# Question inside $\int \tan^2 x dx$

$$\int \tan^2 x dx$$ the question isn't behind on how to do it but rather on its solution, let me go through my solution quickly:

If we do a substitution $$u=\tan(x)$$ then $$du=\sec^2 x dx=(u^2+1)dx$$ hence we get $$\int \frac{u^2}{1+u^2}du=\int\left(1-\frac{1}{1+u^2}\right)du=\tan x-\arctan(\tan x)+C$$ so no problem until here. The solution from the book or WolframAlpha is: $$\tan x-x+C$$ so it raised a question, since $$\tan$$ isn't injective and only has an inverse for $$x\in(-\frac{\pi}{2},\frac{\pi}{2})$$, for which then $$\arctan(\tan(x))=x$$ is only true for $$x\in(-\frac{\pi}{2},\frac{\pi}{2})$$. But then why if we do $$\arctan(\tan x)=x$$ in my solution we then get $$\tan x-x+C$$ which is the same as the correct solution the book or WolframAlpha shows, while we have only just considered the $$x\in(-\frac{\pi}{2},\frac{\pi}{2})$$ in my method? Isn't the indefinite integral considering $$\forall x\in \mathbb{R}$$? Why is this happening?

Note: Not looking at all for an alternative solution or how to get there, rather why I get the correct solution by considering $$x\in(-\frac{\pi}{2},\frac{\pi}{2})$$ while an integral should be $$\forall x\in \mathbb{R}$$

• The result in your textbook/Wolfram arises since $\tan^2 x = \sec^2 x - 1$, so the integral becomes $\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x + C$. Commented Dec 9, 2022 at 0:10
• @moofasa I really appreciate you but that's totally out of the scope of my question that was raised. What I'm asking is in the very last paragraphs Commented Dec 9, 2022 at 0:15
• I'm not sure if this answers your question, but indefinite integration tends to be loose, IMO. I wouldn't worry about it. The notation $\int f$ just means "all antiderivatives of $f$ on some subdomain of the domain of $f$". There's nothing wrong with using u-subs for indefinite integrals, but without any bounds of integration, your u-sub might not be valid. But as far as indefinite integration cares, it's fine. There is a domain involved, sure, but that's because functions naturally have a domain by definition. Your main goal is to just find the antiderivative and put that $+C$ at the end. Commented Dec 9, 2022 at 1:24
• You might be interested in looking at this, math.stackexchange.com/a/2992231/399263, for an interesting technique to make the C "disappear", or rather on how to express the $C$ from possibly $\arctan(\tan(x))$ itself (I didn't solve it, so I cannot be sure) without going for the floor function as shown in Ninad Mushi's answer.
– zwim
Commented Dec 9, 2022 at 2:05
• @Aley20, I understand that you were asking about something different - hence the comment rather than posting it as an answer. My point is that you don't need to do $u$-substitution to get the antiderivative of $\tan^2 x$. In fact, you use the same exact trig identity to perform the change of variables anyway. Commented Dec 9, 2022 at 17:36

This is a phenomenon that happens with all indefinite integrals. The original integral

$$\int \tan^2 x\:dx$$

has singularities at $$\frac{\pi}{2} + \pi k$$ for $$k\in\Bbb{Z}$$, therefore, no, the original integral was not for all $$x\in\Bbb{R}$$. Whenever you have discontinuities like that, they are tucked away in the $$+C$$. For example, take the textbook case of

$$\int \frac{dx}{x} = \log|x| + C$$

$$f(x)=\begin{cases}\log( -x) +7 & x<0 \\ \log (x )-2 & x>0\end{cases}$$

Taking the derivative, we see that it is still $$\frac{1}{x}$$. That is because $$f$$ can still be written as

$$f(x) = \log|x| + C$$

but in this case, the constant $$C$$ changes values across the discontinuity

$$C = \begin{cases}7 & x<0 \\ -2 & x>0\end{cases}$$

and that is allowed for $$+C$$. In your answer, we have that

$$x = \arctan(\tan x) + C$$

where the $$+C$$ hides the arbitrary constant (s)

$$C = \pi\left\lfloor\frac{x+\frac{\pi}{2}}{\pi}\right\rfloor$$

which makes the two answers equivalent, since they are only off by an arbitrary constant.

• that sounds good but still is unclear to me how $\arctan(\tan(x))$ and $x$ differ by an arbitrary constant, will plotting into a graph help me see it? Commented Dec 9, 2022 at 0:32
• @Aley20 It is as you say. One graph is a triangle wave, and the other is a triangle wave but with the beginning of each triangle lined up with the end of the previous (aka a sloped line) Commented Dec 9, 2022 at 0:32
• the thing is that I already did and it confused me because the $C$ then varies for each different $x$, but now I remember that you said it is true for the $C$, isn't it? So then it should be fine? And as a second last question, does this phenomena ocurr also with the $\arcsin(\sin x)$ and $\arccos(\cos x)$ with $x$ as well? Commented Dec 9, 2022 at 0:37
• In general, $\arctan(\tan(x)) = x + k\pi$ where $k$ is some integer. That $k\pi$ is a constant, so that's why you can just say $\tan(x) - (x+k\pi) + C = \tan(x) - x + C$. I'm not sure what you mean when you ask if $C$ can vary for each $x$. No matter what $x$ is, the $C$ is just some constant. @Aley20 Commented Dec 9, 2022 at 1:32
• I am deeply uncomfortable with a "constant" which seems to depend on $x$. I get that you have structured things so that on any connected subset of the domain of $\tan$, the expression will evaluate to a constant, but this seems notationally dangerous. Commented Dec 9, 2022 at 1:45

