 So let's take a look at a few more trigonometric substitutions. So how about this integral of 1 over x squared square root of x squared minus 4? So since the square root is the problematic part of this integral, we'll want to draw a right triangle where x squared minus 4 occurs naturally. And since x is x squared and 4 is 2 squared, we'll want a triangle with sides of x2 and square root of x squared minus 4. If you don't play, you can't win, so let's try to draw such a triangle. Setting up our Pythagorean relationship, making x and 2 the lengths of the legs of the right triangle does not give us a right triangle where x squared minus 4 appears naturally. So let's switch which sides have which lengths. For example, we might make our hypotenuse have length 2. Find the Pythagorean relationship, do a little algebra, and we have something that's not x squared minus 4. So we'll try again, this time making x the hypotenuse. We'll write our Pythagorean relationship, do a little algebra, and finally we have a triangle where x squared minus 4 appears naturally. We need our angle. Since the problematic part of this integral is the square root of x squared minus 4, we should look to a trigonometric function that incorporates it, and that's either going to be sine or tangent. However, tangent, because it's opposite over adjacent, will give us the easier expression. So we'll start with that. The integrand also includes x squared, which means that somehow we want to incorporate the hypotenuse, and we have our choice of functions, but the easiest is going to be relating the hypotenuse to the adjacent side. So that'll either be cosine or secant, and in this case, secant is easier. So let's do a little bit of algebra, and let's find our differential. So we'll substitute x squared is going to be 4 secant squared theta. Squared of x squared minus 4 is 2 tangent theta. And finally, dx is, we'll do what algebraic simplification we can, and after all the dust settles, we're ready to do some calculus. So let's integrate. And we have one-quarter sine theta plus c. Since we started in x, we end in x. So we'll pull up our picture, and from our picture we see that sine of theta is opposite over hypotenuse, and making that final substitution gives us our answer for the indefinite integral. Sometimes we have to do a little bit more work. Here we want to find the integral of 1 over squared 6 minus 2x minus x squared. And again, we're looking for a triangle where square root 6 minus 2x minus x squared appears naturally. Since our Pythagorean relationship is expressed in terms of squares, it's easiest if our variable terms can be reduced to a perfect square. So let's do that. 6 minus 2x minus x squared is 6 minus 2x plus x squared. We'll complete the square by adding one on the inside of the parentheses. But since we don't want to change the expression, that also means we'll have to add one on the outside of the parentheses. And so our expression is 7 minus x plus 1 squared. And so that means we want a triangle where one side is square root 7, another side is x plus 1, and the third side is square root 6 minus 2x minus x squared. And one possibility is, and we note that sine of theta x plus 1 over square root 7, and why not cosine theta square root 6 minus 2x minus x squared over square root 7. We do want to be able to solve for an expression in x so we can find dx. So let's go ahead and work that sine expression. And we'll use our simpler substitution, square root 7 sine theta equals x plus 1, and so differentiating root 7 cosine theta d theta equals dx. Since we want to do something with this square root 6 minus 2x minus x squared, we might want to solve our cosine relationship, making our substitutions. We can simplify, then integrate, we get theta plus c, and we start it in x, so we should end in x. Theta is going to be arc sine of x plus 1 over root 7.