WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to introduce a technique called solving integrals by substitution.
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This is a very powerful technique.
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For all practical purposes, for the rest of you, the work that you do as an engineer or physicist or mathematician,
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this solving by substitution is probably the one technique that you are going to use more than any other,
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if you are going to do anything by hand.
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Most of the time, you are going to be using software to do it.
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But this technique is really what we use, most of the time.
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Let us jump right on in.
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I’m going to do this lesson not by doing theory because the theory will make sense by doing the examples.
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It is the examples that I think make it clear, what it is that is going on.
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If we just sort of launch into theory, this is one of those things
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where I think the symbolism gets in the way of what is actually happening which is very simple.
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I do not want to off you skate that.
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Let us take a look at what it is that we do know, as far as antiderivatives and solving integrals.
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This gives us a list of the antiderivatives and integrals that we know how to find.
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Pretty straightforward.
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We know how to do polynomials, things with powers.
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We know how to do basic trigonometric functions, things like this.
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However, we are going to be dealing with functions that are a lot more complicated than this.
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We want to be able to integrate those functions.
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In order to integrate those functions, what we are going to be doing is using the substitution rule,
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in order to transform these complex integrals into something that looks like one of these.
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This is our home base, we are going to transform them into something that looks like one of these, by using the substitution.
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We are going to integrate it because we know how to do these.
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And then, we are going to substitute back and get back our final answer.
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Having said that, let us jump right on in with some examples and I think it will make sense.
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Evaluate the following indefinite integral, the integral of tan(x) dx.
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We know that we do not have a basic integral for tan x dx, it is not in that list.
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Let me go ahead and write that out.
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I think I will work in red here.
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We do not have a basic integral or antiderivative whichever you prefer, the derivative for the tan(x).
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The question is how can we transform this and turn it into something we recognize, that we can integrate.
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How can we transform this into something we recognize, something on our basic integral list?
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A lot of this substitution technique is going to be based on just start trial and error.
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You remember back in pre-calculus, when you are doing those identities,
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when you have to prove that two identities are equivalent.
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Sometimes you might start on a path and you end up hitting a wall, not a problem.
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There is no way of knowing beforehand until you gain some experience,
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what to choose for your substitution how to go about this.
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You are just going to try things and follow the mathematical logic.
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If you end up at some place which is good, you are good.
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If you end up hitting a wall, just go back to the beginning, like everything else in life.
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If you lose your way, just go back to the beginning, that is it, nice and simple.
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Do not think that you have to just look at it and know exactly what to do.
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You are going to try different things.
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Calculus is sophisticated mathematics.
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We do not just look at it and see, we know what to do.
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Let us go ahead and change this around.
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We know what tan x is, in terms of the basic functions sin x and cos x.
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I’m going to write this as the integral of sin x.
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Let us make this a little clearer.
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The integral of sin x/ cos x dx.
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I’m going to do a substitution, I'm going to call cos x and I’m going to call it u.
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Let me do this in red.
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I'm going to call cos x u.
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I'm going to take du dx, that equal cos x, I'm sorry.
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That is not equal cos x, the derivative of cos x du dx is -sin x.
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I’m going to write du is equal to -sin x dx.
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All I have done is I have moved this over here.
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This is a ratio, I can just move that over.
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Now I have du = - sin x dx.
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I’m going to move the negative sign over and I’m going to write sin x dx is equal to –du.
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I’m going to plug those back in.
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Sin x dx, that is right there, that is sin x dx.
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This is the integral of, I’m going to put this in for that – du.
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The denominator, I have cos x.
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Cos x is just u, that is that.
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I will pull the negative sign out and I will get du/u.
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I have transformed this into something I recognize.
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I recognize that in the list in the previous page, I know that dx/x, the integral of that is just ln x.
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The variable does not matter.
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du/u, dx/x, dz/z, dp/p, it is the same thing.
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It is just a symbol.
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I have transformed it by making an appropriate substitution.
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I have called something, something, I have taken the derivative of it.
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By doing that, I ended up having something which looks like,
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Now I put that back in, I have transformed it into something I recognize.
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I just go ahead and integrate this.
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-the integral of du/u = the natlog of the absolute of u + c.
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I substitute back in, u is equal to cos(x).
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ln of the absolute of cos x + c, then I have my negative sign, and I’m done.
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That is all I have done, taken something, I have transformed it and then I have made a u substitution.
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Transformed it into something I do recognize, for which I do have an antiderivative, an elementary antiderivative.
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I have taken the antiderivative and I have substituted it back to turn it back into a function of x.
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Do not worry, we will do plenty of these.
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Evaluate the following indefinite integral, let us go to blue.
