 Let's continue our discussion with trigonometric substitutions in the following way. In general, the most difficult part of integration by trigonometric substitution is going to be the very last step, which is putting everything back into place. So we'll have made our substitution x equals trigonometric function of theta. We'll have done a whole bunch of things, so then we'll end up with a final answer that is expressed in terms of theta, but now we have to go back and put everything in terms of x. And for that, it's helpful to remember the fundamental triangle. Anytime we have a trigonometric function, there's some right triangle that we can draw, where we have our angle theta, we have our sides, a, b, and our hypotenuse, c. And in this case, we have sine is the opposite side, a, over the hypotenuse. We have cosine is adjacent side, b over hypotenuse. Tangent is opposite over adjacent, a over b. And then finally, c-count is reciprocal of cosine, that's c over b. And these are going to be the four important trigonometric relationships that we'll want to have at our disposal. And if we combine these with our Pythagorean relationship for any right triangle, a squared plus b squared equals c squared, then we'll have a good and easy way of substituting for theta in the final step of the trigonometric substitution. For example, consider the anti-derivative one over square root one plus x squared. Now the problematic portion of this integral is this radical right here, square root one plus x squared. And so we might reason as follows. If we're going to use a trigonometric substitution, then this radical is going to correspond to one side of a right triangle. So let's play around with that. A little bit of analysis is going to go a long way. If we make this side length have length s, then one side of the right triangle looks like this. And let's see, if I square both sides to get rid of the radical, then I end up with s squared equals one plus x squared. And this looks an awful lot like the Pythagorean theorem. In particular, it looks like we're talking about a right triangle with sides one, side x, and third side is going to be hypotenuse s, but rather than writing s as a new variable, I'll substitute in what that is, square root one plus x squared. And so I can record it that way, and this s here doesn't ever enter into our picture. And now I have this nice right triangle, and here's the important thing. Part of that right triangle is part of the integrand. Now in general, what we're going to try and do is we're going to try and make the substitution x equals some trigonometric function of theta, and here's my x, and the thing that suggests itself, the simplest trigonometric function I can use here is tangent theta, which is, here's theta, tangent is opposite over adjacent. So tangent theta would be x over one, and so I make the substitution x equals tangent theta, dx secant squared theta d theta. The other thing that's helpful about drawing the triangle first is that we can use it to simplify the integrand. So let's consider this expression one over square root one plus x squared. Well, that's actually going to be cosine of theta because that is adjacent over hypotenuse. So my integrand, I could go through the procedure of let x equals tangent theta, substitute in applied trigonometric identity to evaluate simplify, but I can look at my picture, and I can see that this one over square root one plus x squared that's adjacent over hypotenuse, and I recognize that as cosine. So I can use these substitutions, integrand as cosine, dx secant squared theta d theta, drop that in, do a little bit of algebra. This is antiderivative secant. We do have to remember certain antiderivatives and secant is log secant plus tangent, absolute value thereof. Well, that's our constant. Well, again, we have to put everything back where we found it, so we need to express everything in terms of x. And again, our fundamental triangle helps us do that because what is secant theta? Well, secant is hypotenuse over adjacent side. So secant root one plus x squared over one, tangent opposite over adjacent is going to be x, and so I can substitute those back in, and there's my final answer.