Here’s the graph of cosine and secant on .

Like all of the trig functions, secant is not one-to-one. So we need to restrict the domain to get a one-to-one function and then invert. Which pieces should we use though? It seems reasonable, and in analogy with inverse cosine, to restrict our domain to . These are the green and yellow branches. Let’s call this restricted secant . It turns out, unfortunately, that this natural choice isn’t perhaps the nicest choice for calculus purposes. So we will also consider , the restriction of secant to . So consists of the green and orange branches. We will refer to these functions as when it doesn’t matter which choices we’ve made.

In either case is now one-to-one with range . Thus we have with domain . What is the derivative of ? Well

where we’ve used the identity .

We now need to worry about our choice of domain. In the case of our belongs to and here . Thus

But in the case of our belongs to . Now on and on . Thus for these . Therefore

These formulas agree on of course.

on

we substitute for to get

on

and anti-differentiate to find

on .

Now since we see that . Thus we have

good for all .

As an example of how to calculate with this we remember and so

.

Thus we have

.

Now if, for whatever reason, we wanted to calculate we could use the well-known estimate from the world of alternating series to find

where the sum of an alternating series of the form with and decreasing to differs from its -th partial sum by at most .

Now Machin’s Formula states

.

Given our series expression for , the niceness of the numbers and , and the easy estimate provided because our series is alternating in the right way, Machin’s Formula gives a very efficient way to calculate .

How do we prove this formula? The proof is actually an easy, though tedious application of the tangent addition formula:

Here goes:

and so

Now if we have .

Ok, now consider

and if and we have

and therefore

as desired.

Here’s a too clever solution: Since we have

Using the Weierstrass substitution here would be a less clever, more general way to solve this integral. Remember we set yielding , , and . Applying these here we find

We complete the square: and go on.

We make the substitution , and find

I’ll leave it to you to reconcile the two answers.

This can be evaluated in a number of ways. Here are two of them along with a nice consequence.

First we’ll treat this as a straight-forward partial fraction decomposition question. We have

where are real numbers to be determined. We have then

.

Letting we find ; letting we find . Thus

Done and done.

Now we’ll treat this using hyperbolic substitutions. Remember the fundamental hyperbolic identity: . From this we can derive the identities and by dividing our fundamental identity by and respectively.

Remember the graph of . We see that the domain of is all real numbers and the range is . Further is one-to-one and so we have a with domain and range all real numbers. Here’s a graph if your memory of hyperbolic tangent is a little fuzzy.

So in the case where we can make the substitution . Then and .

Combining this with our first solution (and setting ) we see that

and since and we see that . This formula for can be derived in other ways of course.

When we make a substitution and the reasoning is similar.

Thus we have

on

and

on

where we calculated from .

Here are a pair of examples.

Consider the integral . We apply the Weierstrass substitution to find

We considered the integral in an earlier post. Let’s do it again with the help of the Weierstrass substitution.

This answer has a very different form from the ones given in our earlier post. We can reconcile this answer with the older ones through algebra and trigonometric identities.

We have

as expected. (In the penultimate equality we used the usual , , and identities.)

We start with the curve on the interval . Here is a part of the graph:

At it has a height of and as the height goes to zero. We then take this curve and wrap it in three-dimensions about the -axis. This gives us a surface of revolution, part of which looks like:

(The light green line represents the -axis.) The shape is something like a horn of infinite length. At the wide end it is a circle of radius and if we cut the horn perpendicular to the -axis the exposed end is a circle of radius .

It has finite volume. Why? By the usual formula, the volume is

where is the cross-sectional area of the shape when we cut it with a plane perpendicular to the -axis. In this case we get circles of radius and so . Thus

It has infinite surface area. Why? The formula for surface area of a surface of revolution is

where here so

(That because of course.)

.

The line through the point with slope is given by . This line intersects the unit circle in one other point and as we vary we strike every point on the unit circle. Here’s an illustration for a few values of :

What are the coordinates of this point? Well, the point satisfies both and so we have

corresponds to the point . When we have

and we want the to be so that positive slopes correspond to the upper half of the circle as we illustrated above.

