My calc professor once taught us a fallacious proof. I'm hoping someone here can help me remember it.

Here's what I know about it:

  • The end result was some variation of 0=1 or 1=2.
  • It involved (indefinite?) integrals.
  • It was simple enough for Calc II students to grasp.
  • The (primary?) fallacy was that the arbitrary constants (+ C) were omitted after integration.

I'm not certain, but I have a strong hunch it involved a basic trigonometric identity.


4 Answers 4


It's probably the classic $$\int \sin 2x \;dx = \int 2\sin x\cos x \;dx$$

  • Doing a $u=\sin x$ substitution "gives" $$\int 2u \;du = u^2 = \sin^2 x$$

  • Alternatively, using $v = \cos x$ "gives" $$\int -2v \;dv = -v^2 = -\cos^2 x$$

Since the solutions must be equal, we have $$\sin^2 x = -\cos^2 x \quad\to\quad \sin^2 x + \cos^2 x = 0 \quad\to\quad 1 = 0$$

As you note, the fallacy here is the failure to include "+ constant" to the indefinite integrals.

Note that there's also the substitution $w = 2x$, which "gives" $$\begin{align} \int \frac12 \; \sin w \; dw = -\frac12 \; \cos w = -\frac12\;\cos 2x &= -\frac12\;(2 \cos^2 x - 1 ) = -\cos^2 x + \frac12 \\[6pt] &= -\frac12\;(1 - 2 \sin^2 x) = \phantom{-}\sin^2 x - \frac12 \end{align}$$ that leads to the same kind of apparent contradiction when compared to the other integrals.


Here is my favourite: integrating by parts with $u=1/x$ and $v=x$, we get $$\int\frac{dx}{x}=\frac1xx-\int x\Bigl(\frac{-1}{x^2}\Bigr)\,dx =1+\int\frac{dx}{x}$$ and "therefore" $0=1$.

Admittedly there is no trigonometry and so it's probably not the one you were looking for, but still...

  • $\begingroup$ One I learned in calculus class was basically exactly this, except on $\cot x\,\mathrm{d}x = (\sin x)^{-1}\,\mathrm{d}(\sin x)$, so it technically used trigonometry, but was otherwise exactly your answer. $\endgroup$
    – Stan Liou
    Jan 23, 2015 at 6:41
  • $\begingroup$ I used to know a similar one, which I'm pretty sure involved natural logs and proved that $0=1$ using integration by parts, but I can't seem to find it... $\endgroup$
    – lokodiz
    Jan 23, 2015 at 18:21

Here is one that fits your description, but there are many possibilities. We integrate $4\sin x\cos x$ in two ways, incorrectly leaving out the constant of integration.

Way 1: Let $u=\cos x$. Then our integral is $-2u^2$, that is, $-2\cos^2 x$.

Way 2: We have $4\sin x\cos x=2\sin 2x$. Integrate. We get $-\cos 2x$. But $\cos 2x=2\cos^2 x-1$, so the integral is $-2\cos^2 x+1$.

"Thus" $0=1$.

  • $\begingroup$ You beat me by 8 seconds! I'm leaving my answer up, anyway. :) $\endgroup$
    – Blue
    Jan 22, 2015 at 4:52
  • $\begingroup$ We organized things differently. $\endgroup$ Jan 22, 2015 at 5:01

The simplest one I have is not actually 0=1 but $\pi=0$. This is one of my favourites,the most shortest and has confused a lot of people.

$\int \frac{dx}{\sqrt{1-x^2}} = sin^{-1}x$

But we also know that $\int - \frac{dx}{\sqrt{1-x^2}} = cos^{-1}x$

So therefore $sin^{-1}x=-cos^{-1}x$

But also, $sin^{-1}x+cos^{-1}x=\pi/2$

$\implies \pi/2=0$ $\implies \pi=0$.

I'm so evil. :)


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