 It will be helpful to rewrite a rational function using a trigonometric substitution. The rational function can then be simplified using a trigonometric identity. For example, use the substitution x equals 2 sine theta to simplify 1 over square root 4 minus x squared. And we'll assume theta is a first quadrant angle. So remember, equals means replaceable, so we'll replace x with 2 sine theta and simplify. Now both of our terms have a factor of 4, so let's remove that common factor. And this expression 1 minus sine squared theta, well that corresponds to the identity sine squared theta plus cosine squared theta equals 1. And so that means we can simplify it. Now ordinarily when we take a square root we have to worry about something being positive or negative. But because we're assuming theta is a first quadrant angle, the square root of cosine squared is just going to be cosine. And so this expression simplifies. And we won't get into the details now, that's for a later course. This expression 1 divided by 2 cosine theta is much easier to work with than 1 divided by square root 4 minus x squared. We don't have to restrict ourselves to simple substitutions. We can use the substitution x plus 2 equals secant theta to simplify this rational expression. Now we might begin by noting that if x plus 2 is secant theta, remember all of our trigonometric identities are based on the squares of the trigonometric functions. So let's square both sides. And what we get is almost our denominator. So let's change it, subtract 1. And now we have our denominator equals secant squared theta minus 1. But wait, secant squared shows up in our trigonometric identity. And secant squared theta minus 1 is tangent squared. So our denominator will be tangent squared theta. And so using our substitution allows us to rewrite our rational expression as 1 divided by tangent squared theta. Now in the two examples previously you were given a particular substitution. So the question is how can we decide which trigonometric substitution to make? One suggestion is going to come from the domain. So we typically will make a trigonometric substitution in one of three cases. We might have the square root of 1 minus x squared, or the square root of 1 plus x squared, or the square root of x squared minus 1. Now since our radicans have to be non-negative, these different expressions have different limitations on the value of x. In the first case, x has to be between negative 1 and positive 1. In the second case, it doesn't actually matter what x is, x can be anything from minus infinity to positive infinity. And in our third case, x has to be greater than or equal to 1, or less than or equal to negative 1. And here's the big hint. Our trigonometric functions have these same restrictions of the range. So comparing these restrictions to our trigonometric functions, we know that sine of theta must be between negative 1 and 1, tangent of theta could be anything, and secant theta must be greater than or equal to 1, or secant theta must be less than or equal to negative 1. Now we could say similar things about the other trigonometric functions, cosine, cotangent, and cosecant, but no one uses them. Now we do have to introduce one more thing. Since we want to be able to use our Pythagorean identities, we note that the two fundamental identities, sine squared plus cosine squared equals 1, and tangent squared plus 1 equals secant squared, if we multiply through by a squared, this suggests our substitution should be of the form a sine theta, a tangent theta, or a secant theta. Let's see how that works. So let's try to find an appropriate substitution for x divided by square root 9 minus x squared. And we may assume that all trigonometric functions are restricted to non-negative values. So if we look at our radicand, square root 9 minus x squared, we note that x squared has to be less than or equal to 9. And we won't worry about the fact that it's in the denominator. We'll focus on the radicand itself. That means that x has to be between negative 3 and 3. Now we know that sine, and to be sure, cosine are limited in this way. And since sine has to be between negative 1 and 1, then 3 sine of theta is between negative 3 and 3, which is exactly what we want x to do. And so this suggests the substitution x equal to 3 sine theta. So we'll replace x with 3 sine theta. Now the radicand 9 minus 9 sine squared theta will factor a 9 out of it. And remember our Pythagorean identity. 1 minus sine squared theta is cosine squared theta. We're assuming our trigonometric functions are restricted to non-negative values. So when we take the square root, we can just write this as 3 cosine theta. And we'll go one step farther. The 3 can be removed as a common factor. And sine theta divided by cosine theta, well, that's just tangent theta. And there's our final simplification.