Therefore every point on the unit circle (other than ) is of the form

for some . As we have .

We now have two parameterizations of the unit circle. How are they connected? For every there is a such that

which gives

.

This calls to mind the tangent addition formula . This suggests and . The usual calculations show that this is correct: If we start with we get the desired and .

The complex cubic equation has roots , , and where

and

are chosen to preserve ,

, and .

Suppose now that the coefficients of our cubic are real. The behavior of the roots is largely determined by the sign of the expression . (The discriminant of a cubic is . Our is . We’ll discuss discriminants some other time.) There are two cases to consider.

First, suppose . Then are real. It follows that there is one real root and two complex conjugate roots. Why? The roots of our cubic are (real) and and (a complex conjugate pair).

Second, suppose . Here there are three real roots. Why? We have

where

and can be calculated if we really need it. Now set and and note we have as required. Thus our roots are

,

, and

.

Note that when we have and in this case (at least) two of our roots above will coincide.

where the coefficients are any complex numbers and . Before doing this we need to recall a couple of facts from our earlier work.

First, we saw in an earlier post that every complex number has -th roots and that if is any one -th root of the complete set is

where . There isn’t a natural way to choose one of these from among the others. (When is a positive real number we can always choose the unique positive real -th root.) For this reason the symbol is ambiguous. Despite this, when convenient, we’ll use the symbol to mean some -th root of .

Second, to solve a quadratic equation we only need to extract a and then the roots are . Since every complex number has a square root it follows that every complex quadratic equation has complex roots. We’ll use this observation a couple of times in what follows.

Now to the cubic equation. Let where the coefficients are complex and . Since we’re interested in the roots of we may suppose . Then by substituting we find

for some complex numbers .

Thus is suffices to find the roots of . This is called a depressed cubic, by the way.

Now

so if we can find complex numbers such that and then will be a root of .

Notice that implies . Then we see that are roots of the complex quadratic equation

Therefore, by the quadratic formula, we have

Now we extract cube roots of these quantities to find

and

where we are careful to choose our cube roots to preserve .

Thus is one root of our cubic. What are the othe two roots? There are three cube roots of (namely the already chosen and ) and three cube roots of (again these are ). In order the preserve the condition we must choose them in the pairs and .

To summarize: The complex cubic equation has roots , , and where

and

are chosen to preserve and

.

Example: Find the roots of . To depress the equation we instead consider . Here and so

and thus

and . Now these numbers are real and so have unique, unambiguous real cube roots

and

and these satisfy . Thus the roots of are . Therefore the roots of are .

“Your goal is to reach every single student and make sure they acheive the material to the best of their individual abilities. My goal is to fail the weakest students in the first month.”

There was a great round of laughter and the conversation moved on.

His comment, though perhaps a joke, does point to a real difference between the goal and focus of teaching at various levels. Regardless of the level of instruction, we have the material to be presented and the group of learners to receive it. Now assuming the teacher is competent and the learners are willing, what are the differences between teaching in elementary school and teaching in college?

In the lower grades the teacher’s primary loyalty is to the student. Yes, there is important and fundamental material to be taught but it is the individual student receiving this material which is the focus. The good teacher will try various techniques and styles to communicate. She may even vary these from student to student. She has an entire school year and many opportunities within it to get each student to master the material to the best of their ability, readiness, and willingness. (Moreover, the same material will be presented in a slightly more sophisticated fashion the next year.)

At the college level things are very different. There is limited time to cover large amounts of material. The material must be mastered on this exposure so that the student can go on to the next section of this course and to the courses which follow. A good teacher here keeps the students in mind but his primary loyalty is to the material itself. The teacher of Calculus I, say, must teach the material at a certain pace and level for the class to cover what is required and for the course to be at the appropriate level. The teacher can try to give additional help to individual students — office hours! — but there can be little flexibility for individual needs.

The material taught in elementary school is crucial for the basic education of everyone. So here any failure robs the student of something essential for his later life. We give the students a half-dozen years or so to learn to read, to calculate, and to understand a little of the workings of the world. College is different; College is entirely optional. A student who fails Calculus may take it again. And if the student is ultimately unable to pass, the consequences are minimal.