calculus - 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.

Answer

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.

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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.