I can't remember a fallacious proof involving integrals and trigonometric identities. 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.

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.
