 So by now, we might be getting used to this notion of a trigonometric substitution. We feel like we're getting pretty confident, pretty good, life is wonderful. And so now I have to show us that. Turns out we haven't seen the worst of it yet. So let's consider the integral where we're gonna integrate from zero to three, root three over two. That might seem like a weird choice, but it'll be very clear why. That's actually a pretty, that's a generous choice for upper bound here. So we wanna find the definite integral, the area under the curve, y equals x cubed over four x squared plus nine, raised to the three halves power, that's in the denominator there. Now to get started with, what does this have anything to do with trigonometric substitutions? Well, let me remind you that if you take four x squared plus nine to the three halves, that's the same thing as the square root of four x squared plus nine, raised to the third power. And so we do have this square root of a quadratic polynomial inside this integral. That's something that indicates to us that we would wanna use a trigonometric substitution. But what's a little bit different this time is that notice we don't have an x squared, we have a four x squared. You still do have a difference, or I should just excuse me, a sum of squares right here inside of the square root. What do you do with that coefficient right there? Well, the way you should be thinking of this is instead think of this square root instead of the form u squared plus three squared like so. We're in this situation, u equals two x. And so if you saw something like this, the square root of u squared plus nine, that would tell you you wanna do the substitution that u equals three tangent theta. Because it's a sum of squares, we wanna do a tangent substitution. And then you run from there. But u is not the variable we care about, we care about x. And so what I'm gonna tell you is you're gonna actually start off with the equation two x equals three tangent theta, all right? And so this is sort of like the gold standard how you get this to go. And so looking at this original square root, we have a square, we have a square. One is a function, one is a variable. Here's the function, here's the variable. And so whatever happens, take the square root of the function side that goes on the left. Take the square root of the constant that goes on the right. And then plug in the appropriate trig function, sine tangent or secant, based upon whether this is a plus or minus and the order does matter there. So take this as your format two x equals three tangent. Everything in this problem will be based upon this substitution right here. Now if you wanna know what x is, you're gonna solve for x and get three halves tangent theta. Now if you need to know what theta is, you would actually solve and get tangent theta equals two thirds x. And this will be helpful for us like if we're trying to write our right triangle. So we got this triangle right here. And this is a triangle that's gonna be with respect to the angle theta. Well, what this tells me right here is two x over three, opposite over adjacent is two x over three. And then the hypotenuse, if you go through the Pythagorean equation, will be the square root of four x square plus nine, the exact same square root that we had before. So you'll notice that in one situation, we have a three halves and another situation, we have a two thirds. Now, if you just try to memorize the formulas, it's not gonna be much different than alphabet soup. You'll see a bunch of symbols floating around and you're gonna make mistakes. It's like, okay, does the, there's like a coefficient in front of the x in front of the constant right there, where does it go? Don't worry about where they're being placed here or here, instead focus on this identity. You have the function square root and then you have the constant square root that are separated. The tangent's gonna go with the square root of the constant there. And then you can manipulate this equation as is appropriate. So for example, we wanna know what tangent theta is that we can construct our triangle. But we also wanna know, and we should also mention, well, I'll get back to that in a second. We also know that from this right here, x equals three halves tangent theta. If we wanna find dx, we take the derivative of this, we get three halves secant squared theta d theta. We're gonna need that for the dx up right up here. But also be aware that you could have actually taken the derivative of this equation implicitly. You get that two dx is equal to three secant squared theta d theta and you can divide both sides by two. You have that liberty of doing so, excuse me. And so the last thing we have to deal with is what is the square root, right? The square root of four x squared plus nine. What does it turn out to be? Well, if you trust in the triangle over here, connect the square root side with the constant side, you're gonna make some type of cosine argument here or a secant actually works out really much better here. Secant is gonna equal the square root over three times the both sides by three. You end up with the square root is equal to three secant theta. You can drive that from the triangle. You can also use the identity approach if you want to. Or if you've been paying attention in these lectures that we've been doing, I told you that whatever we set to the right-hand side of this original one, the two x, that three tangent will switch to its buddy which is three secant. We could have done that without any calculation whatsoever but calculations are always good because we can double check that our intuition matches up. All right, so let's rewrite the integral using the information we've now discovered. All right, so we have an x cubed on top. I'm gonna drive a big fraction bar here. We have