The integration series has resumed! In Part I I broadly discussed the idea of using a \(u\)-substitution first to simplify the integrand. Then I discussed the idea of choosing \(dg = dx\).

Now here are a few more techniques within the realm of integration by parts (and integration in general.

Working The Equal Sign

The vast majority of students go through their early math career with a “left-right” bias when it comes to working with the equal sign. In other words, students tend to understand an expression like \(v = 3z\) as “\(v\) equals \(3z\)” but not as “\(v\) and \(3z\) are equal to each other”. And there is a distinction here. Students feel that the “answer is on the right” while the “variable being solved for is on the left”. This is a typical convention. But what’s forgotten or perhaps not in the forefront in a student’s mind is that the equal sign works both ways — the left is equal to the right and the right is equal to the left.

Consider this staple integral found in practically all Calculus texts.

$$\int e^{x}\sin(x)\ dx$$

We may as well go with \(f = e^{x}\) and \(dg = \sin(x)\ dx\) yielding
$$\int e^{x}\sin(x)\ dx = -e^{x}\cos(x) + \int \cos(x)e^{x}\ dx$$

At this point, students have one of two reactions:

• Give up because \(\int \cos(x)e^{x}\ dx\) looks just as bad as the starting integral. or …
• “Do I have to integrate by parts again?

In either case, yes, another integration by parts is necessary. Thus, skipping a bunch of steps, we have

$$\int e^{x}\sin(x)\ dx = -e^{x}\cos(x) + e^{x}\sin(x) – \int e^{x}\sin(x)\ dx$$

Now, the majority of students tend to say something like “Oh no, I have \(\int e^{x}\sin(x)\ dx\) again!” (Also, there are going to be mix-ups with minus signs.) This is, in some sense where “the Calculus stops and the Algebra begins”, if we deign to classify things as such. But really, what do we have here? We have this:

$$ThingIWant = -e^{x}\cos(x) + e^{x}\sin(x) – ThingIWant$$ where \(ThingIWant\) is just \(\int e^{x}\sin(x)\ dx\), so now it’s just a matter of performing a basic algebraic step to obtain $$2\cdot ThingIWant = -e^{x}\cos(x) + e^{x}\sin(x) $$ and then dividing by two we obtain $$\int e^{x}\sin(x)\ dx = \frac{-e^{x}\cos(x) + e^{x}\sin(x)}{2}$$

Easy-peasy, right? If you’re teaching integration by parts, you will undoubtedly show this standard example. I recommend giving students the following $$\int e^{ax}\sin(bx)\ dx$$

with sensible \(a\) and \(b\) (or just work it out in generality). It is the same mechanic as with \(\int e^{x}\sin(x)\ dx\), the difference is now they have to work with fractions. (Then replace \(e^{ax}\) with an exponential of a different base.)

One of the reasons I like the integration portion of an undergraduate Calculus sequence is that I view it as a “total math workout”. There’s going to the gym and working on each muscle group in isolation. And then there’s exercising several muscle groups all at once. That’s what integration is — it’s going to work several math muscle groups. Students will have to use elements from Trigonometry, Algebra, Geometry, Arithmetic, etc. And this is why Calculus is “really hard”. Students have been training with isolated workouts. Calculus (and Calculus II, especially) can feel like bootcamp in that way. The truth is, from here on out, that’s how mathematics works (and it should have been taught that way from the start — the only reason it isn’t is that the other “basic” subjects are more atomizable into tidy bite-size packets).

Anyway, I’m digressing into rant-y mode and it’s unbefitting of an uncle. (Get it? Get it? ranty? auntie? (pronounced in that annoying way).)

The idea of obtaining the integral in question on both sides of the equation is something that comes up often when working with exponentials and trigonometric functions because their derivatives don’t “go away” and have some type of periodicity to them.

But …

\(0 = 1\)

Ok, given the above, now consider integration by parts of $$\int \frac{1}{x}\ dx$$ We know that \(\ln(x)\) differentiates to \(\frac{1}{x}\) and thus \(\int \frac{1}{x}\ dx = \ln(x) + C\) where \(C\) is the constant of integration.

But suppose we tried integration by parts with \(f = \frac{1}{x}\) and \(dg = dx\), then we have \(df = -\frac{1}{x^{2}}\) and \(g = x\) which gives $$\int \frac{1}{x}\ dx = 1 + \int \frac{1}{x}\ dx$$ and now … molasses. We have \(0 = 1\) since $$\int \frac{1}{x}\ dx \ – \int \frac{1}{x}\ dx = 0 = 1$$

Why did this happen? Have your students chew on this!

Working The Equal Sign … Part Deux

Typically integration of trigonometric functions raised to non-negative integer powers is the next sub-unit in the integration series coming after integration parts. So it’s a nice time to give a little prelude to what’s in store.

Spoiler alert for students reading this:

Consider $$\int \cos^{2}(x)\ dx$$
Some students may be tempted to say $$\int \cos^{2}(x)\ dx = \int 1 – \sin^{2}(x)\ dx$$
and then will ultimately come to $$\int 1\ dx = \int 1\ dx$$ which is useless! Or they may try to substitute \(1 – \cos^{2}(x)\) back in for \(\sin^{2}(x)\) yielding the ever frustrating \(0 = 0\) or \(\int \cos^{2}(x)\ dx = \int \cos^{2}(x)\ dx \).

It’s easy to go in circles. I know that when I first learned Calculus decades ago, I went in circles like this. It’s natural. I didn’t understand how to substitute. I just substituted blindly and exercises like these gave me a better sense for what correct substitutions should be.

Another reason to consider working with $$\int \cos^{2}(x)\ dx$$ before delving into trigonometric integrals is to introduce reduction formulas. And that will be Part III!