Welcome everyone. For today's prep, we're going to start looking at a new integration technique called trigonometric substitution. And well, this is going to be one of our main applications of trig integrals.
So trig integrals show up all over the place and kind of get to use them right away. All right, so let's start with the three main trig substitutions we're going to be using. And the idea behind these trig substitutions is they give us a way to integrate square root functions.
So square root functions are kind of notoriously difficult to integrate. Sometimes we can get lucky and just do a plain old substitution or things like that. But generally, they can be tricky.
So the idea of trig substitutions... is one way to help us integrate these square root functions. All right, so the kind of the three main square root terms that show up quite a bit are right here.
So when we have the square root of some constant squared minus x squared, we might see square root terms that have some constant squared plus x squared. And then we might have another square root term that has x squared minus a constant squared. And for each of those, there's a trig substitution that we would select. So if we see something like this in our integral, the trig substitution that we would use is right here.
All right, so the first example we're going to look at is for that first square root term where we have a constant squared. minus x squared. And if that's the square root term that we see, the choice for x that we're going to make for our substitution is we're going to let x equal a times that constant times sine theta.
So we're letting x equal this trig function. And once we have x kind of selected here, Then we're going to calculate the differential dx. So in this case, we differentiate a times sine theta, and we get a times cosine theta, and then we multiply by d theta.
So we see how our differential is going to change. And where this technique is going to shine is we're able to simplify this square root term so that we don't have a square root term. And this part here... is going to involve our trig identities. So for this first one, let's actually show how this square root term simplifies to just a times cosine theta if this is our choice for x.
All right, so let's start. So we're going to prove, we're going to prove the first one, and you know if you want to have some fun you can try all three of them. But we're going to do the first one.
They kind of feel the same. All right, so let's suppose that we're starting with the square root a squared. Again, just a constant minus x squared.
And let's suppose that we're going to use this for our choice for x. So let's fill that in. So wherever we see an x, we're going to replace that with a times sine theta. All right, so. We're just doing our trig substitution there.
All right. Well, we've got a constant times sine, all being squared. So if we distribute the squared, this becomes the square root of a squared minus a squared times sine squared.
And now underneath the square root, both of those terms have an a squared, so we can factor out an a squared term. And what we have left is 1 minus sine squared. Okay, so far so good. And now, kind of where the magic happens, we have a trig identity that allows us to simplify that.
And, whoop, made my square root too big. Let's shave off a little bit. We know that 1 minus sine squared becomes cosine squared. So a little bit of magic there.
Trig identity magic. All right, so there's our trig identity. And now this is a perfect square underneath the square root.
So we can simplify. And it just becomes a times cosine theta. All right, and we proved it. We got it there.
All right, so... In practice, we can, you don't necessarily have to show this every time. Good to kind of go through it once.
And for the other two, if we're going to let, if we have the square root term a squared plus x squared, well, we're going to let x equal a times tangent theta. And the trig identity for this one that you would use is that secant squared is equal to 1 plus tangent squared. So the plus sign.
kind of matches up with the tangent there. And you could run through it, very similar, and you'd use that trig identity, and you would see that this square root term gets converted into a times secant theta. And same thing for if we have a square root x squared minus a squared, well you're going to let x equal a times secant theta, and you would use that same trig identity to convert the square root term, and it would become a times tangent theta.
So a way to kind of remember, once you make the choice for x, the square root term is just going to kind of switch over to the opposite trig function relative to those trig identities. So we know that we have sine squared plus cosine squared is equal to one, kind of the main trig identity here. So you want to think of, based on that trig identity, sine and cosine are kind of opposite functions. So if we're going to let x equal a times sine theta...
Well, then the square root term is just going to become the opposite one. It's going to become cosine. If we're doing the trig substitution where we're letting x equal a times tangent theta, again, let's remember we have that main trig identity that says secant squared theta is equal to 1 plus tangent squared theta. So based on this trig identity, you can kind of think of, in some way, secant and tan as opposite trig functions based on that trig identity. So if we're going to let x equal tangent, well, then the square root term becomes secant.
And if we let x equal secant, well, then that square root term when we simplify is going to become tangent. That way, kind of a quick way, you don't necessarily have to run through all of this business here. All right, so enough with the description here.
Let's actually try an example. And we'll probably talk about the process, the detailed process in class. a bit and, you know, some of this other stuff. We're going to look at example one.
Give this one a try. So we've got the integral of 1 over square root x squared minus 9. So when we look at this, we want to start by identifying the square root term. So the square root term here is x squared minus a constant squared. Alright, so we want to decide what is x. And then once we decide on x, we want to calculate how our differential is going to change.
So we want to calculate dx. And then we want to calculate and see how that square root term is going to change. So we want to really calculate these three things.
And how we choose x is based on the square root term that we see. Alright, so we have x squared minus... a constant. So let's go back up.
