TITLE: "Proof" that $0=1$ using an integral.
QUESTION [3 upvotes]: 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!

REPLY [3 votes]: 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.
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