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For the integral 6, 7 sin" (z)cos(r)dz, if we use the substitution u = sin(z), what will be the resulting ~integral?duWhere aAnd b...

Question

For the integral 6, 7 sin" (z)cos(r)dz, if we use the substitution u = sin(z), what will be the resulting ~integral?duWhere aAnd b

For the integral 6, 7 sin" (z)cos(r)dz, if we use the substitution u = sin(z), what will be the resulting ~integral? du Where a And b



Answers

Use the method of substitution to find each of the following indefinite integrals. $$\int \sin (6 x-7) d x$$

And this problem, we want to evaluate the integral, uh, co sign of wide d y over sign square life plus sign of why, my ass six. Um, so by first clients, looking at the top coastline why do white? This kind of sets us up for a substitution. Where if you was equal Teoh Sign of why do you would equal co sign of Y d A y? So it's, uh what we want to dio is get this fraction into a form of polynomial als instead of trig functions. Because if we did that, then what we could do is split the fraction up into smaller pieces of partial fractions. Let's go ahead and use the substitution. So it's gonna equal. The top is just you and the bottom is going to be you squared. Plus you, my six. Okay, so let's go ahead. And we want to look at the fraction one over you squared plus U minus six. So if I factored, this six is two times three and this is positive one. So we want negative. Two plus three is equal. Teoh Positive one. So that will be one over you, my ass. Two times you two plus three. Okay, so I want Oh, um, perform partial fraction before I want to convert one over U minus. Two times you plus three and a partial fractions of some constant over a U minus. Two plus some constant over u plus three. Okay, So since these are linear in power, well, the nominators air linear. So I can use the heavy side cover of method to find these really quick. So have you. Equal to two and you equal to negative three. So for a recovering up U minus two So than any is gonna equal 1/2 plus three, which is equal to 1/5. And B, we're covering up you plus three. So that's gonna be one over negative. Two minor. Sorry. Negative three. My next to your skin vehicle. Negative one best. That's enough fives Here. Here we go. Okay. So we can rewrite our role as, ah 1/5 times one over U minus two minus 1/5 1 over U plus three deal. So this is gonna be 1/5 l an absolute value of U minus two minus 1/5 Ellen. Absolute value of you plus three good eso From here, we can combine the ISS and to 1/5 Ln, uh, absolute value U minus. Chew over. Absolute value. U Plus three constant close concept Post constant. Uh, looks good. Okay, So what was you? You was equal. Teoh Sign of Why So then this becomes 1/5 Ln of absolute value. Sign of why minus two over absolute value. Sign of why most three plus a constant so negative one is less than or equal to sign of. Why this is what's very cool toe one for all values of why reels So we can rewrite these absolute values. Fifth Ln of so we'll have to minus sign of why Over sign of why plus three plus a constant since. So this is always gonna be negative on the inside. So that's why didn't the gates. But on the bottom, it's always gonna be positive. His three is always greater than negative, right? That completes our problem.

Eso here we can define a function you g you equal to G Its end is four plus three signs said It's not the numerator denominator. I've been two grand is just from you and then we have do you is three co signs at eso No signs that he said then is due divided by me. So end up interviewing a suspect to you. Uh, just one over three times the square root of you buying the are forming invented rib is the year olds Ah, 2/3 use Where? Rubes. It's her general anti derivative. And now, uh, for party so we can substitute. We have that G 80 is four plus three sign zero shines the reserves That's four plus become 042 pie for the same reason we have signed two pies, you know? So, uh, G at zero was Julia. True pie is four. So we're evaluating integral from 44 which eyes, therefore is, you know, realign in case B. We have that sign and a minus pi is, um minus zero is signed by that It's signed and plus minus pi zero s o. We have both husbands. I dri is foreplay Street on Jewish For eso again. Weren't you going to comport for an interview? As you zero in this case falls, we consult. Its true value is worrying me both, you know.

