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.

1. 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?
2. 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 $$f(x) = -\frac1{x^2}$$.