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Find the 1 indefinite TanApCalc9 6 2.044. integral Hint: LetNeed Help?1(Remember use absolute values where approprlate...

Question

Find the 1 indefinite TanApCalc9 6 2.044. integral Hint: LetNeed Help?1(Remember use absolute values where approprlate

Find the 1 indefinite TanApCalc9 6 2.044. integral Hint: Let Need Help? 1 (Remember use absolute values where approprlate



Answers

Evaluate the following integrals. $$\int_{0}^{\pi / 6} \frac{d x}{1-\sin 2 x}$$

Soon in this problem, we are asked to evaluate this integral. And so the first thing that we notice spotted it's that it's an integral off the power eggs and were derived integrating with respect to X. So we the first thing that we're going to recall this tower pour into girl. So for integrating X lay still power and where in has to be different from one. Uh, this tentacle will be ex raced and plus one and we divide bye and plus one to this is a indefinite And the growth we are, eh? A constant in this case, we don't really need it. But so in our case and ISS minus sixty five. So first thing that we want to do to integrate this pendant in plus one. So since this is a fraction of going to do it separately, yeah, and plus one isthe over six from seven hundred six over five plus one. But instead of range one since we want to be able to add a two two numbers were one of them is a fraction they have to have the same the number denominator. So we want something with sign on the bottom Since we're talking about one, we want same number asked in a greater and greater. So instead of writing one working to write five five. Still, this is equal to five days right here and now top arguably. Haps my six plus fine. And that is people too. Minus one over. So this isthe class. Fine. Which is what we're gonna do. Okay, uh, look. And here and here are integrated. So and this first up, we don't need this practice. We're already integrating, and we have ex place to n plus one, which we've figured out this minus one over pi Delighted by the scene number Mendes one over. And this expression, it's going to be when x equal one and up to thirty two. So before, without practice, huh? Function in thes and points of the integral. I'm home to reeling to this so that it's easier to work with. So we have, uh, this minus saying we'll keep it here. We could put it instead over here instead of here. And this five since it's in the numerator after the narrator, we can bring it back here to bring it up floor here. All right, so we can rewrite it like minus five eggs. Fruit base, too. Mice one. Over and again. We don't lose this X equal to one, two, thirty two. All right, so now, before we do this next part, which issue seemed the fundamental Purim of calculus where we evaluate this when ex house and one x thirty two minus the same expression on except one we are going to I remember some of that fact. So here we have. When we raise one to any power, get one and thirty two, we can write that us to the park soon. I am going to first one X equals to this first part is when X equals thirty two, which we're going to raid us two to the fifth power and then we're going to use the same expression when X equals one. All right, so let's move this downtown. So we have Manus fire. That's a constant Tim Specs, just two to the five. And this text is going to be raised to one. Men's went over and bring to this is the expression on X equals thirty two and three Subtract the same expression on X equals y. So the constant does not change eggs. This one and the power. All right, so you know, it's all constants. We can no work on getting this a little bit. One similar simplifying this right? So we have. And this power, it's five times who won. Minus whatever fight the chain. Since this is denominator, the numerator and denominator thes cancel out. So we're left with two races to the minus one, which is people to one half. And in this case, since we have one race to some power, this is going to be equal to just so this simplifies too. Minus five times one half between kin, Right? Like this. And this place no, outside this minus and minus turned into positive. And this is five times one. The chains. It's just five. Now again, tow up to numbers we have. This is a fraction and a hostin a mayor too. So we're going to put it to here. And since we do it on the bottom of this number, we're goingto have to do it the same. No matter. So think of this five as five over one. That's what it is. And then we're just multiplying by to both the numerator generator. Right? So now we can actually at these numbers till you have minus O on the bottom. We have to. That does not change. And we have minutes five plus times two. And that's equal to five. Number two seven. That isthe our answer when we evaluate this tibia.

We're gonna get a girl from 1 to 32. Have x the negative 6/5 d X. So when we do an anti derivative, we add one to our exponents. So I'll have X. So I'm adding 5/5. So that's why it gets the negative 1/5 and you divide your new exponents. Negative 1/5 when you evaluate that from 1 to 32 32 is kind of a suspicious number. I'll bring that make up here 32 to the negative one. Oh, come on. One fifth divided by negative 1/5. That's F A B minus F of a one to the negative, 1/5 over. Negative 1/5. That's that way. Um, remember that the fifth power or the 1/5 power is the same as doing like the fifth root. And so I can actually think of this as as asking me to do the fifth root of 32 to the negative one over negative 1/5 over here. It doesn't matter what the power is. For one, it's just one over negative 1/5. The fifth through 32 is two. So if two to the negative one in the over negative 1/5 minus. There's two negatives over years. We'll make that into a plus. And one divided by 1/5 is five. If you bring that down here, we actually just have Ah, two to the negative. One is 1/2 times, uh, negative 5/1 plus five. So we end up with negative five house plus five, which is positive five house.

