part, we will get the co sign off by over two times assigned to a polo fan minus one. Never would play over too, minus the same function, but now evaluated at the lower limit that this case is serious plus and minus one over, and they interrupt from Syria to pay well, too off signed to the power of friend minus two x t X and we know that course. And if I were to zero and sign off, zero is zero. That means that this whole expression here in this the whole term it&#x27;s going to be serial. So basically, we get that thing to grill from zero to buy over too upside to a power. A friend of X dx. It&#x27;s equal to n minus one over and vein tumor from seal to play over too. Sign to the power of Finn minus two affects you for part B. So we need to evaluate first. You seem too well thing too well from Syria to play word too. Signed to power three off X dx and we need to use the formula. We get in part a meaning this formal here, and in this case, we need to use it with n equal to three. So if we replace in the right hand side with a three minus one over three name to her from zero by over two, sign off three minus two off X GX. That&#x27;s going to be to 1/3 thing to grow from cereal to play, too. No sign of X t X and us to third and the interviewer off sign is minus school. Sign off X, and we need to evaluate this in between Ciro and Pirate Joe. So we get to third times minus the coal sign off. Fly over to the upper limit. Minus minus. CO sign of Syria. Co Sign off Baylor to a serial and co sign of Seo is one and minus time. Minus is once we get just to over three. Also, we need to evaluate the integral from zero too. But you were too off signing to the floor. Five Health, ext. D X. And again. We can use the formula that we&#x27;re getting park, eh? In this case with and equal to five. So it would take a look. Here. We will get five minus one over five. Thing to go from Cyril. Toe player too up. Sign off. Five minus two. Thanks, Deeks. So this will give us four over five times interior from Syria to pay over to off. Sign to the power of three off X D. X. So we need to evaluate this integral. This will give us for over five. See, you will to pay over too. And we can rewrite the sign and kill us. Sign off pecs time saying square effects DX. So we get for over five things over from Syria to play word too inducing. So, trigonometry identities. We&#x27;ll replace science where? One minus close and square Fix T X. So we get four over five. I think they&#x27;re from Syria to play word too. Oh, sign off. X minus. Sign off. X coursing square. Both x ttx. And this is for four or five the integral of sine. We know that it&#x27;s minus clothes and effects. And if we use this substitution, you equal to cosign effects. We can solve this integral to get course and cube off X over three. And we&#x27;re below eighties from zero to pirate too. So that will give us for over five. First, we will await in pine over two and we&#x27;ll get minus course I by over two plus course and cure flavor too. Over three minus on this course in at zero plus course in Cuba. Zero over three. Okay, so we know that this is zero and this is also zero. And closing off suit was one only. This one. Here&#x27;s one. So we get four or five times and minus 10 miners will give us one minus one over three, and that will give us four or five times toe over three. It is the same. That eight over 50. And that&#x27;s part B. Okay? And now we need to finally do part see, and went to use the formula in a I&#x27;m gonna call before me today, so we have it in hand. We need to use this formula to prove the following the being together from Cyril to play word tool off sign to the power of 20 plus one affects E x A simple too. Two times, four times, six times, two in over three times, five times, servant times two plus one. So if we start with the integral and we applied for me that we&#x27;re getting a ones while we will get peace 20 plus one minus one over 20 plus one time Seeing to grow from Sewell Tau pi over too. Sign off tour plus one minus two off x d. X. So wasted the every right, this formula by changing and with toe and plus one and we&#x27;ll get this. Okay, So this is the same us to win over to Empress one times integral from serial tau pi. Over too. In here with that sign off. Tow n minus one off x t x and we&#x27;re gonna play this formula one more time. Now, Toby seem to go here. Who would do it first? Now and what we get now, Uh, we use the same formula, but with an replaced by twin minus one. So we get 12 minus one minus one over twin minus one time. Seem to grow from Ciro to buy over too. The sign off too. And minus one minus two off X dicks. And that&#x27;s just to win times to one minus two time toe in my plus one, too. And minus one in this times integral from cereal by. Over two off. Sign off to the power of twin now minus three off x d x. So we see that every time that we apply the formula to the scene to grow we have uneven power tow the nominator and the north power to eliminator and scenes original exponents off the sign is too and plus one we can apply this formula end times and if we apply it in times we will get that thing to grow from Syria to play over to off sign off 20 plus one effects T X is going to be well, too. And times two and minus two times tool and down We will have in the denominator 20 plus one toe and minus y, and we will follow it to work it to three here. And he would get into well, to pi over too, the sign off to end. So since we apply end times of formula tow the exponents here that originally was to n plus One, we need to substructure two times. And since each time that we apply the formula works of threat to that means that here the exponents should be equal to one x d x, and we already compute this in part B and we know that it&#x27;s one. So we got that desire formula by using the formula that were obtaining

