"Proof" that $0=1$ using an integral.

Question

I saw the following:

$$\begin{align} \int \tan x \ \mathrm{d}x &= \int \sin x \sec x \ \mathrm{d}x \\ \int \tan x \ \mathrm{d}x &= -\cos x \sec x - \int - \cos x \sec x \tan x \ \mathrm{d}x \\ \int \tan x \ \mathrm{d}x &= -1 + \int \tan x \ \mathrm{d}x \\ 0 &= -1 \end{align}$$

I figure the mistake has to do with constants of integration, but I can't quite point it. Can someone explain to me what happens, please? I browsed a bit around here, looking for $0=1$ and the tag fake-proofs, but I didn't found it, so I apologize in advance in case this is a duplicate of something (just provide me the link and I'll delete this, no problems).

Thanks!

Looks good. The results we have proved with great effort now have much simpler proofs. — André Nicolas, Nov 12 '14 at 19:24
There is a very similar error in another post here that I can't find, which involves integrating $\sin^2 x$ on one side, and $1-\cos^2 x$ on the other, and coming out with $\frac\pi2 = 0$ as a result. — MJD, Nov 12 '14 at 19:28
@DanielFischer: I would say that the question here is different/specific enough. — Thomas, Nov 12 '14 at 19:28
Here's another similar one: $−1=0$ by integration by parts of $\tan(x)$ — MJD, Nov 12 '14 at 19:30
@Thomas Up to a sign, this is the special case of $f = \cos$. And for the special case, we have two more duplicates. — Daniel Fischer, Nov 12 '14 at 19:32
Seeing the answers and comments, it's very clear now. Thanks for the attention. I'll take a look at all these links too (: — Ivo Terek, Nov 12 '14 at 19:33
@DanielFischer: I know about the discussion about abstract duplicates. But in general I am opposed to closing questions that aren't "clear" duplicates. I vote to close as a duplicate of one of the other questions. — Thomas, Nov 12 '14 at 19:33

score 3 · Accepted Answer · answered Nov 12 '14 at 19:23

3

Yes, integral equations like this are equal only up to a constant difference. Therefore you need a + C on one of the sides, and then there is no contradiction, as it is true that $0=-1+C$ for a suitable value of C, namely $C=1$

answered Nov 12 '14 at 19:23

Alan

16,582

score 3 · Answer 2 · answered Nov 12 '14 at 19:26

3

You, in fact, have $$ \int f(x) \; dx = c + \int f(x)\; dx $$ for any constant $c$. The problem is that you can't subtract the integral on both sides. Remember that the indefinite integral really represents(is) a family of antiderivatives.

answered Nov 12 '14 at 19:26

Thomas

43,555

"Proof" that $0=1$ using an integral.

2 Answers2

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