 Welcome back to our lecture series Math 1220 Calculus II for students at Southern Utah University. As usual, I will be your professor today, Dr. Angel Nisselein. This is the first lecture of our lecture videos about what we would call Lecture 12, which actually begins Section 7.3 about Trimmetric Substitution from James Stewart's Calculus textbook. In the past, we've used the method of substitution to simplify integrals so that our anti-derivatives will be more recognizable. Against all intuition whatsoever, the goal of this section will actually be to use substitution to complicate the integral so that the anti-derivative will be more readily recognizable. I like to try to think of, in this situation, the analogy of my shoelace is tight and really tight not. We have my shoelace, it's all tangled up right here. Now the problem is if your idea is to try to flatten the knot right here so you pull tightly and pull tightly right here, then what happens is your knot is going to get tiny, tiny, tiny, like this, right? In terms of area, we're taking far less space with this tight or not, but it's actually less likely to be untied right now. In fact, the more preferable way of untying the knot would have the looser knot, even though it takes up more space, will have a better chance of unraveling it in that situation. So in particular, what I mean in terms of this technique, this forthcoming technique of Trigonometric Substitution, let's suppose we have some 1 to 1 function g, right? And then we can make the following change of variables. What if we make the change of variables that x equals g of t, right? And then dx equals g prime of t dt. We've seen this type of substitution all the time, and we might use it in this type of situation right here. If x equals g of t, then the integral f of x dx would then transform into the integral of f of g of t times g prime of t dt. So we can do that. And so in the right context, the right integral may actually be more favorable than in the left. And some examples of such are going to be some that I want to show you right down here, right? So consider some examples right here. With the first one, right, if you had the integral of x times the square root of a squared minus x squared dx, where a is just any constant right here, in this situation, the standard u substitution would be great here. You take u to be a squared minus x squared, then du would equal a negative 2x dx, for which case we then need a negative 2 right here. So we'll divide by negative 2 to compensate. Using this technique of u substitution, you end up with negative one half the integral of u to the one half power du, for which case then you can start doing, you can keep on finishing this thing, we're not going to do it, but it would be very much a cinch. U substitution works really great by simplifying things in that regard. But in this, the second example here, if you were taking the integral of the square root of a squared minus x squared dx here, this time you don't have, you don't have an x to do a u substitution on the outside. What we could actually do instead is the following, again, this is going to seem quite bizarre. But what we're going to do is we're going to set theta equal to be arc tangent, sorry arc sine of x over a, like so. Again, it's like, what did you just say? Where in the world did that come from? Now, to explain why this would ever be the substitution one would consider, notice that if you take sine of both sides, this becomes sine theta, which is equal to x over a. And then if you times both sides by a, you actually get that x equals a sine theta. And this last statement is actually the intuitive, is the intuition behind this one. But just, just give me a little bit of faith in this situation here. x would equal a sine theta, in which case then dx would then equal a cosine theta d theta. And so if you make those substitutions into this integral right here, well, let's see what happens, right? So you take the integral, you're going to take the square root of a squared. And then the x becomes an a sine x theta squared, all right. And then the dx becomes an a cosine theta d theta. Again, just have some hope on what's going on here. Trying to simplify the square root there, you're going to end up with an a squared times sine squared. This all sits inside the square root. You have an a cosine theta d theta. Remember, we're still taking the integral of this thing. We can factor out this a into the front. So you get a times the integral. And then inside the square root, there's a factor of a squared that leaves behind one minus sine squared inside the square root. And then there's this cosine theta d theta. I can take the square root of a squared, that is this one right here. That's also going to be an a. So you get a times the integral of a. And we're left with the square root of one minus sine squared times cosine theta d theta. In which case, again, it's like hope beyond hope. How is this making life better for us? And since we do have these trigonometric functions, we can use the right trigonometric identity. Remember that sine squared plus cosine squared equals one ergo one minus sine squared equals cosine squared for which that's a perfect square. The square root of cosine squared is equal to cosine. This thing would simplify to be an a squared times the integral of cosine times cosine d theta. And then that would simplify to be an a squared times the integral of cosine squared, for which I'm going to stop the calculation right there because I think I've gotten sufficiently far to prove the point is it took a little bit of work. Absolutely. The right trigonometric identity at the right moment, wonderful. But we are the term, the function, the square root of a squared minus x squared into a squared times cosine squared for which cosine squared is a function we can recognize as we're able to calculate its antiderivative. If we use the half angle identity, we would be as good as gold right here. I think we've actually done this exact one before. And if not, we could do it. So we're able to make a substitution that changed the integral into something we, the integral into something we didn't know it's antiderivative into something that we actually have a possibility of doing so. And so what, what I want to summarize in this table here, this is one, this will be on the screen at the end of the video, you're going to want to copy this one down. This right here, what we saw in the last example demonstrates this method of trigonometric substitution. And you can think of trig, trig subs as an analog of translating some ancient language, like perhaps we were to discover are be presented with this, this ancient codex that represents the, the communications of an ancient civilization, right? Unfortunately, we have a dictionary that helps us translate this ancient tongue. And if we translate this into modern language, we can make this sentences more translatable. Now for us as calculus students, we can think of algebra as our native language, right? That's the language that we want to converse in. And trigonometry is sort of like the foreign tongue. Now this, this right here, this ancient codex you see in front of you, like a Rosetta stone, it helps us translate from, from algebraic statements into trigonometric statements. It also helps us translate from trigonometric statements to algebraic statements. So what's going to happen is we're going to run across trigonometry, we're going to run across integrals that involve algebraic expressions of the form, the square root of a squared minus x squared where x is a variable, a is a constant. We're going to run across ones that involve the square root of a squared plus x squared or the square root of x squared minus a squared. Three different things we're going to see here. Now if you see these type of square roots, what we're going to do is we're going to, this tells us what substitution to use. If you see the square root of a squared minus x squared, you're going to make the substitution x equals a sine. If you see the, the square root of a square plus x coordinate integral, I want you to do the substitution x equals a tangent and the last one for secant as well. Now why, why do you, was one want to use this substitution? Like we saw in the example, it comes down to the appropriate trigonometry, or Pythagorean identity. If you substitute sine in for x here, you're going to end up with a one minus sine square, which is a cosine square, which if you take the square root of a perfect square, oh, you'll just get a cosine. It's going to work out really nicely. In this situation, if you replace x with a tangent, then you're going to get, you can utilize the one plus tangent square to get a perfect square for which then the square root cancels out the square. And likewise, if you plug in a secant here for x, sorry this one, you're going to end up with a secant squared minus one at one point, which will become a tangent squared. And the square root of tangent squared, of course, is equal to tangent. So using the Pythagorean identities, we can simplify the square roots in the algebraic expression using trigonometric substitution. And so we'll see in the forth coming videos some more detailed examples that we actually go to the end of the problem as well and show you exactly how to use a trig substitution and when, when does one want to use a trig substitution. So stay tuned for those.