And which one of those square root terms do we have? Well, this is a constant minus x squared, so no. This one has a plus, so no. x squared minus a constant squared. So we're going to let x equal secant here.
And we just got to be a little careful. How do we choose that constant? Well, whatever constant that's inside our square root, we just do the square root. And that's what's going to go in front.
So let's kind of take a look at that. And for that first example we're doing, we have this x squared minus 9. So how do we choose what a is? How do we decide what a is?
Well, you see what constant you have. So in our case, it's a 9. And the a value is just the positive square root of this number. So a is going to be the positive square root of nine.
So three. All right, so once you look at your whole big square root term, you can figure out the a value by just looking at the constant and doing the square root of it. All right, so that means for our example here, we're going to let x equal 3. And again, why are we doing 3? Well, because the a value here...
is the square root of 9. And based on our table, we're going to let x equal secant theta. So 3 times secant theta. All right, so now once we've made this decision, the rest of it just kind of comes along.
So from here, we calculate the differential. So the derivative of 3 times secant is going to be 3 times secant tan. with our new differential d theta.
And our square root term just becomes the opposite trig function. So we have that trig identity again. secant squared is equal to 1 plus tangent squared. So secant and tan are kind of like opposite trig functions in a way, based on that identity. So our square root term is going to become the opposite.
So it'll be 3 times tangent theta. All right, so this is really the first part of this. Kind of get all the pieces calculated.
And once we've done that, then we're going to make our substitution. So it's called the trig substitution. We've got the trig part.
Now we're going to do the substitution part. So when we plug in, we're going to, well, if we did see any X's floating around, we would fill in with this. I don't really see any X's. Um, we do have our differential dx, so we'll fill that in at the end.
And then we have the square root term, which we'll be able to replace with our three times tangent theta. And again, sometimes you might see some loose x's floating around, so you'd fill those in as well. But here we don't have any.
So if we... Kind of fill in the pieces. What are we going to have? Well, instead of that square root, we're going to end up with just 3 times tangent theta.
And instead of our differential dx, we're going to have 3 times secant theta, tan theta, d theta. And now we have a trig integral with no square roots. So now we can use our trig integral strategies.
And before we start trying to integrate this thing, usually there's some simplifying that you can do. And this is kind of the fun part. A lot of canceling normally happens. So the 3 cancels off with the 3 in the denominator.
And then the tangent cancels off with the tangent in the denominator. So this just becomes the integral of secant theta d theta. All right. So now that's just a nice plain.
trig integral. And integral of secant is something that, well, we would kind of know what it is. Maybe I wouldn't say, I don't want to go so far as you'd have this one memorized, but be aware that this is something that you would just kind of have quickly to reference.
So if there were some other powers of secant and tan here, you would use your our strategies for the powers of secant and tan. But for plain old just secant, probably, maybe you have this one memorized, but just be aware what the formula is. So this is one that we don't normally kind of go through the full integration. So the integral of just secant is the natural log of absolute value secant theta plus tangent theta.
And then we got our plus c. So integral of just plain secant. Just be aware that this is something you would just do.
You would just have the formula for. All right, so we're almost there. We've got the antiderivative. Now we just have to convert back to x. And we can kind of use some of our pieces here to help us with this.
So we want to know, we want to get rid of the secant theta here. We want to get back to x. And then we also want to get rid of this tangent theta. We want to get back to x there as well. So we want to get rid of all these trig terms.
So look at the pieces that you have to work with and see if you can solve any of those. And it turns out you can. So in our first x equation, We have a secant theta here, we just have this extra 3 in front, so divide it.
So if we start with the equation x equals 3 secant theta, we can solve for secant theta, and we get x divided by 3. And we can do the same thing with our square root equation right down here. So we want to know what tangent theta is. Well, we've got tangent theta, but then we've got this extra 3 in front, so... Again, just divide it.
Get it over to the other side. And when we do that, we get tangent theta is the square root term, x squared minus 9 over 3. So then we just fill these values in. We'll put the x over 3 in for secant, and then we'll put the square root over 3 for our tangent. And... That'll do it.
Now, sometimes this conversion process takes a little bit more work, so we'll get some practice with that. Sometimes we've got to use right angles and right triangles and things like this. But you want to start by just looking at the pieces that you have to work with. Sometimes the pieces that you have from the original substitution, they're enough, and you can just substitute back.
All right, so this is going to be the idea. And we're going to, there's a lot of things going on with this. So we're doing substitution.
We're dealing with trig functions. We're dealing with trig integrals. So these are tricky, and they take a little bit of time.
So don't worry if you're not feeling like you're getting it right away. So these take practice. This is probably one of the... most difficult topics that we do in the course. So if you can make your way through these trig substitutions, you'll make it.
All right, so we'll practice more in class. We'll see you then. Bye-bye.