### Elementary Explanation

Most elementary textbook authors are fairly careful to define things in such a way that the situation you are describing is not an issue. Because I have been teaching out of it recently, here is what Thomas' Calculus (instructor's 13th ed) has to say:

Definition: (p. 232) a function $$F$$ is an antiderivative of $$f$$ on an interval $$I$$ if $$F'(x) = f(x)$$ for all $$x$$ in $$I$$.

and

Definition: (p. 237) The collection of all antiderivatives of $$f$$ is called the indefinite integral of $$f$$ with respect to $$x$$, and is denoted by $$\int f(x)\,\mathrm{d}x.$$

Notice that the indefinite integral of $$f$$ is defined in terms of antiderivatives of $$f$$, and the antiderivative of $$f$$ is only defined if $$f$$ is defined on an interval.

This implies that before you even start to ask what the indefinite integral of a function is, you first have to be clear about the definition of that function. What does the collection of symbols $$\tan(x)^2$$ represent?[1]

Barring any other information, I would assume that $$\tan$$ represents the tangent function, which is defined on the domain $$\mathbb{R}\setminus \{\pi/2 + k\pi : k\in\mathbb{Z}\}$$ (that is, the tangent function is defined everywhere that the cosine is not zero). BUT there is a problem here, since the question is about the antiderivative of the square of the tangent function, and the tangent function is naturally defined on a set which is not an interval.

This implies that $$\int \tan(x)^2\,\mathrm{d}x$$ does not, in fact, refer to the square of the tangent function on its natural domain, but only on some connected component of that domain, say $$(-\pi/2,\pi/2)$$, or $$(\pi/2, 3\pi/2)$$. Before you can evaluate the indefinite integral, you must first specify the domain of the function which you are integrating.

### But...

In their answer, Ninad Munshi argues that one can discuss antiderivatives of functions which are defined on disconnected domains, and that the "disconnection" is swallowed by the constant of integration. I will not argue that they are wrong, but I do think that this is a wrongheaded approach, which could, potentially, lead to pain later on—being able to "split out" the behaviour of a function across singularities is often helpful.

Looking at the example of $$x \mapsto 1/x$$, we have two facts:

1. if we assume that $$x > 0$$, then $$\int \frac{1}{x}\, \mathrm{d}x = \log(x) + C,$$ where $$C$$ is an arbitrary constant of integration, and
2. if we assume that $$x < 0$$, then $$\int \frac{1}{x}\, \mathrm{d}x = \log(-x) + C,$$ where $$C$$ is an arbitrary constant of integration.

If we want to talk about an "antiderivative"[2] or "indefinite integral"[3] of $$x \mapsto 1/x$$ on the set $$\mathbb{R}\setminus \{0\}$$, then I think that it is wise to keep track of the various domains of definition. Hence I would argue that the "correct" antiderivative is something like $$\int \frac{1}{x} \,\mathrm{d}x = \begin{cases} \log(x) + C & \text{if x > 0, and} \\ \log(-x) + D & \text{if x < 0,} \end{cases}$$ where both $$C$$ and $$D$$ are arbitrary constants of integration.

In the case of $$\tan(x)^2$$, it would not be unreasonable to write $$\int \tan(x)^2 \,\mathrm{d}x = \begin{cases} \tan(x) + x + C_k & \text{if x \in \left(\dfrac{\pi}{2} + (k-1)\pi, \dfrac{\pi}{2} + k\pi\right),} \end{cases}$$ where $$C_k$$ is an arbitrary constant of integration for each $$k\in \mathbb{Z}$$.

[1] Minor pet peeve: I work in an area of mathematics where $$f^2$$ means $$f \circ f$$, i.e. $$f$$ composed with itself. Because of this, I find the notation $$\tan^2(x)$$ somewhat ambiguous. Yes, it is common notation, and no, you aren't going to be misunderstood if you use it. But my choice is to use $$\tan(x)^2$$, as I think that it is less ambiguous. This explains my choice above. Please make your own choices when writing.

[2] "Antiderivative" is in quotes because, per the definition at the top, an antiderivative is only defined on an interval, and we are about to talk about a function which is not defined on an interval, so the definition we are working with doesn't really apply.

[3] Ditto.

When you apply a substitution u=f(x) for an integral, the function f(x) must be bijective so that the inverse $$f^{-1}$$ exists. In your case, you should let $$u=\tan x$$ for $$x\in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ so that $$\arctan(\tan x)=x.$$

For otherwise, if we fix an integer $$n$$ and let $$u=\tan x$$ for $$x \in\left( n \pi-\frac{\pi}{2}, n \pi+\frac{\pi}{2}\right)$$,

then $$\arctan(\tan x)=n \pi+x$$

and $$I=\tan x-\arctan(\tan x)+C= \tan x-(n \pi+x)+C=\tan-x+C’,$$

where $$C’=C-n\pi$$ absorbs $$-n\pi$$ despite of the choice of principal values for the function $$\arctan x$$.

Wish it helps :-)