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.
 A: The simplest one I have is not actually 0=1 but $\pi=0$. This is one of my favourites,the most shortest and has confused a lot of people.
$\int \frac{dx}{\sqrt{1-x^2}} = sin^{-1}x$
But we also know that $\int - \frac{dx}{\sqrt{1-x^2}} = cos^{-1}x$
So therefore $sin^{-1}x=-cos^{-1}x$
But also, $sin^{-1}x+cos^{-1}x=\pi/2$
$\implies \pi/2=0$ $\implies \pi=0$.
I'm so evil. :)
A: 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.
A: Here is my favourite: integrating by parts with $u=1/x$ and $v=x$, we get
$$\int\frac{dx}{x}=\frac1xx-\int x\Bigl(\frac{-1}{x^2}\Bigr)\,dx
  =1+\int\frac{dx}{x}$$
and "therefore" $0=1$.
Admittedly there is no trigonometry and so it's probably not the one you were looking for, but still...
A: Here is one that fits your description, but there are many possibilities. We integrate $4\sin x\cos x$ in two ways, incorrectly leaving out the constant of integration.
Way 1: Let $u=\cos x$. Then our integral is $-2u^2$, that is, $-2\cos^2 x$.
Way 2: We have $4\sin x\cos x=2\sin 2x$. Integrate. We get $-\cos 2x$. But $\cos 2x=2\cos^2 x-1$, so the integral is $-2\cos^2 x+1$.
"Thus"  $0=1$.