Robert Daugherty · Answer

section 6.1 Problem number three. So they&#x27;re asking us to use the reduction formula that we found in the previous problem for the sign to the power to violate the definite integral zero party to sign of N x DX. So using this formula integral from zero to pi over two. Uh, this signed it in the power of X dx is going to be minus one over in co sign of X signed to the N minus one of X. This is evaluated from zero two pi over two plus the integral in minus one over in in a row from zero, however, to signed in the end, minus two of X DX. So what we just need to do is to look at this evaluation, you see, right here. So So when I substitute X equal to pi over two, the key factor is gonna be the coast aren&#x27;t in this case, you know that the coast on a par for two is going to be zero. So this term is going to be zero when you substitute and zero, the key term is the sine function because, you know, the sign of zero is going to be zero. So no matter what the powers are, this is going to go away. And so you&#x27;re going to see your final answer. Is that the integral from zero to pirate, too, Of the power of sign of X D. X is equal two in minus one over in the integral from zero to harbor to sign. And this is to the n minus two of X dx. So that is the formula for the definite and a road zero departure YouTube have signed to the power. Now where they&#x27;d like us to do is let&#x27;s go ahead and use this formula to evaluate the integral of sine cubed and signed to the fifth. So in part B, they want us to evaluate this interval for sign cubed of X DX. So I&#x27;m just going to use the formula that we just came up with here in this case in is equal to three. So this is going to be three minus 1/3 in a row, from zero to pirate too, in a sign of in minus 23 monies to is one. So this becomes 2/3 the integral from zero power to sign of X DX. The integral of the sine function is just minus co sign. So this is minus 2/3. Could a sign of X from zero to pi over two. So this is minus 2/3 substitute and a pirate too. You get zero minus one. This answer is just 2/3. So that&#x27;s the integral of sine cubed Now they asked us to go ahead and also find what is Thean devoe from zero to pi over two of sign to the fifth of X dx. So according to our formula, this is where now in is equal to five. So this is going to be five minus one over five in a row from zero to pi over two. Sign to the N minus two. So sign cubed of X DX. Well, sign cubed of X. T X is what we just solved to beat the 2/3. So this answer is 4/5 times 2/3 which is what, 8/15? Now, hard to continue doing this. That didn&#x27;t ask for this. But if I were to go and say well, what would be the integral from Ju to part two of signed with seventh power of X DX where that&#x27;s gonna be seven minus 1/7 and then this integral of sign to the fifth of X DX. So there&#x27;s going to be six sevenths times for fifth times, 2/3 and what to see? So you start to see a pattern with these numbers that you see here. So they&#x27;re calling that out there saying, Hey, let&#x27;s go look at that and see if we can come up with a formula that works for odd powers of sign. So that is the final part. See, in this particular question, So in part C, they say, Can we come up with a general formula for the integral of odd powers of sign from zero Harper two of X DX. Okay, we already have our reduction formula, which is in minus one of her in zero. The pirate. Too sign of sorry about this. I need a race this thistles for in. So I&#x27;ll take that in just a moment. They want us to come up with a formula is what we already have. Pardon me is we already have the Formula zero to pi over two for assigned to the power of X DX is in minus one over in, and then the integral of sign to the n minus two of X DX. So what we&#x27;re asked to do is to find a formula for odd powers of sign. So for odd powers of sign when I substitute the value in. So let&#x27;s just look at the interval from zero two pi over two signed to the two n plus one, which is an odd number. So according to a reduction formula, it&#x27;s going to be two n plus one minus 1/2 n plus one and the integral from zero to pi over two of sign to the two n plus one minus two of xcx. And so this just becomes too in over two n plus one and then the in a row, and that&#x27;s gonna be sign ah, to the two in minus one of X dx. Now what you see is I&#x27;m still every time I do this, I&#x27;m going to end up with If you start with an odd power, the integral gets Decker minted by two. So it stays an odd power. So what would happen when I apply the reduction formula to this next interval? So If I keep trying to work with this, I&#x27;m gonna have to in over two n plus one. And now I&#x27;m going to have to end minus one minus one over to end minus one and the interval. Zero to pi over two of the sign of two in minus one monies to of X dx. So this gives me to in two n plus one. And then you have, what, two in minus 2/2 in minus one and in the integral from Gerald Empire itude sign of to the two in minus three of X dx. So what you can see here is we know that when I&#x27;m integrating sine cubed, I end up with 2/3. When a integrate sign fifth, I end up with 4/5 times. 2/3 signed with seven to see this. What&#x27;s happening is so every time on the top, I&#x27;ve got even numbers increment id by two on the bottom of God. Odd numbers increment it by two. If you look at what we&#x27;re seeing here in this tube, in even number two in monies to even number look at the bottom two in minus two M plus one Odd number two in minus one. Odd number. So you&#x27;re going to see that same pattern? So what&#x27;s gonna happen here is the integral from zero to pirate, too. Sign of two n plus one of X dx is going to be. Well, where will this end up? It ends up. The last odd number is going to be, um, in equal three. So the sign cube. So it is going to end up with a 2/3 so it&#x27;s going to be 2/3. And then what was It was signed fifth sign to the fifth Power was 4/5 times 4/5. And then after that, with signed to the seventh, um, we had 6/7. And so it&#x27;s going to continue until we get to two in over two in plus one and then the integral of zero. We&#x27;re sorry. So this this integrates. And so it starts out here. So you this is what would happen when n is equal to two in his equal to three. Excuse me. Sign Cube, when in is equal to three in is equal to five and is equal to seven. And this is in equal to n plus one. So this is the value of this definite integral. So it starts with even numbers on the top, all the way up to the degree minus one. And then it started with odd numbers of the bottom all the way up to the degree, and that&#x27;s the value of the definite integral.

