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

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 $C=1$