an x cubed on top. x is the same thing as three halves tangent. So put that in for x and we're gonna cube it. The dx becomes a three halves secant squared theta d theta. And then on the bottom, we have the square root of four x squared plus nine and that's all cubed. So the square root of four x squared plus nine is just three secant. Three secant theta and then we're cubing it. So you'll notice that some of them have fraction coefficients and some did not. Why is that? Again, if you're trying to memorize nonsense, you're never gonna remember it but if you follow the arguments we used here, the geometric intuition, you can see exactly why some have a denominator of one half and some do not. We also do wanna change the bounds, right? As we switch from x to theta, our bounds remember we have zero on the bottom and three root three over two on the top. Well, using our identity that we had before, so we can't see it on the screen but I'll just sketch it out one more time, we had that two x equals three tangent theta, right? When x equals zero, you're gonna get zero equals three tangent theta, divide both sides by three, you'll just get tangent equals zero, which that happens at zero. So the angle correspond also to be zero right there. But then when x equals three root three over two, you're gonna get the left hand side as three root three, which is equal to three tangent theta, divide both sides by three, you get tangent equals the square root of three, that actually is one of the special angles we know as pi thirds. So like I said, that angle was chosen to be very generous for us. We get theta equals zero and theta equals pi thirds. Now as this is a definite integral and I changed the bounds, I don't have any need to ever change back to x at the very end. So we're gonna finish this calculation using theta, which has the benefits, we don't have to use the triangle to translate back to x, I really would recommend that when you do these trig substitutions. Let's try to simplify things. We have two secants here and we have three on the bottom, so I'll leave one behind. Let's see, in terms of threes, we have a tangent cube, which has a coefficient of three halves cubed. These threes will cancel with all of those threes. And so what do we then have left over? We integrate from zero to pi thirds. We have a one half that's cubed here plus or multiplied by another one half. That should give me a one sixteenth for the coefficient of the denominator. We have just a three that's left over in the numerator. We have a tangent cubed in the numerator and we have a secant in the denominator, the theta, scanning over this object that does, yeah, that's right there. So what are we gonna do with the tangent cubed in the secant right here? Secant and tangents are good friends with each other. We might be looking at some type of U-substitution or something. What I actually wanna try to do is I'm gonna replace tangent squared with a secant squared theta minus one. So you could explore how to proceed with this integral at this moment but this is the substitution I wanna use. Two of the tangents we're gonna switch into secant squared minus one. If we do that, I'm gonna pull the three sixteenths out front. If we do that, we're gonna have a secant squared minus one times that by tangent theta d theta. This all sits above secant, of course. We're gonna distribute the tangent through and then cancel with the secant when appropriate, right? What we end up with is something like the following. For the first one, we distribute the tangent, there's a secant squared on top, there's a secant on the bottom. That'll simplify to be a secant theta tangent theta. We really like that one because that is the derivative of secant. And then with the other one, we're gonna end up with a tangent theta over secant which remembering that secant is sine over cosine and secant is just a cosine. This ratio will simplify just to be a sine. And remember, it's our goal to find antiderivatives and we know the antiderivative of negative sine. That's a cosine d theta. And remember our bounds zero to pi thirds. So using the right trig identity at the right moment, the Pythagorean identity saved us right here, we can then find the antiderivative three sixteenths. The antiderivative secant tangent is a secant like we mentioned earlier. Antiderivative of negative sine is a positive cosine. And we evaluate from zero to pi thirds like so. And so plug these things in here, make sure the three sixteenths is distributed on to everything. So we're gonna get secant of pi thirds. We get cosine of pi thirds. And you might not know what secant pi thirds is off the top of your head and use a calculator I guess, but you could also intuit what it is based upon we know cosine. We'll come back to that one. You get a minus secant of zero minus cosine of zero. Now be cautious. Cosine of zero is not zero nor a secant of zero. Instead what happens is cosine of zero is the number one. Secant is its reciprocal. So this is also gonna be a minus one. Cosine of pi thirds, that turns out to be a one half. That one you might have memorized from your unit circle diagrams. And then secant which is the reciprocal will be a two. So we get two plus one half minus one minus one. The minus ones cancel off with the two. So we have three sixteenths times one half and gives us the final area which is three over 32. So what I want you to get out of this video is that even if we have something of the form u squared plus a squared, we can still do a trig substitution. It doesn't matter how complicated u is. In this case, u is two x but it could be much more complicated. And the idea of u substitution would still work just as, sorry, the trig substitution would work just as great in this situation as it would on the previous examples we saw where u was just equal to the variable x.