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.