 One final topic on trigonometric substitutions involves the situation where we're working with a definite integral. Here's where we make an important distinction. If I'm working with an indefinite integral, then I have no choice. I have to state the final answer in terms of the original variables. So if I'm working indefinite integral one over one plus x squared dx, and they write the answer theta plus c, this is a wrong answer because there is a new variable here theta that was not one of the variables that we started with. And so this answer is wrong. It is incorrect. So instead what I have to do is I have to take my value of theta and replace it with some appropriate expression of x. So I might replace it with inverse sine of x, and here is my answer. Well, it's still wrong. Actually, this one doesn't work that way. But on the other hand, it's still a better answer because my answer here is at least in terms of our original variable x, whereas this answer here is not. Well, if we're working with a definite integral, we actually have a choice. We can restore the original variables and use the original limits of integration, but we don't have to. We can actually change all of our variables to whatever the new system of variables we used. So, for example, let's take a look at this integral, 0 to 1 half, square root 1 minus x squared dx. And so one of the things I can do is I'll find the anti-derivative root 1 minus x squared, and that's going to give me a trigonometric substitution. And at the end of the day, after all the dust settles, it'll look something like this. And because I've restored my original variables, I'm allowed to use the original limits of integration. And so substituting in 1 half gives me this mess. Substituting in 0 gives me this mess. And after all the dust settles, I get as my final answer pi over 12 plus 1 quarter square root of 3 fourths. On the other hand, as long as we remember that the limits of integration correspond to the values of the differential variable, then I can work with a new variable. So here these limits 0 and 1 half, they really say x equals 0, x equals 1 half. And if I change to a new variable theta, as long as I change the limits of integration appropriately, I don't have to change back. So again, if I were to use my standard approach to this type of integral, this trigonometric substitution, this I make my substitution equals theta over 2 1 quarter sine of 2 theta. And the limits are in terms of x, so I need to find the corresponding values of theta. Now, in this particular case, the substitution we would have made is going to be x equal to sine theta. And so I know that x equals 0, so 0 is sine theta. And that tells me that theta must be 0. The upper limit of integration, x equals 1 half. Well, again, sine theta equals 1 half. And that's going to correspond to a theta value of pi over 6. And what's important here is that I can use these limits, theta going from 0 to theta going to pi over 6, instead of the old limits, x equals 0 to x equals 1 half. So I can evaluate this integral as normally at pi over 6. I substitute pi over 6 into this. I get this mess. I substitute theta equals 0 to this. And I get this. And after all the that settles, I end up with the answer pi over 12 plus 1 quarter square root of 3 over 2.