Willis James · Answer

Yeah. Hello. Welcome to this lesson in the first place. I would like to show that the integral from zero to pile into science. An ex d ex cult too and minus one of our end integral from zero to pi on to sign and minus two DX. Okay, but the whole of that Okay, The whole of the anti golf sign N x. Yes. Without any restriction without having the bounce is equal to negative one on n course X sign in on one x the x last 10 minus one on and integral sign and minus two x dx. Okay, so he realized that, uh, realize that if you&#x27;re taking the integral with bounce Yeah, set that we have this 20 then? Yeah, the first place that we put we put down to connect course piling 20 in a second place where we put zero in it. Sign zero is equally zero. So for whatever values of X, that is the pie on two and zero. One of these would always resort to zero, thereby making the whole product zero. Okay, So that means that this is ignored outwardly. Okay, then we have only this path that works. Yeah. Mhm. So we can say that that is n plus one and minus one on end than the integral sign and minus two x dx. Yeah. All right. Okay, So the second part is to use the first party value the integral zero to buy on two Sign three x, the eggs. So this is equal to the end. Here is three. So we have three minus one on three, then integral Sign three minus one, which is the third minus two third, minus two. That is one mhm. So we have X dx, and this is to over three. So that is Yeah. So to over three. Negative, of course. Yeah, X. So you take it from pi on 2 to 0, and this becomes three. 22 on three. They get to Of course, I own two, then minus costs zero because minus minus. Mix up. Last. So that is two on three. Cause pound to zero. Then that is CO zero. Because there is one. Yeah, so last one. And that gives us two on. Sorry. Okay, So let&#x27;s use the same thing to evaluate. Integral from zero to pi on two. Sign five X, The X. So this would be five minus one on five, then the integral from zero to apply on to sign five minus two. That is three x dx. So that is four on five. Then the whole of the integral from zero to pound to sign three x signed to the past three x. The X has been evaluated US two on three. So this gives us eight on 15. Okay. All right. So, looking at the pattern, looking at the pattern we the CPAP will be next to look at this. Mhm. Okay, so here we have two. And last one or minus 1/2 and last one, then the integral for pie from zero to pi on to sign two n plus one minus two. Mhm. Okay, so this always yelled to n last one, minus 1 to 2 n over two and last one. Yeah. Then it&#x27;s multiples. Okay, So you always find out that you have the end starting form the end that is greater than or equal to two. Okay, so this test from to us in a case of any calls to three. So this start from two king, then it goes to four. As you saw in ex, uh, And when n was five. And it goes to 61 n is seven. Okay, Was the nature we saw start at three. Then 25 If it goes in, that sequence would have a seven and a nine. So have s three, five, seven and on and on. Okay, Mhm. So this proves what we are looking for. Thanks for your time. This is the end of the lesson, Yeah.

