I had a calculus final yesterday, and in a question we had to find a primitive of $\tan(x)$ in order to solve a differential equation.

A friend of mine forgot that such a primitive could easily be found, tried to integrate $\tan(x)$ by parts... and then arrived to the result $0 = -1$. The kind of thing you're pretty satisfied to "prove", except during an important exam. :-°

So afterwards I tried to do the same :

$$\begin{align*} \int \tan(x)dx &= \int \sin(x) \times \frac{1}{\cos(x)}dx \\[0.1in] &= -\frac{\cos(x)}{\cos(x)} - \int - \frac{\cos(x) \times \sin(x)}{\cos(x)^2}dx \\[0.1in] &= -1 + \int \tan(x)dx \end{align*}$$

And therefore we get :

$$ \int \tan(x)dx = -1 + \int \tan(x)dx \implies 0 = -1$$

What? The reasoning sounds about right to me. Could someone explain where something went wrong?

Thanks, Christophe.

  • 14
    $\begingroup$ And yet some students complain when instructors require writing "$+ C$" after indefinite integrals... $\endgroup$
    – GEdgar
    Dec 18, 2013 at 1:46

3 Answers 3


Without even reading I answer: the antiderivatives of a function are equal only up to a an additive constant, that is any two antiderivatives will always differ by a constant on an interval.

Edit: Ok, having now read the question I confirm my suspicion, note that the symbol $\int f(x)dx$ is not a well defined function. You should interpret the symbol $\int f(x)dx$ as being an undertermined differentiable function which, once you differentiate, yields $f(x)$. Although more formally I believe it's more common to define $\int f(x)dx$ as the set of functions described above, that is, for some non degenerate interval $I$, $$\int f(x)dx=\{F\in \Bbb R^I: \text{F is differentiable and }(\forall x\in I)(F'(x)=f(x))\}.$$

Using this definition one has to very careful about what one means with $\int f(x)dx=g(x)$, because it doesn't mean what one would initially suspect.

  • $\begingroup$ This answers my question, thank you! $\endgroup$ Jun 19, 2013 at 20:27
  • $\begingroup$ @Christophe Glad I could help. You should remember your example to explain it to other people if you ever get the chance. $\endgroup$
    – Git Gud
    Jun 19, 2013 at 20:29
  • $\begingroup$ @Christophe Besides it being a thorn in your side if you forget, you might remember it because of what it does to the integral of 1/x. $\endgroup$
    – Loki Clock
    Jun 19, 2013 at 20:49
  • 3
    $\begingroup$ There are some indefinite trigonometric integrals that have two or more different- looking anti-derivatives that turn out to all be the same function, differing only by a small numerical constant. Since these are indefinite integrals, the "integration constant" $ \ C \ $ "absorbs" these numerical constants, so there is considered to be just one anti-derivative for each of the integrals. $\endgroup$ Jun 19, 2013 at 20:50

Let's try using "real" primitives. Suppose $-\pi/2<x<\pi/2$; then \begin{align} \int_0^x \tan t\,dt &=\left[(-\cos t)\frac{1}{\cos t}\right]_0^x -\int_0^x (-\cos t)\frac{\sin t}{\cos^2 t}\,dt\\ &=\bigl[-1\bigr]_0^x+\int_0^x\tan t\,dt\\ &=(-1)-(-1)+\int_0^x\tan t\,dt\\ &=\int_0^x\tan t\,dt \end{align} which of course doesn't say much, does it? ;-)

Fixing the lower limit of integration is choosing a well determined primitive (or antiderivative). Always do it in case of doubt.


I would like to say that you were able to prove that $-1 = 0$, but unfortunately, you used some (or really a lot of) bad calculus. :(

While integrating the equation
you must separate it into two parts, like you did to get $\int \frac{\sin(x)}{\cos(x)}dx$,

Then, you must find a $u$ and do $u$ substitution. $u = \cos(x)$, and $du = -\sin(x) dx$

then replace the $u$ and $du$ to get $\int \frac{du}{u}$, which differentiates to $\ln|u|$ where $u = \cos(x)$, so the answer is really $\ln|\cos(x)|$, not anything else then $-\ln|\cos(x)|$ or $\ln|1/\cos(x)|$ or $\ln|\sec(x)|$ with C.

In addition, it is bad calculus to put the bounds into the equation before actually differentiating, so the entire $\int \tan(x)dx =$ stuff is completly incorrect


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