 So, there's a few more integrals that are useful to know, the integral of tangent x and the integral of secant x. Let's take a look at those two. Tangent isn't too bad. And remember, don't memorize formulas, understand concepts. And the useful idea here to remember is that tangent is the quotient of sine and cosine. So the integral of tangent is the integral of sine x divided by cosine x dx. So we can evaluate this integral. Well, remember, try the easy things first. And the easiest thing to try first is a u-substitution. And we'll try a u-substitution with u equal to cosine x. Then du is negative sine of x, and so our u-substitution gives us, which we can integrate, put things back where we found them. And we can simplify this a little bit. Negative log cosine, well, that's really the log of 1 divided by cosine. And we know that 1 divided by cosine x is really secant x. What if we wanted to evaluate the integral of secant x? Well, suppose a voice in our head told us to multiply and divide by secant plus tangent. Well, since we always listen to the voices in our head, we have. And if we let u equal secant x plus tangent x, we have du equal, and so our u-substitution gives us, which we can integrate, then put everything back where we found it. And this all makes sense, as long as we obey the voices in our head. Let's change the formula, why not eat more chocolate? Now if you didn't have voices in your head to listen to, how could you get there? Well the answer comes from what seems to be a completely unrelated problem, but this illustrates an important idea in mathematical research. It's useful to solve the same problem in different ways. So if I want to evaluate the integral of 1 divided by x squared minus 1, then I can do this in a couple of different ways. And the obvious way is to evaluate it using partial fractions. But the other way we can do this, because this denominator has the form x squared minus 1, we could try using a trigonometric substitution. And because the denominator is an x squared minus 1, the substitution we'd use would be x equals secant theta, which gives us dx equals, and so our integral would become, and we can simplify our integrand, and remember equals means replaceable. And so if x equals secant theta, then this integral, 1 divided by x squared minus 1, is the same as integral of cosecant theta, and so the integral of cosecant theta is equal to one-half log x minus 1 divided by x plus 1 plus c. Of course this is an example of what we should never do, start with a problem in one variable and answer in a different variable. And so we need to do something to simplify the two. And so we can do a little bit of simplification. This one-half log is the log of the square root, and we'll do a little bit of algebra. Let's rationalize the numerator, and that allows us to split up this rational expression. But if x is equal to secant theta, then we can draw our right triangle, where we can make our hypotenuse x and our base 1. That makes the other side square root x squared minus 1, and we find this quotient x divided by square root x squared minus 1, well that's cosecant theta, and this other rational expression is cotangent theta. And so the integral of cosecant theta is log cosecant minus cotangent theta. Of course the problem we were actually looking at was to find the integral of secant theta, and in general remember that a trigonometric function is equal to the co-function of pi-halves minus the same value. And so we'll substitute theta equals pi-halves minus x. So let's say I want to evaluate the integral of cosecant x, so we've just determined the integral of cosecant theta, so we'll let x equals pi-halves minus theta, then dx equals negative d theta, and we substitute we get, and we know this integral. Now since we started with x, we have to put everything back, so since x equals pi-halves minus theta, then theta equals pi-halves minus x, and so cosecant and cotangent become, and so there's our integral. Now you might notice we get a different integral than the one we got when we followed the voice in our head. And what this means is that sometimes we get different answers when we think for ourselves instead of following what the voices tell us to do. However, since the integral of cosecant is usually given as log cosecant plus tangent, I'll probably get my mathematician card ripped up if I don't show how the two answers are actually the same answer. So let's take a look at that. So in this case negative log cosecant minus tangent, well again that negative log is the same as log of 1 divided by, and remember that cosecant is 1 divided by cosine, and tangent is sine divided by cosine. So we can simplify this if we multiply numerator and denominator by cosine, and that gives us, and we can simplify this further if we multiply numerator and denominator by 1 plus sine, and that gives us, and the reason we did that is this 1 minus sine squared is cosine squared. And so simplifying, simplifying some more, simplifying some more, and that gives us our log cosecant plus tangent. So at this point let's talk about some real mathematics. Well it's easy enough to simply memorize. The integral of cosecant is log cosecant x plus tangent x plus c. The reason that we derived it is we learned or reviewed quite a few important things. And in particular when we derived it we had to review and practice partial fractions, trigonometric substitutions, the rules of logarithms, and trigonometric identities. And this is true in general, memorization in mathematics almost always replaces practice, and so you should avoid memorization.