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Let us do just a little bit of trial and error here.
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I need a substitution, I need to set u this variable, substitution that I’m making to equal something in here.
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Let us do a little trial and error.
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I’m going to let u = cos x.
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My choices are, I can let u = cos x, I can let u = sin x, or I can let u = sin³ x.
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Just to see if I get something that works.
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If I let u = cos x, du is equal to -sin x dx.
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-sin x dx = du, is there a -sin x dx in here?
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No, there is a sin³ x.
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This does not really work too well because we need sin³ x to be some differential du.
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Let us try something else.
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Let us try u = sin x.
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du = du dx = cos x which means that du = cos x dx.
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Do I have a cos x dx?
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I do, there is a cos x dx.
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I’m going to let that equal u.
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Now I have the integral, cos x, that is this part.
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I’m going to call that du, I’m going to put that in here.
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u is equal to sin x.
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This is sin³ x, that is just u³.
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I’m going to rewrite it in the order that I’m accustomed to.
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The function and the differential, u³ du.
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I recognize that, I can integrate that.
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The integral of that is u⁴/ 4 + c.
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U is equal to sin x, it becomes 4x/4 + c.
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That is it, nice and simple.
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I choose my appropriate u, I see if it gives me something that I can convert into, that I recognize.
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I chose u = sin x, I got down to du = cos x dx.
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Yes, cos x dx is in the integrand, I will call that du.
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For this part, I call it u³.
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This is now an integral that I do know how to solve.
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It is an elementary integral.
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I solve it then I substitute back.
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Evaluate the following indefinite integral.
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We have a couple of choices, we can let u = x, we can let u = ln x, we can let u = ln³ x of ln(x)³.
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Again, experience will dictate what choices you are going to make for your u.
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In this case, I’m just going to go ahead and go directly.
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u = natlog(x).
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du dx is equal to 1/x, which implies that du is equal to dx/x.
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dx/x, here is my dx/x, this is equal to, the integral of,
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this is my dx/x is my du, I will put the du there.
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I have ln x³, ln x is u, it is just u³.
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I ended up with the same thing, u³ du.
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I will write it in a form that I’m used to, which is the function first and the differential afterwards.
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This is just u⁴/ 4 + c, u is ln x, it becomes ln x⁴/ 4 + c.
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I’m done, this is the integral of that via this process of substitution.
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Turning it into an elementary integral that I do recognize.
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Evaluate the following indefinite integral.
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This time I'm going to let u equal to x ^½, that is just √x du dx.
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I will just go ahead and do it straight.
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I will put the dx over there.
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du dx = ½ x⁻¹/2 which means that du = ½ × x⁻¹/2 dx, which is equal to 1/ 2 × √x dx.
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I’m going to bring this 2, multiply it out over here.
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This is going to be 2 du = dx/ √x.
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Here is my dx/ √x, that is equal to 2 du.
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I’m going to write 2 du.
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Cos(√x), √x is equal to u, this is cos(u).
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I’m going to pull the 2 out, 2 × the integral of cos u du, that is an elementary integral, I know how to do that.
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That is equal to 2 × sin(u) + c.
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I know what u is, u is equal to √x.
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Therefore, my final answer is 2 × sin √x + c.
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The integral of cos x/ 1 + sin² x.
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u could be cos x, u can be sin x, u could be 1 + sin² x.
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Let us see what works, I’m going to go ahead and take u is equal to sin x.
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Again, there is no way for you to know beforehand, what u was supposed to be.
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Try things, if you try it and it works out well, then you can convert it an integral of the elementary type.
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You are good, if not, then start again, pick another u.
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It is that simple, that is all.
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You are not supposed to just look at it and know.
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Experience will dictate, looking at it, and knowing eventually.
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Which is why I just immediately chose sin x.
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du = cos x d(x), cos x d(x) = du.
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Here is my cos x d(x), I’m going to rewrite that as the integral of du/ 1 + sin² x.
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Sin is u, this is 1 + u².
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This is an elementary integral, du/ 1 + u², that is equal to the inv tan(u) + c.
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u is sin x, I substitute back in, the inv tan of sin(x) + c, I’m done.
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Evaluate the following, we do the same thing, except now we actually evaluate the integral of the n.
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Because now it is a definite integral from 0 to 2.
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Do the exact same thing that you did.
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Now instead of c, you are just going to evaluate the integral, f(b) – f(a).
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u is equal to x² + 2, du is equal to 2x dx.
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I’m just going to do the whole thing.
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du/dx is equal to 2x which means that du is equal to 2x dx.
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I have an x and dx, I’m going to move the 2 over.
