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

Solution 1:

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.

Solution 2:

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...

Solution 3:

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" $0=1$.