I recently noticed on Youtube a solution to an indefinite integral that I think is a useful chance to clarify an important point about antidifferentiation of a given function.

The integral I am talking about is \[\mathcal I = \int \frac1{\cos^2x(1+\tan^2x)}dx.\]The proposed solution is based on the substitution \(\tan x = u\), that leads to \[\mathcal I = \int \frac1{1+u^2} du,\]that is\[\mathcal I = \arctan u + C,\] where \(C\) is an arbitrary constant. Replacing back the original function yields \[\mathcal I = \arctan(\tan x) + C.\tag{1}\label{eq3177:1}\]

Now, what is suggested is that \(\arctan(\tan x)\) is simply equal to \(x\); so that the solution proposed by the author of the video is \[\mathcal I = x + C.\]

Here are now a couple of questions.

- If the proposed solution is correct, then the RHS of \eqref{eq3177:1} should represent all possible antiderivatives of the integrand. But derivating that expression just leads to \(1\). How is this possible?
**Is it true that \(\arctan (\tan x) = x\)**? At first sight it looks as if we were applying a function and then its inverse, but*this is not the case*(recall in fact that \(\tan x\) is not a bijective function). Note for example that \(\arctan (\tan x)\) is*periodic*, and, also, it is not defined for \(x = \frac{\pi}2 + k\pi\)!

Starting from the first question, note that, **for \(x\neq \frac{\pi}2 + k\pi\) we have that the integrand is equal to**\begin{eqnarray}f(x) &=& \frac1{\cos^2x(1+\tan^2x)} =\\ &=& \frac1{\cos^2x\left(1+\frac{\sin^2x}{\cos^2x}\right)}=\\ &=&\frac1{\cos^2x + \sin^2x}=1.\end{eqnarray}

So it is correct that the integrand is equal to \(1\), **where it is defined**. But can we actually ignore the domain of \(f(x)\)?

The answer is **no,** and analyzing why this is the case brings us to the second question. Below is a plot of \(F(x) = \arctan (\tan x)\).

Note that in the range \(|x| < \frac{\pi}2\), this function coincides with \(G(x) = x\). Also, **whenever it is defined **(that is again for \(x\neq\frac{\pi}2+k\pi\)) the derivative of \(F(x)\) is equal, too, to \(1\).

So **both** \(F(x)\) and \(G(x)\) are valid antiderivatives of \(f(x)\) in its domain.

They **do not differ by a constant, though**. And this is the most important observation! As a matter of fact, when the integrand is defined over a **disjoint union of two or more open intervals**, then **a different additive constant can be chosen in each interval**, thus generating a much ‘larger’ set of antiderivatives then those represented by either\[F(x) = \arctan(\tan x)+C\] or\[G(x) = x + C.\] For example, if restricted to the usual domain, \[H(x) = \begin{cases}x & (|x|<\frac{\pi}2)\\ x-\frac{\pi}2&(x<-\frac{\pi}2)\\x+\frac{\pi}2 & (x> \frac{\pi}2)\end{cases}\]is another valid antiderivative of \(f(x)\).

In conclusion, taking into account the domain of the original function leads to a correct definition of the set of all its valid antiderivatives .

As a further exercise, in the domain \(\Bbb R – \{0\}\) find all the antiderivatives of \(f(x) = -\frac1{x^2}\).