21
$\begingroup$

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.

$\endgroup$
  • 12
    $\begingroup$ And yet some students complain when instructors require writing "$+ C$" after indefinite integrals... $\endgroup$ – GEdgar Dec 18 '13 at 1:46
26
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This answers my question, thank you! $\endgroup$ – christophetd Jun 19 '13 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 '13 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 '13 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. $\endgroup$ – colormegone Jun 19 '13 at 20:50
7
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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
$$\int\tan(x)~dx$$
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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.