 What makes trigonometric substitutions particularly useful is the following. As a general rule, any quadratic equation, any quadratic expression that we have can be transformed into some trigonometric expression. And what this means is that we can generally take anything that involves some horrible square root functions or reciprocals or whatever. As long as the expressions involve a quadratic, we can transform that quadratic expression into a trigonometric expression and then have some hope of integrating it. The key here is you want to be able to transform that quadratic expression into a perfect square and some leftovers. And this goes back to the general problem of converting, of completing the square. So in general, our perfect square is going to correspond to a trigonometric function, so leftovers will be some constant that we'll have to deal with. So let's take a look at this integral, square root six minus four x minus x squared. Again, we go through the easy things first. Do we recognize this as the derivative of some function? Uh, no. Can we do a u substitution that will make this easy to integrate? Uh, no. Do we see an integration by parts problem here that will be easy to work with? Uh, no. So we've exhausted the easy thing, so let's try the next thing. We'll try a trigonometric substitution. And here we need to convert this into a square plus other stuff. And so that means we're going to need to complete the square on the radicand. And that's going to give us something like this. I'll split off that constant term and then whatever's left over. And let's see, to make this a perfect square I need to add four. And because I'm adding four inside of the subtraction, I add four again. And this expression on the left is exactly the same as this expression on the right. Except the expression on the right, I can write as a perfect square x plus two squared and then some leftovers. And I'll rewrite that because ultimately I do want to see how this fits into our fundamental triangle. I'd like to write this as a bunch of things squared. So this expression here is the square root squared. Ten is square root of ten squared and this is already a square. And rearranging this a little bit so I'm adding two squares to get a square gives me something like this. And the reason we did this is that now this looks like a Pythagorean theorem relationship. And so what this suggests is that I have a right triangle where my sides are going to be square root six minus four x minus x squared x plus two. And my hypotenuse is going to be square root of ten. So there's one side, there's the other side, and there's my hypotenuse. Now you should convince yourself that I could actually have a right triangle with sides this, this, and hypotenuse that. Now the triangle that we've produced again we want to do some sort of trigonometric substitution. And what it suggests is we can either let this thing be some trigonometric function or this thing be some trigonometric function. I'm inclined to let this thing be some trigonometric function because that seems to be easier to work with. That's messy, that one's much nicer. And the trigonometric function that we might want to use here is, well, I could use sign because that has me having this over square root of ten. If I use tangent that's this over a messy thing and that could be a lot more difficult. Sign, on the other hand, is relatively square root of ten is just a number. In fact, I'll multiply across to get rid of it. And there's my substitution and I'll find my differential, root ten cosine theta d theta. Now again we can use our triangle to allow us to figure out what to do with this thing. And the thing to notice here is that this thing is going to be part of our expression for cosine. Cosine theta is adjacent over hypotenuse. That gives us this thing. And so I now know what to do with this square root 6x 6 by 4x minus x squared. It is the same as root 10 cosine theta. So now I have the integrand root 10 cosine theta, the differential root 10 cosine theta again. And so my new integral becomes 10 cosine squared theta. And at that point I could use my trigonometric identity that allows me to expand cosine squared theta one half plus one half cosine two theta. And I can do a little bit of algebra, a little bit of calculus, do the integration. And again because my starting variable is x, I want to make sure that theta gets expressed in terms of x. So let's find theta and what I might do is from here it's on the one hand the easiest way of finding theta. I can look at this. It's the arc tangent, the inverse tangent opposite over adjacent. I could have solved it from here if I wanted to, but I can also solve it from here by appealing directly to our fundamental triangle. So I know what theta is. Sine of two theta is a little bit more complicated, so I'll use my trigonometric identity. Sine of two theta is two sine cosine. And if only I knew what sine of theta was. Oh wait, sine of theta is opposite over hypotenuse. And cosine is adjacent over hypotenuse. So I know what sine of theta is. It's going to be two. There's my sine of theta opposite over hypotenuse. There's my cosine theta adjacent over hypotenuse. And there's my sine of two theta. And so now I can put everything back into place. And I have one half theta, this thing, one quarter sine of two theta, this thing. And don't forget my plus constant of integration, which I unfortunately fell off the screen here.