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du/2 is equal to x dx.
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My x dx is right here, this is going to be the integral from 0 to,
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I’m not going to put those in yet.
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I’m just going to do the integral, first part.
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x dx that is my du/2 × u which is x² + 2.
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x² + 2 ⁺15, it is u ⁺15.
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I'm going to pull the ½ out and I’m going to write this as u ⁺15 du, switch the order,
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in order to turn it into the order that I’m accustomed to which is integrand times the differential element.
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It will always be that way, there are books that actually put the du first, especially in science.
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Do not worry about the order, the order itself does not matter.
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This is just multiplication.
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Now I solve it, nice and straightforward.
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This is just equal to ½ × u ⁺16/ 16.
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I put the u back in, I get ½ of x² + 2 ⁺16/ 16.
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I evaluate that from 0,2.
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That is it, I just put my values in.
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I’m going to leave the ½ out for a minute.
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I put 2 into here, it is going to be 2 × 2² is 4, 4 + 2 is 6.
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It is going to be 6 ⁺16/ 16 -, then I put a 0 in.
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It is just going to be 6 ⁺16/ 32, nice and simple.
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Evaluate the following, what do we pick for u?
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This is the integral from e ⁺e³ of 1/ x × 3√ln x, this is not x³.
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This is x × 3√ln x.
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I’m going to let u equal to ln x which means that du dx is 1/x, which implies that du is equal to dx/x.
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There we go, I have got my dx/x, right here.
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This equals the integral of, I will write my du.
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This 3√ln x, the 3√ln(x) is just ln(x)¹/3.
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1/ ln(x)¹/3 is nothing more than ln x⁻¹/3.
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Therefore, 1/ ln x¹/3 is ln x⁻¹/3.
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Ln x is equal to u, we are dealing with u, this is just u⁻¹/3.
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I will write in a form that I’m accustomed to, u⁻¹/3.
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du is going to equal u²/3 / 2/3 which is going to be 3/2 u²/3, which is equal to 3/2.
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u is equal to ln x, it is going to be 3/2 ln x²/3.
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I'm going to evaluate it from e ⁺e³, that is going to equal 3/2 ln(e³) is just 3.
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It is 3²/3 – ln (e) is just 1 - 1²/3.
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It is just equal to 3/2 × 3²/3 – 1.
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Evaluate the following, nice and simple.
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u can equal inv sin(x), u can equal 1 - x², or u can equal √1 - x², which one do we choose?
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This is going to be trial and error.
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That will save a little time, in terms of the trial and error, I'm just going to go ahead and say u is equal to inv sin(x).
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du dx = 1/ 1 - x² under radical which means that du is equal to 1/ 1 – x² under the radical × dx.
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There we go, I have my dx × √1 – x².
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This is equal to the integral of my du, this is du.
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Inv sin(x) is u.
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This one is easy, u du this is equal to u²/ 2.
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u is inv sin(x)²/ 2.
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I’m going to evaluate that from 0 to 1.
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This is going to be inv sin of 1²/ 2 - inv sin of,
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I will slow down and write everything out properly.
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Inv sin(0)²/ 2.
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The inv sin(1) is π/2.
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It is going to be π/2²/ 2.
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The inv sin(0) is 0.
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My answer is just π²/4/ 2.
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This is going to be π²/ 8.
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That is it, nice and straightforward.
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Nothing strange happening here.
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Evaluate the following, sin x × cos of cos x dx.
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In this case, I'm going to let u equal cos x.
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I'm going to let du dx = -sin x, which means that du = -sin x dx,
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which means -du = sin x dx.
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Sin x dx, here is my sin x dx.
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I’m going to write the integral, this is my –du.
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I have u is equal to cos x, it is cos(u).
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I rewrite that by pulling the negative out, putting the cos u first, du.
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I know how to do that, that is just equal to –the integral of cos is sin u.
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This is going to be -sin of cos x because u is cos x.
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I get –sin(cos x) evaluated from 0 to π/2.
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I will go ahead and leave it like this which is equal to -,
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cos(π/2), I put this in for here.
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cos(π/2) is 0, the sin(0) is 0, that is 0.
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cos(0) is 1, sin(1) is 0.841.
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I hope that make sense.
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The whole idea behind u substitution is to find some appropriate u.
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Take the derivative of it, move the dx over and see if you can find something that actually matches this,
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so we can start substituting back in and turning the variable into everything in terms of u,
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some elementary function of u.
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Once we have an elementary function of u, we just integrate it.
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Once we get our final answer, we substitute back and turn it back into a function of x.
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At that point, we can either have a c, stop, we are indefinite integral, or evaluate the integral itself.
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