# Monthly Archives: September 2014

## Trigonometric and related substitutions in integrals

$\DeclareMathOperator{\csch}{csch}$
$\DeclareMathOperator{\sech}{sech}$
$\DeclareMathOperator{\arsinh}{arsinh}$

1. Introduction

In many ways, a first semester of calculus is a big ideas course. Students learn the basics of differentiation and integration, and some of the big-hitting theorems like the Fundamental Theorems of Calculus. Even in a big ideas course, students learn how to differentiate any reasonable combination of polynomials, trig, exponentials, and logarithms (elementary functions).

But integration skills are not pushed nearly as far. Do you ever wonder why? Even at the end of the first semester of calculus, there are many elementary functions that students cannot integrate. But the reason isn’t that there wasn’t enough time, but instead that integration is hard. And when I say hard, I mean often impossible. And when I say impossible, I don’t mean unsolved, but instead provably impossible (and when I say impossible, I mean that we can’t always integrate and get a nice function out, unlike our ability to differentiate any nice function and get a nice function back). An easy example is the sine integral $$\int \frac{\sin x}{x} \mathrm d x,$$
which cannot be expressed in terms of elementary functions. In short, even though the derivative of an elementary function is always an elementary function, the antiderivative of elementary functions don’t need to be elementary.

Worse, even when antidifferentiation is possible, it might still be really hard. This is the first problem that a second semester in calculus might try to address, meaning that students learn a veritable bag of tricks of integration techniques. These might include ${u}$-substitution and integration by parts (which are like inverses of the chain rule and product rule, respectively), and then the relatively more complicated techniques like partial fraction decomposition and trig substitution.

In this note, we are going to take a closer look at problems related to trig substitution, and some related ideas. We will assume familiarity with ${u}$-substitution and integration by parts, and we might even use them here from time to time. This, after the fold.

## A bit more about partial fraction decomposition

This is a short note written for my students in Math 170, talking about partial fraction decomposition and some potentially confusing topics that have come up. We’ll remind ourselves what partial fraction decomposition is, and unlike the text, we’ll prove it. Finally, we’ll look at some pitfalls in particular. All this after the fold.

1. The Result Itself

We are interested in rational functions and their integrals. Recall that a polynomial ${f(x)}$ is a function of the form ${f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0}$, where the ${a_i}$ are constants and ${x}$ is our “intederminate” — and which we commonly imagine standing for a number (but this is not necessary).

Then a rational function ${R(x)}$ is a ratio of two polynomials ${p(x)}$ and ${q(x)}$, $$R(x) = \frac{p(x)}{q(x)}.$$

Then the big result concerning partial fractions is the following:

If ${R(x) = \dfrac{p(x)}{q(x)}}$ is a rational function and the degree of ${p(x)}$ is less than the degree of ${q(x)}$, and if ${q(x)}$ factors into $$q(x) = (x-r_1)^{k_1}(x-r_2)^{k_2} \dots (x-r_l)^{k_l} (x^2 + a_{1,1}x + a_{1,2})^{v_1} \ldots (x^2 + a_{m,1}x + a_{m,2})^{v_m},$$
then ${R(x)}$ can be written as a sum of fractions of the form ${\dfrac{A}{(x-r)^k}}$ or ${\dfrac{Ax + B}{(x^2 + a_1x + a_2)^v}}$, where in particular

• If ${(x-r)}$ appears in the denominator of ${R(x)}$, then there is a term ${\dfrac{A}{x – r}}$
• If ${(x-r)^k}$ appears in the denominator of ${R(x)}$, then there is a collection of terms $$\frac{A_1}{x-r} + \frac{A_2}{(x-r)^2} + \dots + \frac{A_k}{(x-r)^k}$$
• If ${x^2 + ax + b}$ appears in the denominator of ${R(x)}$, then there is a term ${\dfrac{Ax + B}{x^2 + ax + b}}$
• If ${(x^2 + ax + b)^v}$ appears in the denominator of ${R(x)}$, then there is a collection of terms $$\frac{A_1x + B_1}{x^2 + ax + b} + \frac{A_2 x + B_2}{(x^2 + ax + b)^2} + \dots \frac{A_v x + B_v}{(x^2 + ax + b)^v}$$

where in each of these, the capital ${A}$ and ${B}$ represent some constants that can be solved for through basic algebra.

I state this result this way because it is the one that leads to integrals that we can evaluate. But in principle, this theorem can be restated in a couple different ways.

Let’s parse this theorem through an example – the classic example, after the fold.