Geena Pullo · Answer

Hi. So, in this question, we are going to be using the Wallace signed formula, which I wrote down all ready for us to solve these equations. So for our first integral, we have, that is from zero to over to of these Stein of abs Huge de axed. So using our patient, then for when are and is odd, um, and equal to three. In this case, actually, it would just be equal to 2/3. Um, because we were to plug it in for equation, we would just end up getting the first values right here and for our next. When we have the integral from syrup, you pot it over Teoh, sign before I start before. So this is gonna be equal to be part over two times, often times three over two times for, um, because you plug that in to the equations. That&#x27;s really the question is an venous one over. And, um, so we have to plug it in four. All right, two even numbers. So that would be equal to three. It over 16 on. Now, if we were to see our next questions, we have zero. Do you over two of the biggest sign of if it this is going to be equal to to over three. Um, but for three times a day or 2 to 4 and are B times five, this is going to be equal to 80/50. And for our last equation, we&#x27;re going tohave the integral from zero you probably do of six x d. X. And this is equal to I over to one time. Three times eyes over. Shoot happens set is going to be equal to high over to times 15. Oh, over 40. Aged. Uh, from here we can simplify that to being equal to 15 already over 96. Andi, this actually will simply because they could both be divided by three. So this will be fun. Hot it over 32.

Brian Ketelobeter · Answer

Okay, ladies and gentlemen, today we&#x27;re looking at a problem number, uh, 90 from section 4.5. And, um, now, I know in the book, they sort of break this into two parts and be, but we&#x27;re just gonna capture it all in one, um, fell swoop stealing. It&#x27;s true for all. And, um And okay, So before we begin, though, I have to make a couple. Uh, we have to do a couple make a couple of comments here. First off, uh, I&#x27;ll need the trick Identity A here, which becomes important. It&#x27;s just that dive you could sign Axis, sort of. Ah, shift of no sign. And the 2nd 1 is a little more interesting even. Is that, um, the coastline and of exes, even. And you could see this through my little proof here, which is just observing that, uh, all I&#x27;m really doing is rewriting co sign and of negative acts into co sign of negative acts to the end. Um, once I&#x27;ve done that, then because co sign is even I get that this is equal to that. Um, and and there There we go. So, um, so with these two comments made, then I began on the actual proof of the problem. And to do that I start with the left hand here, um, the left hand side here, which is zero to, um, pie over, too. And, um, now, I&#x27;m just using part a here. Just making the shift. Um, making the shift here, um, you know, to get X minus pi over, too. And then, um, if you substitute you equals two acts minus high over too. Of course. Then we&#x27;re just going to get do you, uh, to be the axe. And in doing so, then we get this in a girl down here. Now, again, this is from negative pi over too. Um, to zero. Of course, because the upper bound here was just by over two, and basically you switched. You switched us back to here. Um, so matters equal to that. And now, of course, we&#x27;re gonna use the evenness of it. Um, which allows us to just sort of switch back, if you will, to zero the pie over two. And this is really just using the evenness of it. Um, and the fact that you know that the inner basically we can switch the interval um, And now, once we have that, now we have to do is go back to the co sign, and we get that. So So that&#x27;s really all you have to do. Um, and the key to the whole problem really comes from the first, The first, um, identities here, which is the trick identity in the co sign, uh, and is even And then we just go through the algebra and we get our answer. So this is sort of a sort of a straightforward problem, but on the other hand, I have to admit, it kind of struck me as a little surprising. Still, um, I wasn&#x27;t actually aware of this identity, so it&#x27;s something, you know, interesting, actually. Prove it. And to see to see how it works. So it is kind of an interesting thing, but, uh, yeah, Anyhow, um, thank you very much. Have a nice day

