 Let's take a look at the integral of dx over the square root of x minus a, the square root of x squared minus a squared here, where a is just some unspecified positive number. The fact that it's unspecified, we still are going to treat it like a constant, but we don't know what that constant is. Now, when we look right here, we look at the square root of x squared minus a squared. We have a quadratic polynomial inside of a square root. That tells me that we want to do a trigonometric substitution. Which substitution do we use? Well, if we refer to our code x, we have a function squared minus a constant squared. That tells me I want to say x equal to a secant theta. If I had a squared minus x squared, I want to do a sine for x. Or if I had x squared plus a squared, I'd want to do a tangent. Those are basically your options. The only other options, we could try to use hyperbolic functions, which behave very similar to trig functions. I'm not going to go down that path because I think the trig functions will be sufficient for our purposes. Taking the derivative of x here, we get a secant theta, tangent theta, d theta. As we've seen before, if we were to draw our triangle here, the angle associated to this triangle, we're going to call it theta. Playing around with this friend right here, we see that secant theta equals x over a, which is to say cosine theta equals a over x. So adjacent over hypotenuse, and then the other side's going to be x squared minus, the square root of x squared minus a squared, just from the Pythagorean identity. We see that the square root of x squared minus a squared is equal to the best friend of secant, which is tangent, a tangent. You can use the identity one plus tangent squared equals secant squared to derive this, or you can get it from the triangle, or I'm really just coming from just observation that we keep on seeing this over and over again. That whoever we choose as the substitution here, its best friend will be equal to the square root right there. So then rewriting the integral in terms of this substitution, the dx becomes a secant theta, tangent theta, d theta. The denominator just becomes an a tangent theta, d theta. You can make some cancellation, the a's cancel, the tangent cancels, and so this just becomes, give myself a little bit more space here. Oh, what happened? Come back here on the right slide. We get the integral of secant theta, d theta. That's all that we have to do. Now, we did do the integral of secant previously. I'm just going to cite the reference. It's a little complicated, but the antiderivative of secant is the natural log of the absolute value of secant theta plus tangent theta plus a constant. And so then we want to translate it back. We need to know secant and we need to know tangent. Now secant, we talked about before, right? Secant is going to be, well, x equals a secant, so therefore secant is x over a. We're going to make that substitution, natural log of the absolute value of x over a. What about tangent? Well, we saw that above as well. You can use the triangle to help you translate back, but some of this information we already know, right? Remember that a tangent is equal to square root, so therefore tangent is the square root over a. And so then writing that down, we get the square root of x squared minus a squared over a plus a constant, like so. Now, this is a perfectly good answer, but it turns out we can simplify it a little bit, and I want to demonstrate that to us because otherwise we might not have noticed it otherwise. So notice that both of these fractions have a common denominator of a, so let's add them together. Doing so gives us the natural log of the absolute value of x plus the square root of x squared minus a squared. This all sits above a. Now this is sitting inside of a natural log and a fraction inside of a natural log can be turned into a difference of logarithms. So we end up with the natural log of the absolute value of x plus our square root of x squared minus a squared. We then get minus the natural log of the absolute value of a, which admittedly as a is assumed to be positive, the absolute value is redundant here, and then we get plus c. Now in this situation, I want us to remember our plus c as this large gelatinous cube, right? Just like in the movie Onward, it absorbs everything that comes near it and then any constants, right? So our plus c is gonna devour this plus or this minus natural log of a as that it's a constant. And so we can actually simplify our anti-derivative form as the natural log of the absolute value of x plus the square root of x squared minus a squared plus our gelatinous cube, aka plus c. And so this gives us a simplified anti-derivative that's a lot cleaner than the previous one. I do want you to keep your eye open for these things because as you're comparing your answers with calculators or backs of the books or maybe you have a software like WebAssign or some other computer program out there that's checking your answer for you, be aware that different answers can be equivalent to each other because this plus c does kind of compensate for a constant that could be added or subtracted from this thing right here. So keep that in mind as you're working through these anti-derivatives.