Nuking the Mosquito — ridiculously complicated ways to achieve very simple results Here is a toned down example of what I'm looking for:
Integration by solving for the unknown integral of $f(x)=x$:
$$\int x \, dx=x^2-\int x \, dx$$
$$2\int x \, dx=x^2$$
$$\int x \, dx=\frac{x^2}{2}$$
Can anyone think of any more examples?
P.S. This question was inspired by a question on MathOverflow that I found out about here. This question is meant to be more general, accepting things like solving integrals and using complex numbers to evaluate simple problems. 

Inspired by your (far too simple!) example in the question.
Let us compute $\int_0^1 x^2\,dx$. We start with the obvious change of variables:
$$
t = \frac{x}{1-x} \quad\Leftrightarrow\quad x = \frac{t}{1+t}
$$
from which we get
$$
dx =\frac{dt}{(1+t)^2}.
$$
and the integral transforms to
$$
\int_0^1 x^2\,dx = \int_0^\infty \frac{t^2}{(1+t)^4}\,dt
$$
which can be attacked using residue calculus. Define
$$
f(z) = \frac{z^2\log z}{(1+z)^4}
$$
where $\log$ is chosen as the natural branch of the complex logarithm and integrate over a keyhole contour:

Standard estimates on the various parts of the contour shows that on $C_R$:
$$
\left| \frac{z^2\log z}{(1+z)^4} \right| \le \frac{R^2(\ln R + 2\pi)}{R^4-1}
$$
so
$$
\left| \int_{C_R} f(z)\,dz \right| \le 2\pi R \cdot \frac{R^2(\ln R + 2\pi)}{R^4-1}
$$
which tends to $0$ as $R \to \infty$. Similarly, on $C_\varepsilon$:
$$
\left| \frac{z^2\log z}{(1+z)^4} \right| \le \frac{\varepsilon^2(\ln \varepsilon + 2\pi)}{(1/2)^4}
$$
(if $\varepsilon < 1/2$) so
$$
\left| \int_{C_R} f(z)\,dz \right| \le 2\pi R \cdot  16 \varepsilon^2(\ln \varepsilon + 2\pi),
$$
which tends to $0$ as $\varepsilon \to 0^+$. It remains to investigate what happens on $I^+$ and $I^-$. On $I^+$ we get
$$
\int_{I^+} f(z)\,dz = \int_{\varepsilon}^R \frac{x^2\ln x}{(1+x)^4}\,dx
$$
and on $I^-$:
$$
\int_{I^+} f(z)\,dz = \int_R^{\varepsilon} \frac{x^2(\ln x+2\pi i)}{(1+x)^4}\,dx.
$$
Putting everything together, using the residue theorem and letting $R\to\infty$, $\varepsilon\to0^+$ (note that the integrals containing $\ln x$ cancel) we get
$$
-2\pi i \int_0^\infty \frac{x^2}{(1+x)^4}\,dx = 2\pi i\operatorname{Res}\limits_{z=-1} \frac{z^2\log z}{(1+z)^4}.
$$
Finally,
$$
\operatorname{Res}\limits_{z=-1} \frac{z^2\log z}{(1+z)^4} =
\frac1{3!} (z^2\log z)^{'''}\big|_{z=-1} = -\frac13
$$
(omitting tedious algebra), and we reach the amazing result
$$
\int_0^1 x^2\,dx = \int_0^\infty \frac{t^2}{(1+t)^4}\,dt = -\operatorname{Res}\limits_{z=-1} \frac{z^2\log z}{(1+z)^4} = \frac13.
$$