Robert Daugherty · Answer

Section 61 Problem 31. So they&#x27;re asking us to use this generalized reduction formula for the sign of in X integral, um, and to prove that for sine squared. So let&#x27;s use our reduction formula that they give us here. We&#x27;re gonna be using the case where in is equal to two because I&#x27;m integrating sine squared. So it would tell me that the integral of the signs great of X dx just substituting in in equal to B minus 1/2 co sign of X, and then the sign in is too so in minus one would be one. So it&#x27;s minus 1/2 co sign X sign of X and then plus in minus one of 1 1/2 the integral. So in minus two, that&#x27;s gonna be the sign to the zero power of X DX. So you know that raising that to the zero power is just going to be one. So this is minus 1/2 co sign of X sign of X plus 1/2 That&#x27;s just the integral of DX. So this is minus 1/2 co sign of X sign of X plus 1/2 axe. Now what? You just need to look at is using our trig identities. I knew that the sign of two X is equal to to sine x co sign X So let me just make that a little bit cleaner. So double angle formula. Sign of two X. It&#x27;s just too sign of x co sign of X so co sign of Exxon of X, just 1/2 of the sign of two x So this is minus 1/2 one. Have sign of two x plus 1/2 X, So this final answer is just X over two minus sign of two x over four, plus a constant of integration. So that&#x27;s what I was after. So I was out to prove the integral of sine squared is X over two minus sign of two x over four, plus a constant of integration. Now what they ask us to do in the second part of this problem is to use this form, um, to figure out the integral of sine to the fourth. So let&#x27;s use our reduction formula again that they gave us and that was the integral of the sign of in to the end of X. It&#x27;s equal to minus one over in co sign of X sign of in minus one of X plus in minus one over in and the integral signed to the end minus two of X DX. So now I&#x27;m using the case where in is equal to four. So if n is equal to four, this is minus 1/4 co sign of X sign cubed of X plus 3/4 um, the integral of sine squared of X dx now sine squared of exits the interval that we just came up with a moment ago. So a moment ago, we figured out that the integral of the sine function is given right here. So I can now write this as this is minus 1/4 co sign of X sign cubed of X and then plus 3/4. And this is going to be X over two minus the side of two x over four, plus a constant of integration. And that&#x27;s just the formula we got from what we did a moment ago in the first part of this problem. So this becomes minus 1/4 co sign of X sign cubed of X and then plus 3/8. Thanks. And then minus 3/16 sign of two x plus a constant of integration. And so that is your final answer for the integral of the sign to the fourth power of X dx. So again, we just used our reduction formula that we were given and the reduction formula turn this sign fourth and two sine squared integral, which we had already achieved by using the formula to begin with.

Problem

Prove that, for even powers of sine, $$\int_{0}^…

Problem 45 Hard Difficulty

Answer

a. $\frac{n-1}{n} \int_{0}^{\pi / 2} \sin ^{n-2} x d x$
b. $\int_{0}^{\pi / 2} \sin ^{3} d x=\frac{2}{3}$ $\int_{0}^{\pi / 2} \sin ^{5} d x=\frac{8}{15}$
c. $\int_{0}^{\pi / 2} \sin ^{2 n+1} x d x$

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a. $\frac{n-1}{n} \int_{0}^{\pi / 2} \sin ^{n-2} x d x$b. $\int_{0}^{\pi / 2} \sin ^{3} d x=\frac{2}{3}$ $\int_{0}^{\pi / 2} \sin ^{5} d x=\frac{8}{15}$c. $\int_{0}^{\pi / 2} \sin ^{2 n+1} x d x$

Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Caleb E.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

James StewartCalculus Early Transcendentals

Catherine R.

Anna Marie V.

Caleb E.

Joseph L.

a. $\frac{n-1}{n} \int_{0}^{\pi / 2} \sin ^{n-2} x d x$
b. $\int_{0}^{\pi / 2} \sin ^{3} d x=\frac{2}{3}$ $\int_{0}^{\pi / 2} \sin ^{5} d x=\frac{8}{15}$
c. $\int_{0}^{\pi / 2} \sin ^{2 n+1} x d x$

James Stewart

Calculus Early Transcendentals