How to evaluate $\int \cos{x}/\cos{x}\,dx$ without cancelling?

Question

In my (mostly physics based) education, when confronted with an integral like the following: $$\int \frac{\cos{x}}{\cos{x}}\,dx$$ I've typically been taught to cancel functions that appear in both the numerator and denominator, so that it can be evaluated as follows: $$\int \frac{\cos{x}}{\cos{x}}\,dx$$ $$=\int\,dx$$ $$=x+c$$ where $c$ is a constant of integration. However the integrand is undefined at infinietly many points. The above clearly holds for all $x\neq\pi/2 \pm n\pi$, but does it hold for all (or any) $x=\pi/2 \pm n\pi$? If it does, then how should an integral like this be evaluated rigorously?

Answer 1

To do things rigorously, we first need to formulate a precise definition of "antiderivative". I propose the following definition:

If $E\subseteq \mathbb R$, and $f:E\to\mathbb R$ is a function, then an antiderivative of $f$ is a function $F:E\to\mathbb R$ such that $F'=f$.

In other words, an antiderivative of a real-valued function $f$ is a function $F$ with the same domain as $f$, and which satisfies $F'(x)=f(x)$ for all $x$ in its domain.

In your case, we are looking for the antiderivatives of the function $f(x)=\cos x/\cos x$, where the domain of $f$ is tacitly understood to be the set of real numbers $x$ such that $\cos x\neq 0$. In what follows, all of the proposed antiderivatives of $f$ should also be understood to have this domain, meaning that they are undefined at $\pi/2+n\pi$ for every integer $n$.

Clearly, the function $F(x)= x$ is an antiderivative of $f$, as is the function $G(x)=x+C$ for any $C\in\mathbb R$. Is every antiderivative of $f$ of this form? As it turns out, the answer is no. The function $H(x)$ which equals $x$ when $x<\pi/2$ and $x+1$ when $x>\pi/2$ is also an antiderivative of $f$.

In general, an antiderivative of $f$ is a function $F:\mathbb R\setminus\{\pi/2+n\pi:n\in\mathbb Z\}\to\mathbb R$ such that $F(x)=x+C(x)$ where $C$ is a function which is constant on each connected interval of the domain of $F$, i.e. for every $n\in\mathbb Z$, there is a $C_n\in\mathbb R$ such that for all $x\in(\pi/2+n\pi,\pi/2+(n+1)\pi)$ we have $C(x)=C_n$.

To establish that we have found every antiderivative of $f$, we can use the mean value theorem to prove that if $F$ is an antiderivative of $f$, and $a$ and $b$ belong to the same connected interval, then $F(a)=F(b)$.

Answer 2

One can only evaluate the antiderivative on an open set that does not contain any point where $\cos x = 0$. The calculation given provides the antiderivative on an interval. If you want a function defined on the set $\{x\mid \cos x\neq0\}$, you can assign each interval a constant value.

Answer 3

1. $$\forall n{\in}\mathbb Z\;\left(x\in\left(\frac{\pi}2+n\pi,\frac{3\pi}2+n\pi\right)\implies\int \frac{\cos{x}}{\cos{x}}\,\mathrm dx=\left\{x+C_n\mid C_n\in\mathbb R\right\}\right).$$
2. $$\forall n,a,b\;\left(n\in\mathbb Z\;\text{ and }\;[a,b]\subset\left(\frac{\pi}2+n\pi,\frac{3\pi}2+n\pi\right)\implies\int_a^b \frac{\cos{x}}{\cos{x}}\,\mathrm dx=b-a\right).$$ If we plug the countably many undefined points of the bounded function $\dfrac{\cos{x}}{\cos{x}}$ with arbitrary values, then $$\displaystyle\int_a^b \frac{\cos{x}}{\cos{x}}\,\mathrm dx\equiv b-a.$$