 Welcome back to our lecture series Math 1220, Calculus II for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misselvein. We are starting lecture 13 today, which will be a continuation of our lecture we started in lecture 12 about trigonometric substitution, which will be based upon examples and content we can find in section 7.3 of James Stewart's Calculus textbook. Now, in the previous lecture, we introduced this ancient codex, this Rosetta stone that helps us translate from our modern language of, and this is of course just an analogy here, our language of algebra to the ancient cryptic language of trigonometry for many of us. And so what we've seen so far is we've seen examples where we've done a sine substitution and a tangent substitution. If you want to see some examples of those, click the links that you can see in front of you. But now I want to consider what happens if we do a secant substitution. So coming back to this right here, when we see this integral, the integral of the square root of x squared minus 4 over x dx, the fact that we have a square root with a quadratic polynomial inside of it indicates to us that we want to do some type of trigonometric substitution. For which case the secant substitution is what we want to do. Here, a equals 2, and so we have the square root of x squared minus 4. And so we're going to want to do the substitution that x equals 2 secant theta. Now, the reason we want to do this is because we have this difference of squares, if we utilize the Pythagorean identity, secant squared theta minus 1, this is equal to tangent squared, we can turn a difference of squares into a perfect square for which case the square root will then dissolve itself. Now, one has to be cautious here because the secant substitution is very similar to the sine substitution. You will notice with the sine substitution, you have a constant minus a variable. But with the secant substitution, you're going to do a variable minus a constant. The order of the difference makes a difference. And these two things are not the same difference. I know those are horrible puns there, but you're going to have to deal with me. You don't have any other choice right now. So taking the square root of, given that we have this square root of x squared minus 4, we want to do a secant substitution. And this square root is going to turn into a 2 tangent theta from this identity right here. All right, let's go back to the problem here. So getting rid of this loop-de-loop for a second, how is the substitution here going to work? Well, based upon what our codex said, we want to take x to equal 2 secant, that's what we agreed upon, 2 secant. Then dx, when you take the derivative, you get 2 secant theta, tangent theta, just using the derivative of secant there. And then we have to take care of this square root, the square root of x squared minus 4. You can use the identity that was on the previous slide, as we've seen before. What I'm instead going to do, as usual, I prefer this triangle approach, right? We have this right triangle, and inside the right triangle, we're going to label the angle theta. It's a right triangle. And because we have the secant ratio, we know that secant of theta is equal to x over 2. Now, if you don't like secants, we can always switch it to cosine by taking the reciprocal. Cosine theta is equal to 2 over x. However you prefer, it doesn't matter to me. Cosine is adjacent to over hypotenuse, excuse me, 2 over x. Just make sure things go in the right spot. Using the Pythagorean identity, we can see that the other side is going to be none other than the square root of x squared minus 4. Honestly, I don't calculate it because I know every time you do this argument, the third side is the square root that started the whole process. But if you have any doubt in that, please work it out yourself there. So if we try to combine together the square root side with the constant side, we can relate that using a tangent, right? Tangent theta is going to equal the square root of x squared minus 4 over 2. And if you clear the denominators, you're going to see that the square root of x squared minus 4 is 2 tangent, like we predicted earlier. All right? So sort of a rule of thumb is the reason we choose the substitutions we do is that whatever you set x to equal, the square root will equal the best friend of that function. So sine and cosines are best friends. Tangent and secant are best friends as well. One could do a cotangent or cosecant, but that just complicates things with negative signs that we don't need. So using this substitution, let's translate the integral into the language of theta. The square root on top is going to become a 2 tangent. Don't forget about the dx because that translates as well. You're going to get a 2 secant theta, tangent theta, d theta. And then this sits above x, which x is 2 secant, like so. There's typically a simplification that happens at some point here. So there's a 2 that cancels. There's a secant that cancels. This thing will simplify to be a 2 integral of tangent squared, theta, d theta. How does one calculate the anti-derivative of tangent squared? I think actually switching it over to secant is probably our best move right here because secant squared is a anti-derivative. It's more obvious. Tangent squared, notice 1 post tangent squared is equal to secant squared. Therefore, tangent squared is secant squared theta minus 1. The same identity we mentioned earlier, d theta. And therefore, the anti-derivative of secant squared is tangent theta. And then the anti-derivative of negative 1 is a negative theta. We have a plus a constant here. And so if we distribute the 2 through, we're going to get 2 tangent theta minus 2 theta plus a constant. We have to translate these theta's back in terms of x. Now, one of them we already know, right? We already know for a fact that 2 tangent is the same thing as the square root of x squared minus 4. We could ascertain that using the triangle right here, but as it's already on the page, I'm just going to use that and save us a little bit of computational effort right there. You're going to get the square root of x squared minus 4, like so. How do we do with the 2 theta? Well, again, come back up to the original identity. We have x equals 2 secant theta, which you can see right here. That translates to secant theta equals x over 2. So we could take theta to be arc secant of x over 2. People don't usually use arc secant so much. You don't see that on your standard scientific calculator. So I'm actually going to employ cosine right here. Cosine theta equals 2 over x. So theta's going to equal arc cosine of 2 over x. That one's a little bit more palatable there. And so we record that down here. We have arc cosine of 2 over x. And don't forget our constant. And this gives us the anti-derivative we were looking for. So notice that doing the secant substitution is not fundamentally different than the tangent substitution or even the sine substitution. Once you get the hang of one type of trigonometric substitution, the other two are very, very comparable. And they work very nicely together. So just give yourself some patience and practice and you'll get this trigonometric substitution stuff, that I'm sure of.