We want to use thes substitution formula in theory. Um, seven. Help us evaluate these two intervals here. Now, normally, when I'm thinking about what to make my use of its normal everything, that makes it look complicated and in this case will be have underneath our radical here, the four plus three sign of Z. That's kind of complicating things. And if we take the derivative, we end up with three co signs. He just kind of like what we have in the numerator. Let's go ahead and get our differential. We're gonna let you get to four plus three times sign of Z, and then our differential should be Do you is equal to three. And the derivative of Sign is going to be co signed Z And then we have Deasy. Let's go ahead, just divide each side of this by three. So those three cancel up and then we're just left with co sign Z d Z Is he going to do you over three? Let's go ahead and plug everything it now. So I'm gonna factor that 1/3 all the way out. So it's 1/3 of and we could figure out what our new bounds of integration will be in a moment. But then the co signs e t Z should become that. Do you over three, Which is that one there? We just factored out and then we have square root of you. Now we know how to integrate her. Actually, before we talk about how to integrate one of the square root of you, let's figure out our balance. I don't do that up here. So when C is equal to zero, you is going to be four plus three times co sign of I'm sorry. Sign of zero. Now Sign of zero is zero. So that should just give us four or lower bound is for, and then Z is equal to two. Hi, this is gonna give us you is equal to or plus three sign of too high. And then Simon to pie is the same thing as sign of zero. So this should also give us four. And actually, at this point, we don't even need to integrate this because if we were evaluating at the same point, then the area underneath that is going to be zero. So zero end. That's just because pounds are sane All right. So these for a we don't have to do anything further, but maybe for b will need to do a little bit more. But at least we should still get to the same spot of the 1/3. Integral off. Um, do you over the square root of you, and then we just need to figure out what our balance for this you're going to be. So it doesn't look like we have any overlap. Four bell. So let's just go ahead and plug both of these in. So when z is it a pie? We should get that. You busy too? So four plus sign of pie and then sign up eyes zero. So this should just be able to four again. So our upper bound is going to be four, and then Z is into negative. Hi, but I am one of footprints. Sees there. Well, this is gonna be you, is even to four plus sign off negative pie and then sign of negative pie is also zero. So this is gonna be four. So we actually end up with the same thing we had over here, and we don't even need to integrate anything really? Because we just know since it has the same bounds, the area under detect curb should just be zero.

To evaluate this one. It might be helpful, too. Rewrite ID in the following way. So we know that co Tangent is one over tangent. And so just looking at like this, there's no obvious relations between you know, the derivatives of co tangent with seeking, you know, besides the fact that it's the derivative co tension is minus coast seeking squared. But we do know that seek and squared is the derivative of tangent squared. So if we could just rewrite this so tangent is showing up, then we can use that as the substitution. Let's rewrite rewrite. This is seeking Squared of Fada over six. So I'm just rewriting co tangent as one over Tan Xin. So I have a tangent to the fifth fate over six. Okay. And so now let's make the substitution You is equal A tangent, a fate over six. So then do you is by the chain rule we get a one six factor from the fate over six Seacon squared fate over six dif ada So let's re write this in terms of you now. So Seacon squared Fate over sixty theta is replaced by six Do you tangent of fate over six is just you. And now let's change our limits. So when use equal Sorry when theta is equal to pi, use tangent of pie over six. So if we maybe draw this triangle or this is the singles pyre is six This angle was pi thirds. This is one of our special triangles. Then we know the sides of this triangle are the shorter side is one the longer side of square two, three and high pot nooses too. So let's go ahead and see tangent of pie over six, then is one of her squared of three. It's the lower limit here becomes one over route three and the upper limit. So if we plug in here, we would get tangent of so in fate Ahs three pi over too. We have tangent of three pi over twelve, which is the same as tangent of pyre before, so I could draw my other special triangle forty five, forty five ninety one. So this is the same angle here pie before the sides air one, the high pot nooses route too. So tangent of pirate for is just one. Okay, so now this integral is completely in terms of our new variable. You and lucky for us, we can recognize that this is thie on anti derivative for one over you is just natural log of you. So I have natural log. If we want to be careful, we should put absolute values around you because well, you can check. If you take the derivative of this, You do get one over you. And the argument to natural log should always be positive. So we don't want any negative values here. Anyhow, we're just going to evaluate this between one and one over square two three. And so this is giving us six natural orb of one minus natural log one over square two three. And while I have no idea what exactly natural log of one ever grew three is I do know that log of one zero. This is leaving me with minus six natural of one of her Route three


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