"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!

Answer 1

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 $C=1$

Answer 2

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.