use-the-substitution-formula-in-theorem-7-to-evaluate-the-integrals-int_pi-6pi-4-fraccsc-2-x-d-x1cot

Question

Use the Substitution Formula in Theorem 7 to evaluate the integrals.
$$\int_{\pi / 6}^{\pi / 4} \frac{\csc ^{2} x d x}{1+(\cot x)^{2}}$$

Bobby Barnes · Accepted Answer

we want to use a substitution for me in the room seven to help us evaluate this integral here. Now, the way they have this written, it seems like they&#x27;re leading us tow. Want to use co tangent of X to be you? Because the derivative of co tangent X should give us the differential negative. Cosi get squared X d x. If we just go out and multiply top to bottom by Daggett one or just multiplied by negative one twice. That gives us exactly d&#x27;you. Let&#x27;s go ahead and plug that ends. We have d&#x27;you over one plus u squared and we know integrating that should just give us universe tangent. So let&#x27;s figure out what our new bounds our first, though. All right, So when you is equal to or not, when you believe X is equal to pi forth, then you is going to be co tangent a pie fourth, which is just one start profound, this one. But what about our lower bound? So that&#x27;s gonna be you is even too. So I have to think about this one. So it&#x27;s co sign of high six over sign of pie sixth, cause that&#x27;s just the definition of co Tanja and coast on the pie six is Route three over to Sign is going to be one over to sew the two scouts. Also, we&#x27;re just left with fruit. Three are lower bound is route three. Let&#x27;s go ahead and into right this. So we already said that should give us park tangent. So is negative. Tanja Inverse of you evaluated from 3 to 1. So we have the negative on the outside now plugging in one. So tangent of theta is equal to one when data is pie fourth Then we subtract off where we plug in room three and tangent of theta is equal to route three at pie Third So then pie fourth minus pi Third is negative 1/12 multiplied by the negative And we just left with pie 12th

Dillon Huddleston · Answer

eso for both into ghosts. Here we can define you being JIA decks with gen x equal to the tangent of X Because, uh, course of the ribbon of pageant of exes he can square backs. So letting you be, Gee, it XB data backs branch Groll. Uh, because the interval of you, Um And then, of course, do you you being 10 ecstasy can square dx dy ex s. We just have the integral of you with this back to you which applying power, forming their yields you square by two by two. Ah, no, For the bounds since we have you is 10 of X, We integrated from on the tangent of zero, which is zero to be attended to by four, which is one eso in case this is for case, eh? To a Jake one square zero square. Is this one by two. That&#x27;s three and 1/2 off, eh? And republic be, um well, we know this wool of this central different only in that be order Venture gave order of integration is reversed. S we know automatically. That boy from B is minus that for me. E minus a. Huh? And we continue with the extra chances for a be expectedly plus minus

Bobby Barnes · Answer

we want to evaluate these two Definite Integral is using the substitution formula hero that they give us. So let&#x27;s go in and figure out what are you is gonna be Now it looks like we have the derivative of tangent right next to it because the derivative of Tanja is seeking squared. So we might say, Well, let&#x27;s let you be tangent of X. So then that&#x27;s gonna give us d&#x27;you by D. X is equal to seeking squared X and then multiply it by d x I say in quotes we get D&#x27;You is eager to seeking squared x d x so that right there ends up becoming deep So that&#x27;s nice. So we&#x27;re gonna have you for tangent, then d&#x27;you and then our new bounds of integration are going to be and I&#x27;ll do those up top here when X is equal to zero, then you is going to be tangent of zero, which is zero. So are lower bound is still gonna be zero. Then when X is equal to pi four, you is eager to tangent off pie four, which is just one. Now we can integrate this here, which is going to be 1/2, you square using powerful for integration. And then we evaluate from 0 to 1, which is going to give us 1/2 1 minus Subaru is one square deserves greater just itself. And then that would give us 1/2 for our first send, bro. Now, what about our second? Well, we should still end up at this step here. The only difference is that our balance would be different. So we&#x27;re gonna have 1/2 you squared evaluated from Well, let&#x27;s see. So our upper bound is zero, which already found to be zero, and then our lower bound is negative. Hi. Four. So we&#x27;d let X equals negative. Hi fourth. Now, which gives us you is eager to tangent of negative pie worth. And this is equal to negative one. Now, we evaluated at thieves, so we have 1/2 of zero and minus one because negative one square, this one and that would give us negative 1/2

Bobby Barnes · Answer

we want to use a substitution formula in there and seven to help us evaluate this integral. Now it looks like we have that the derivative of tangent in verse in here smell. Why don&#x27;t we just let you equal to tangent and verse? And that gives us the differential that D U is even to 1/1 plus X squared DX, And that&#x27;s exactly what we have right there so we can go ahead and plug all that in. So it would just be you. Do you and our new balance of integration all solved for up here? Well, when X is equal to zero, we have you, Izzy to Tanja Inverse of zero. And so this is signed over costar. It&#x27;s asking Where is Tangent? What value for tangent gives us zero? Well, it&#x27;s sign of a co sign and sign is 00 star lower bound, still zero now then, when X is a good the one we get, you go to tangent inverse of one. And it is asking what value of tangent outputs one, and that&#x27;s saying where is signing coast on equal? And that&#x27;s gonna be at pie forth. Now we go ahead and integrate this, which would be done by using power rule. So I have 1/2 you squared, evaluated from zero to pie forth. So squaring pie Ford gives us pi squared over 16 and then we multiplied by that 1/2. So it&#x27;s pi squared over 32 and I&#x27;m plugging and zero gives us zero. So you don&#x27;t even need that and that there would be our solution.