Let's use partial fraction to composition to evaluate the definite integral from zero one. Looking at this fraction, we see that there's a quadratic in the denominator. So let's go ahead and try to factor that. So does this denominator factor. Sure does weaken. Take X minus two X minus three. So this is what the book would call case one. We have a distinct linear factors non repeated. So we have hey, over X minus two plus B over X minus tree. So this is the form for the partial fraction to composition, and I will have to use some algebra here to sell for Andy. So looking at this equation here circled in red, that's good and multiply both sides by the denominator on the list. So multiplying by X minus two times X minus three. On the left, that denominators will will cancel. And on the right hand side, we have a minus three plus B X minus two. Let's just go ahead and rewrite the right hand side. Let's pull out the ex term. We have X and then we have a Plan B and then we have minus three, eh? Minus two b. It's on the left hand side. We see that the coefficient in front of the ex is the one on the right. It's a plus, B. So we must have a plus. B equals one. Similarly, the constant term on the right is on the left. Excuse me is minus four, but on the right hand side, the constant storm is minus three minus two B. So this gives us a two by two system of equations too soft for A and B in that last equation there, you could go out and multiply by a negative one many ways to solve these systems. For example, we could take the first equation, solve it for Bea, and then we could plug this value of be into the other equation and in software, eh? And doing so will have three a plus two, one minus a equals four. And this will give us a plus two equals four so equals two and then must take this value of a plug it into this equation up here and we get B equals one minus two, which is minus one. So we found an B. So now we're ready to integrate. Let's come to the original problem the original integral. In the top left, we can rewrite this. So we replaced the fraction with the partial fraction to composition over here in the far right upper right corner. And we've just found A and B. So I have two over X minus two and then for B, we have a minus one over X minus three, and this new integral should be easier than the original. Now, if this X minus two or explain histories bothering you, you can go ahead and do a U substitution. U equals X minus two and for the second one, x minus three. And by doing so after simplifying, we should have to natural log absolute value X minus two minus natural log absolute value X minus three from zero one. Let me separate this from our scratch work. We've evaluated the inner girl. We have the natural logs. Now we just plug in the end points and simplify. So we plug in one first. We have natural log of negative one, but we have absolute value, so that just becomes a one. Then we have natural log of one minus three is negative to take the absolute value you just get into there and then when we plug in zero, the first term will become two times natural log. We have a negative to their absolute value. Makes it positive. And then finally, we'LL have a natural aga positive three. After we take absolute value and then just go ahead and simplify as much as you can. We know natural Aga one zero Combining these national log of twos. We'LL have negative three Ellen, too, and then watch out for this double minus here. That becomes a plus at one three. And there's many ways to simplify this. For example, we can write. This is Ellen three minus ln eight. By taking that using one of the log properties that says you can take the coefficient in front of the log, bring it inside. But then you have to raise it to the exponents. So two to the three is eight, and then I could use another property of logs log. If you subtract the lines, you can rewrite it as a fraction log of a fraction. So it looks like I have any one of these will be the right answer. Let's just maybe narrow down some of these two right here, and that's your final answer

This problem is from Chapter seven section to a problem. Number forty five in the book Calculus Early Transcendental lt's a Condition by James Store. We hear we have a definite integral of the square of one plus coast high to X. So here let's talk about co sign of two X. We can use a double ankle formula to rewrite this as two times co sign squared X minus one. Then from here we have one plus coastline to X. Here we can cancel the ones and we just have to co sign Squared X. Then we could take the square root so we can write. This is I want to times the square room of Coastline Square, which isn't necessarily co sign because technically, the square root of the square is going to be the absolute value. So this case, we should to be safe. You should always write absolute value there, I guess in general, the rule that I'm using here is squirt of a square is absolute value in our problem. However, this will be actually will simplify too, two times called radical two times co sign since coastline eggs. It's positive in our inseparable of the liberation. What that said we can rewrite are integral. That becomes zero time for six time's radical, too times cosine x so you can pull out the rat radical to outside the integral if you like. So Integral co sign a sign so they get radical, too time. Sinus evaluated at zero and pi over six. So sign of viruses. One half and sign of zero zero So we get squared or two divided by two, and that's our answer.


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