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Hal X prove (-1 2 2 Sin(mmx) dx = Mz| (2n+1)3 (2n+2...

Question

Hal X prove (-1 2 2 Sin(mmx) dx = Mz| (2n+1)3 (2n+2

Hal X prove (-1 2 2 Sin(mmx) dx = Mz| (2n+1)3 (2n+2



Answers

Show that $ \frac {d}{dx} \sqrt[4]{ \frac {1 + \tanh x}{1 - \tanh x}} = \frac {1}{2} e^{x/2}. $

Uh If I was tasked with finding the integral of x squared minus one over Uh x -1 dx, I would have a caveat written somewhere, the x cannot equal one because they need to be divided by zero. But as far as how to properly do the problem, you could factor in the numerator of x plus one x minus one because it's a difference of squares All over X -1 DX. Because then you can cancel out those x minus ones and we know how to do the integral of that will be one half X squared plus X plus C. Again with this caveat, that economic one, so this would be the correct answer. But if you try to do each piece individually, like x squared minus one dx all over In the role of X -1 dx and you work through this problem this way. Uh Again each one of them is the power rule. Uh Plus C I guess. Plus he would be in the numerator and the denominator. Uh Nothing cancels. And this is not equal. Like if you took a cubic function divided by a quadratic, you should get a linear. Yeah. Anyway, long story short, what's circled in green? The correct answer in no way equal. So slash through their does not equal this thing down here. Um And I think showing the correct method would be enough, but you might also work through the other just just a double check. That is not equal.

And this problem, we have to use our knowledge of implicit differentiation and the inverse of, um, pardon me, the derivatives of inverse trig functions To prove the derivative of a of a specific function that were given, Yeah, this will make more sense once we go through the process. So we're told that why equals the inverse co sign of X? So what does that mean? That means the coastline of y equals X. That's just a property of the inverse trig functions. So we need to prove the derivative, and we have to use implicit differentiation. So we'll take the derivative of each side of this equation. So that means that the derivative of the coastline of why would be equal to the derivative of X. So what we do, we would get negative sign. Why Times D y d X equals one again. This part was chain rule, so we'd have this d y dx here. So then we can get d y dx by itself because that's what we want. We want the derivative. So do what do I d? X would be equal to negative one over the sine of why now we could do a substitution. The sign of why is something that we know we can use the Pythagorean identity, the main trick identity that we know. So we can say that Dwight D X equals negative one over the square root of one minus the cosine squared of Why, yeah, now this work is a little bit tricky. We defined what? X waas. We said that the coastline of y equals X so we can do another substitution. Do I. D. X would be equal to negative one over the square root of one minus x squared. So then we can say that the derivative of the inverse co sign of X with respect to X is negative one over the square root of one minus X squared. And that was our goal. That's what the problem told us we needed to dio. So I hope that this problem helped to understand a little bit more about implicit differentiation. And also, I hope, is built your intuition of the differentiation rules of inverse trig functions

In order to prove that this integral here ends up going. B squared minus a squared over two will find Delta X, which is our upper limit. Minus are lower limit, all divided by N. The idea is we're going to rewrite these as limit of the some and then simplify from there. So we've got the limit as an approaches infinity, and then we have the some from I There is one thing up to and multiplied by the width of each of our rectangles, which is B minus a and then all over and that multiplied by the next value, which is X actually. And we start with the lower limit a and then we add the F term of Delta X, which is B minus a all over end times I and this is what we want to simplify here, all of that. So the next thing we could do is go ahead and distribute each of these. And so we'll go ahead and write that as a times be, I must say all over and and then we have plus, by the way, it's still the limit in the some. So these two just dropped down I'm just leaving them off for a second. Um okay, so we've distributed those, then we'll go ahead and distribute there, so that will end up killing us. Be minus a squared. Over this time we have m squared and let's go ahead and apply the some toe Both of these. And of course, it's from I was one toe end so that the sum if we completed on the first one, will just give us another factor of n which will actually end up canceling their and the some on the next one. We could replace this. I hear where m and plus one. And you could look up how that substitution works there. It's just one of a given some nations for just I from has won the end. Okay. And then we're going to rewrite this whole thing up here, so let's do that. Maybe green. So what do we have? We've got Let's distribute a here and here. So we've got, um, a B minus a sward. Then we have plus Okay, if we kind of mush this all together here, we're going to have, we'll rewrite it. We're gonna have be minus a all of that weird. Then distributing the end will have and squared and then plus a term with n so that first one would be in squared over the bottom. We have a two and then end square and then finally plus, then we basically have that same thing with, um, end there. So be minus a all squared times and over to and squared. Now, in the very front, we still have the limit as I am approaches infinity. And it's at this point that we're just gonna go ahead and apply that rule. So when the higher power is in the denominator, all of that goes to zero. These ends here divide out. So really now we don't need to worry about that limit. We could just go ahead and substitute, so that helped us simplify it quite a bit. So let's write what we have at this point. So we've got a B minus, a squared still, let's go ahead and expand this out. This will end up killing us plus B squared and then minus two a B and then plus a squared. Except all of these are divided by two. Still, because of that fraction, So it's right that And in the end, let's see what happens here. So the twos cancel that A B cancels with a negative A B does go away a negative, a squared with a plus. A squared over two ends up giving us a negative. When we combine these two, that basically ends up giving us a negative a squared over to and then notice. We already have a B squared over two, which just drops down. And so, in the end, that is the exact same thing as this B squared minus a squared Over two course, we could just go one step further. In fact, with one half out, one half and then we're left with so her empathy, it is there one half B squared, minus X squared. So one of the questions that I had when I was over the years of kind of understanding integration is where does that b minus a come from? And this is ah, good example of how that appears there upon evaluating a and a girl such as this. So I hope that was helpful. There

Right, Let's just start with the correct first answer. So we have x times x squared plus one dx. So if you were doing this problem correctly, you would distribute that X into the problem X cubed plus X, still D X. And then from there, you can do the anti derivative by adding one uh to the exponents, multiplied by the reciprocal of your new exponents. All right, So this would have been correct. Okay. And then the issue comes in is that they decide to do the no, that's not what they do. They just move the X in front of the inner rule of X squared plus one the X. Well, if you do that, the X goes alone and it's one third X cubed and the integral of one is just X plus C. And then if you distribute that in there, You would get 1/3 x to the 4th plus X squared plus C. Which is definitely not equal um to what was circled in green, you know, the coefficients are all wrong. I guess that's the only issue, but definitely it would give you the wrong answer. One there is not equal to 1/4 and one is not equal to one half.


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