Calculus: Trig Substitution to Integrate Radicals

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Techniques of Integration
Introduction to Trigonometric Substitution
Using Trigonometric Substitution to Integrate Radicals Page [1 of 2]
So let's see this idea of a trigonometric substitution in action with a particular integral. So let's take a look at the following exotic-looking creature. Integral of dx. Now this is an integral, let's see, a u du substitution won't be that particularly interesting here, because if you let u equal the very inside thing, the derivative would not be an x^3. It would be like a 2x thing. Other techniques you might think of trying are integration by parts or other things of that sort. They don't seem to really pan out here. But I do notice that this does have a particular piece that I am aware of, namely, I do see a square root and I see a difference of two things that are actually squared. This is like 1^2 and this is like x^2. When I see that, I think of a right triangle. When I think of a right triangle, I think of trig. When I think of trig, I think of trig substitution. So let's see if we can put all those thoughts together and produce a substitution that would be of a trigonometric flavor.
So, what I'll do I'll draw a right triangle and see if we can correspond the sides, the lengths of the sides of this right triangle to this particular number. I'm just focusing on this, by the way. Notice that I haven't even touched the x^3. I'm ignoring that, because who knows how to deal with that. But this square root of the difference of two squares, that appeals to me. That does something for me, and I want to see how I can lay that out on a right triangle. Since I see a difference, if I think about how the Pythagorean theorem goes, this squared plus that squared equals that squared, it makes me wonder if, in fact, that 1 is really the hypotenuse and this x is just some side. Now, using the Pythagorean theorem, I can actually verify that this side must be this thing down below. And the reason for that is that if you take this thing and square it, it lifts the radical. And then take this thing and square and I get an x^2. If I add those two numbers, I see 1 - x^2 + x^2. Well, the x^2's cancel and I just left with 1, and that's the hypotenuse. So, in fact, this really does satisfy the Pythagorean theorem.
So, what kind of substitution can I make? Well, what I'll do is I'll take this angle here, let's say, and call it theta. Now, there are many different ways of doing this. For example, if you were working on this question on your own, you might have put this thing x, and then you could have called this thing . There's no wrong way to place the things down, as long as you realize that this side is going to be a leg, and then the 1 is the hypotenuse. The reason why I was cautious of that was because I saw a subtraction. If I see something squared minus something squared, the first time must be the hypotenuse of some right triangle and the other term must be 1 or either of the two legs, thus leaving the rest of this quantity to represent the other leg.
Okay, so now there is a triangle. And this triangle actually captures in it; hidden inside this, coded in this is the trigonometric substitution that I'm after. Now, let's see it. For example, let's take a look at what the different trig functions are for this. So I'm not proposing that you always look at these questions this way, but for early on it might not be a bad idea. Let's write down, for example, sin , and then figure out what the value of cos is. And you can put down tangent, secant and so forth. Let's just do those two for fun. So sin would be x opposite over hypotenuse, so it would just be x. cos would be , which would be . Now, that's particularly interesting, because what I see here is that x I can now call sin , where that's what is. And , I can actually call that cos , where that's what is. So I can actually make a substitution here. I can make a substitution that looks like this: I can say this integral equals a new integral. In place of the x^3, I can put in this x^3, which would be sin^3 . So I cube this side and cube that side, so I'd see sin^3 divided by - and then, what is ? Well, we said is actually this cos . So I've got cos and I've made a substitution here, dx. Now, this might be a more easily more integrable function, but there's an annoying problem here. And the annoying problem is everything here has 's, but look at this. I'm supposed to integrate with respect to x. Yuck! So I don't like that at all. That is a problem. What I want to have here is a d. So one way to get around that is just to write d here. That's one way. Now, that's a wrong way, but it is one way. You just can't say, "Oh, dx = d," because how do you know that? In fact, dx might not equal d. So what I want to figure out is how can I convert this dx to a d, and here is an easy way of converting. Just figure out, pick any identity you want that has both 's and x's in it, and then differentiate. So I always pick the easiest expression that links and x together. In this case, it looks like that first expression right there.
So let's take a look at that very first expression, and that expression is just sin = x. And now, that's a relationship between and x. I want to find out how their derivatives relate, so let me differentiate this with respect to x. So now here is the method that I'm always going to use. I first convert all this stuff, but then I've got to convert the calculus part. So the dx has to become stuff with d's. I take a simple relationship that links the two together - by the way, you could use this, too, but it'll be harder to take the derivative. This is actually easier to work with. But all roads will lead to the right answer, if we're careful.
Now, if I differentiate this with respect to x, let's do that. Well, in fact, to do that, since I have here, I'm going to have to use a little chain rule. The derivative of sin blop is cos blop. And then I have to multiply that by the derivative of the inside with respect to x. Well, that's d dx, so that's the derivative of with respect to x. I just took the derivative of this with respect to x by using a chain rule. sin blop, the derivative is cos blop, and then multiply it by the derivative of with respect to x, which is d dx. On the right-hand side I just take the derivative of x in terms of x and that's just 1. And if I multiply through by the differential dx on both sides, it will cancel here and appear here, and I see that cos d = dx. So I just multiplied through by dx. It cancels here and it appears here. Or you can think of cross-multiplying. In any case, I now see that dx doesn't just equal d, but equals cos d. So that's the correct substitution now for this dx. So if I make that substitution, what I see is that this equals the integral of . And then in place of dx, I put down its twin, which is cos d.
And now I have everything in terms of , so that's just great news. And so what I can do now is actually see if I can integrate this. Well, notice that the cosines drop out. The cosines cancel and all I'm left with is the integral of sin^3 d. And this can either be looked up, or you could remember how to deal with these odd powers. We talked about this at some point, and you can actually compute this. I'll just tell you what the answer is, this is sort of a standard one. It's + c. So that's sort of a popular one to remember. Do I remember it? Absolutely not. Could I rederive it? Yes, I absolutely could. I would just peel off a sin^2. That would be multiplied by a sine, and then use the trigonometric identity sin^2 + cos^2 = 1. I would solve and do your substitutions. So, yes, blah, blah, blah or look it up, or use a computer, or take my word for it. Okay, by the way, that last technique is the worst of the techniques, but that's okay.
Anyway, there's the answer, and you go, "Wow! We did it! We made great progress!" It's a great triumph, because now we see the integral of that equals that. Well, no, we don't, because we introduced and the original question was in terms of x's. So how do we get the 's out and the x's back in? Well, we have these identities. So, in fact, I know exactly what cos is. Cos is this square root. So in place of cos , I can just insert this square root thing. And so if we do that, let's see what happens. So putting back the substitution as we had it, I see that this will equal , and in place of the cosine, I remind you that we saw that equals . And so what I see here is the , but then I cube all of that. And then I subtract from that cos , which is , and then I have + c. So now, everything is back in terms of x's and that, in fact, is the anti-derivative of this, which we couldn't have done just by looking at it. However, once we realize that, in fact, there's a triangle in the background and we can make a substitution using trigonometry, then we were able to convert this thing to, in fact, an easier integral, evaluate the easy integral and then insert the information in order to get the answer we want.
This is the theme of trigonometric substitution. So the idea is to find the right triangle, convert things to sines and cosines, and then evaluate. I'll see you at the next lecture.