Bobby Barnes · Answer

we want to use the substitution formula in theory and seven to help us evaluate this It to go now I don&#x27;t know how to directly handle tangent, so any time I&#x27;m kind of stuck with trick functions, always just try to rewrite everything in terms of signing coastline. Let&#x27;s do that. So tender is sign over. Co signs will be sign of X over two over Coast Side X over too. Thank you. And now this kind of looks like we took the derivative of natural log of co sign because if we let you equal to co sign of X over too, then we&#x27;d get the differential. D&#x27;You is so the derivative of co sign is negative sign and then chain rule says we need to take the derivative on the inside. And so the derivative of X over two is just 1/2 and so this will be our differential here and this kind of looks like what we have. But we will need to multiply this by negative 1/2 and then also will need to multiply by Negative too Just over. Multiply this by one and now this is exactly d&#x27;you. Let&#x27;s go ahead and plug that in. We&#x27;re gonna have negative to integrated from do you over you and our new bounds will be well, let&#x27;s go ahead and plug these end when X is equal to zero, then you is going to be equal to So it&#x27;s co sign of zero, which is just one. It&#x27;s our lower bound us one of them. What about our upper bound? That&#x27;s X is equal to not zero but pipe house. So we get you is able to Koh Sai of pie half divided by two, which is pi fourth, which gives us route to over two. All right, now we can go ahead and plug everything in to integrate. So I&#x27;m negative too. And then natural law of the absolute value of you evaluated from one to route to over two. So since both of the numbers were plugging in are positive, we don&#x27;t need to worry about the absent value. Let&#x27;s go ahead and move this up here so I get a little bit more room. So we have negative too. Oh, the natural log of the square root of two over too, and then minus the natural log of one. So the natural log of one. Is this going to be zero? And then we could do a little bit of algebra to rewrite this, to make it look a little bit prettier. So you have negative to natural log of we&#x27;ll Square root of 2/2 is really just one over route, too, and then we can use that negative to reciprocate. This would be to natural log of the square root of two, and in the square root of two is just 1/2 power so we can pull that out front. So it be 1/2 times two, which would deceive us with the natural log of two so you could leave it as what we have like right here or even before, but refine it to just be natural to is a little bit prettier.

Bobby Barnes · Answer

we want to use a substitution formula in there, um seven to help us evaluate this into role here. Now, normally, when I&#x27;m thinking about what to do for a substitution, I try to think about what&#x27;s giving me the most issues. And in this case, I would say this E to the co tangent data is giving me the most issues. So maybe if we let you equal to co tangent of faith because we know how to integrate E to the U, we can somehow get that two counts out with that coast. He can&#x27;t square that we help on the outside. So the derivative of co tangent would be negative. Co c can&#x27;t squared data, and then our differential is gonna be just rewrite and get with do you in detail. And the one thing we&#x27;re missing from this is that negative. So let&#x27;s just multiply each side by a negative. And that tells us co Seacon squared data is going to be negative. D&#x27;You Let&#x27;s just go ahead and pull that negative all the way out. And so now we&#x27;ll have one plus e to the u Do you now for our new bounds of integration Let&#x27;s go ahead and do that up here. Well, when data is equal to pi half we&#x27;re gonna have you is equal to co tangent of pie half and not co tender. You&#x27;re not co signed a coat, Tanja. So this is gonna be co sign a pie half over. Sign up I half. So that should be 0/1. So just zero So our upper bound zero and that are lower bound. So it&#x27;s gonna be you, Izzy, go to Ko Tension of pie forth. So signing co signer both route to over two that pie forth. So that&#x27;s just going to be one. So now this is gonna be our new intro here. Now, we would have some negative on the outside. Still integrated want gives us you Each of you is just eat to the U how we evaluate from 1 to 0 now plugging in zero, we should have zero plus one because each of the zeroes one and then we&#x27;re gonna do minus minus, which would give us plus, and then we plug in one, which is gonna be one plus e to the wondrous just he So now this negative one here on the positive one, their counsel, and we&#x27;ll just be left with G for our

Problem

Use the Substitution Formula in Theorem 7 to eval…

Problem 38 Easy Difficulty

Use the Substitution Formula in Theorem 7 to evaluate the integrals.
$$\int_{\pi / 6}^{\pi / 4} \frac{\csc ^{2} x d x}{1+(\cot x)^{2}}$$

Answer

$ \frac{ \pi}{12} $

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Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$ \frac{ \pi}{12} $

University Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Joel Hass, Christopher Heil, Przemyslaw BogackiUniversity Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Joseph L.

Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals