Calculus: Trig Substitution - Definite Integral 1

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02/09/2011

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Techniques of Integration
Trigonometric Substitution Strategy
Trig Substitution Involving a Definite Integral - Part One Page [1 of 3]
All right, let's take a look at another integral together. This one, the integral from dx.
Now, first let's notice that this is actually a definite integral, which means that I actually have points of integration. So when I get my answer, I'm not going to tack on a + c, but, in fact, I'm going to evaluate a 2 and then subtract what I get when I evaluate at . So we have to be careful that we have these end points here. In particular, if we actually make any kind of substitution at all, any kind, whether it's anything at all, we have to make sure that at the end of the day we're going to put x values in for x and not just stick these values in for other variables.
The other thing I immediately notice is that a u du-substitution won't quite work here, even though the derivative of the inside is 2x and I sort of see the x here. The x is in the denominator, sadly, not in multiplying out in front. So, in fact, this is not even a particularly good integral to try really try a u du-substitution. So that's sort of sad.
The next thing I notice is that, in fact, this is the difference of two perfect squares. I see x^2 and you say, "Well, that's not a perfect square." Well, sure it is. It's - . And I'm taking square roots, so maybe this is really an application and an illustration of a trig substitution. To see if this really can be flushed out, let's draw a right triangle and see if we can actually place these pieces somewhere on our picture.
Now, I see a difference here and I see x^2 minus something, which means that it seems to me that this might, in fact, be the hypotenuse of this triangle, since it's x minus something and the Pythagorean theorem, if you wanted to solve for a leg, you have the hypotenuse minus the other leg. So I put the x here and I put a 3 here. That actually is not quite correct, because I need to take the square root of these things. So this would be = x and the = . So I really do need to write that. And then what's this mystery side? Well, let's compute the mystery side right now. So the mystery side, well, we know that the mystery side squared plus this side squared - so, = 3. That will equal this thing squared, which is x^2. Okay, so now what's the question mark? Well, ? would equal - well, I have x^2 and then I have to subtract the 3, and then I need to take the positive square root, only positives since it an actual length, and I see that ? = . And now I'm feeling actually even surer of myself, because that's exactly the quantity that has appeared in the integral that we were handed. So, in fact, that question mark length is actually the numerator here. So this looks like a pretty neat potential substitution. So I'll write that in here. And then if I call this angle theta, and again, it doesn't make a difference how you place these things down, as long as you make sure that these two lengths represent lengths of legs and the x represents the leg of a hypotenuse. Any other combination of putting them down will work just fine.
Now, once I have this picture, think about this picture as a dictionary that allows me to go from one thing to another. It translates from information about x's, namely, the sides, to information about 's, namely, the angle. So, in fact, let's just notice a few of these potential translations. Well, for example, you could write down sine. So what's sin ? That would be , which would be . What about cosine? cos = - oh wait, I'm writing tangent and I'm saying cosine. So let's think about this now. What is cosine? Cosine is . What I was describing was tangent, which is . So we could do one of two things here. We could either write down the right one or we could write down something else, so let's try it again.
By the way, it's not a bad idea just to write down all of the trig functions and make a complete dictionary of everything. I'm just going to write down some of them. So let's start again with sin = . And now let's write down cos , not tangent - you could write down tangent, too - cosine is . You see, it's really easy to make little mistakes here. And you can write down tangent and so forth and so on and list them all. I'll just write down these two for now and, if we need more, we'll come back and we'll get more.
Now, if you look at this, you immediately see that we're in a great position, because, in fact, this whole piece right here is nothing more than just cos . So, in fact, this looks like a fantastic possibility to get rid of all that stuff and replace it with just cos . So if we consider now a new integral - so now I'm actually making a substitution. The new integral - now, I'm not going to actually write an equal sign here, because I don't want to worry about these end points quite yet. So let me just take a look at this integral. I'll write the whole thing here again, dx. That thing is going to equal - so I remove those end points now. Well, this whole bubble gets replaced by cos . But the annoyance still remains. I have this dx and I can't just write d at my own particular whim. I've got to figure out how changes in x correlate to changes in . So what I need is a connection that links x and together. If I have that connection, then I could find out how x is changing in terms of and so forth. So I need a connection, a relationship between x and and any relationship that you can find through this picture will be fine. Any of these would be fine or anything else you can think of. I always try to pick the easiest one I can. The easiest one seems to be the sine one, so let me actually focus on that for a moment and see if we can get a relationship. So if I take the fact that sin = , let me differentiate the entire thing with respect to x.
Now, to differentiate that first part, as always, we're going to have to use a little chain rule, because it's sin , not sin x. So, in fact, I've got to make sure I take into account the fact that is changing in terms of x. So what's the derivative of sin blop? The derivative of sin blop is cos blop, which, in this case, is cos . And then I have to take the derivative of the inside, so that's the derivative of with respect to x, so that's d dx. And then here I have the and what's the derivative of that? That's . If I multiply everything through by dx, what I see is cos d = dx.
Okay, now I'd like to make that substitution right now and say, well, dx equals all this stuff. Well, if I do that - gee, it doesn't look so hot right now, because I have these x's here. So this is a little bit scary, but let's just try to do it and see what happens. If I try to solve for dx, what I see is dx = cos d. Okay, so let's put that in for this. I'm not exactly really happy, because I see that the x's still remain. But I guess we have not choice at this point. Let's just keep going. So I've got cos , and in replace of dx, I'm going to put all this stuff in. So that's going to be that negative over , so that negative over I'll write out way in front, since it's a constant multiple. So the negative divided by I pull out, and so now I'm just left with the x^2. And then I have a cos , and extra cos , which will make this cos^2 . I have one cosine here and then one cosine coming in here, so I have cos^2. And then, unfortunately, I have this x^2 d. So the good news is I got a d in there. The bad news is that somehow I snagged myself on an x^2. So that's a little bit annoying.
So how do I get rid of that? Well, I guess the only thing to do is to go back to my dictionary and see if there's any way of getting rid of an x^2. Well, look at this right here. If I take this sine expression - by the way, notice that it's not a method that always works. We've got to actually be a little bit creative and see if we can make this thing a go. If I now solve this for x, I want to get rid of the x, I see that x = . And so x^2 would equal the square of this, which would be . So, in fact, the x^2 can be replaced by 3 sin^2 . And so let's actually put that in right here. If I put that in, what I would see is , and then I have this integral and I have cos^2 . And now, instead of x^2, I'm going to put in its twin, which is d. Now that 3 is a multiple constant that I can pull out in front, and so I see this equals . And I have the integral of what? I have . But actually has a name. It's not quite tangent, it's the reciprocal of tangent; it's cot^2 d. So now I've gotten rid of all of the x's and I'm just left with integrating this.
Now, integrating this is something that we can just do using other methods or by looking up and what I would see is this would equal , and then what's this integral? This integral works out to be -cot - . So that's the integral, which I just worked out in advance or looked up. So this integral turns out to equal this.
Now, unfortunately, the problem is I still have the 's there. So what do I have to do? Well, what I have to do is I have to convert the 's back to x's using some sort of dictionary, which comes from this picture. And then once I've done that, I've got to plug in these points. And for that I want to stop first and give you the opportunity to try to do this on your own. So really I urge you to sort of just stop, take a few minutes and actually try to convert this back into x's. And once you have that, try to plug in these end points and see if you can figure out what the actual numerical value of this is. Then we'll come back and we'll try it together. So I'll see you there.