# 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.

Below is an excerpt of integrating a helical staircase I submitted as a solution in a vector calculus class years ago on April Fools when I was still an undergraduate. I got full marks and no regrets.

…therefore our integral is
\begin{align*}
&a\int^{2\pi}_{0}\int^{1}_{0} \sqrt{a^{2}u^{2}+b^{2}} \, dudv\\
&= 2\pi a \int_{0}^{1} \sqrt{a^{2}u^{2}+b^{2}} \, du. \qquad (1)
\end{align*}

Now clearly, we should use the substitution $au = b\tan{\theta}$, but we have already demonstrated how to integrate $\sec^{3}\theta$ in Assignment 8, $\S 15.3$ Question 6. So instead, suppose we restrict ourselves to non-trigonometric substitution. Then we should instead let $\sqrt{a^{2}u^{2}+ b^{2}} = t – ua$. Then squaring both sides gives
$$a^{2}u^{2}+b^{2} = t^{2} – 2uat + u^{2}a^{2}$$
and solving for u gives
$$\frac{t^{2}-b^{2}}{2at} = u.$$
Differentiating both sides gives
$$du = \frac{4at^{2} – 2at^{2} + 2ab^{2}}{4a^{2}t^{2}} dt.$$
Substituting this in we have $(1)$ equal to
\begin{align*}
&2\pi a\int_{t(0)}^{t(1)} \left(t – \frac{a(t^{2}-b^{2})}{2at}\right) \left(\frac{4at^{2} – 2at^{2} + 2ab^{2}}{4a^{2}t^{2}}\right) dt\\
&=2\pi a \int_{t(0)}^{t(1)} \left(t – \frac{t^{2}-b^{2}}{2t}\right) \left(\frac{1}{a} -\left(\frac{1}{2a}\right) +\frac{b^{2}}{2at^{2}}\right)dt\\
&= 2\pi \int_{t(0)}^{t(1)} \left(\frac{2t^{2} – t^{2}+b^{2}}{2t}\right)\left(\left(\frac{1}{2}\right) +\frac{b^{2}}{2t^{2}}\right)dt\\
&= 2\pi\int_{t(0)}^{t(1)} \left(\frac{t^{2} + b^{2}}{2t}\right) \left(\frac{t^{2}+b^{2}}{2t^{2}}\right)\, dt\\
&= 2\pi\int_{t(0)}^{t(1)} \frac{t^{4} + t^{2}b^{2} + b^{2}t^{2} + b^{4}}{4t^{3}} dt\\
&= \pi \int_{t(0)}^{t(1)} \frac{t^{4} + t^{2}b^{2} + b^{2}t^{2} + b^{4}}{2t^{3}} dt\\
&= \pi \int_{t(0)}^{t(1)} \frac{t}{2} + \frac{b^{2}}{2t} + \frac{b^{2}}{2t} + \frac{b^{4}}{2t^{3}} dt\\
&= \pi \int_{t(0)}^{t(1)} \frac{t}{2} + \frac{b^{2}}{t} + \frac{b^{4}}{2t^{3}} \, dt\\
&= \pi \left( \frac{t^{2}}{4} + b^{2}\ln |t| – \frac{b^{4}}{4t^{2}}\right) \bigg|_{t(0)}^{t(1)}. \qquad (2)
\end{align*}
Now when $u = 1$, we have $t = \sqrt{a^{2}+b^{2}}+a$ and when $u = 0$ we have $t = b$. Plugging these in, we have $(2)$ equal to
\begin{align*}
&\pi \left(\frac{(\sqrt{a^{2}+b^{2}}+a)^{2}}{4} + b^{2}\ln \left| \sqrt{a^{2}+b^{2}}+a\right| – \left(\frac{b^{4}}{4(\sqrt{a^{2}+b^{2}}+a)^{2}}\right) – \frac{b^{2}}{4} – b^{2}\ln\left|b\right| + \frac{b^{4}}{4b^{2}}\right)\\
&= \pi \left( \frac{(\sqrt{a^{2}+b^{2}}+a)^{2}}{4} – \left(\frac{b^{4}}{4(\sqrt{a^{2}+b^{2}}+a)^{2}}\right) + b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right) \right). \qquad (3)
\end{align*}
Now noticing the similarity of the first two terms, our intuition suggests this is easily simplified, so bringing this under common denominator we have $(3)$ equal to
\begin{align*}
&\pi\left( \frac{(\sqrt{a^{2}+b^{2}} + a)^{4} – b^{4}}{4(\sqrt{a^{2}+b^{2}} + a)^{2}} + b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right) \right)\\
&= \pi \left(\frac{\left(2a^{2} + 2a\sqrt{a^{2}+b^{2}} + b^{2}\right)^{2} -b^{4}}{4(\sqrt{a^{2}+b^{2}} + a)^{2}} + b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right) \right).
\end{align*}
Expanding again we have
\begin{align*}
&\pi \left(\frac{4a^{4} + 4a^{3}\sqrt{a^{2}+b^{2}} + 2a^{2}b^{2} + 4a^{3}\sqrt{a^{2}+b^{2}}+4a^{2}(a^{2}+b^{2}) + 2ab^{2}\sqrt{a^{2}+b^{2}}}{4\left(\sqrt{a^{2}+b^{2}}+a\right)^{2}}\right. \dots\\
&\left.\dots +\frac{2a^{2}b^{2} + 2ab^{2}\sqrt{a^{2}+b^{2}} + b^{4} – b^{4}}{4\left(\sqrt{a^{2}+b^{2}}+a\right)^{2}} +b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right)\right)\\
&= \pi \left( \frac{8a^{4} + 8a^{2}b^{2} + 8a^{3}\sqrt{a^{2}+b^{2}}+4ab^{2}\sqrt{a^{2}+b^{2}}}{4(\sqrt{a^{2}+b^{2}}+a)^{2}} + b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right)\right)\\
&= \pi \left(\frac{2a^{4}+2a^{2}b^{2}+2a^{3}\sqrt{a^{2}+b^{2}}+ab^{2}\sqrt{a^{2}+b^{2}}}{(\sqrt{a^{2}+b^{2}}+a)^{2}} + b^{2}\ln \left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right)\right). \qquad (4)
\end{align*}
Using our intuition we know that the only term that was canceled after expansion was $b^{4}$ so we should examine powers of $(\sqrt{a^{2}+b^{2}}+a$ before using more complicated methods. We know from earlier that $(\sqrt{a^{2}+b^{2}} + a)^{2} = 2a^{2} + 2a\sqrt{a^{2}+b^{2}} + b^{2}$ and by examining the numerator of the first term of $(4)$, we can see that $2a^{2}$, $2a\sqrt{a^{2}+b^{2}}$ and $b^{2}$ all share the common factor of $a\sqrt{a^{2}+b^{2}}$. Therefore, $a\sqrt{a^{2}+b^{2}}(\sqrt{a^{2}+b^{2}}+a)^{2}$ is a reasonable candidate for the correct factorization of the numerator. A quick check to confirm the cross-terms match shows that
\begin{align*}
a\sqrt{a^{2}+b^{2}}(\sqrt{a^{2}+b^{2}}+a)^{2} &= a\sqrt{a^{2}+b^{2}} (2a^{2} + 2a\sqrt{a^{2}+b^{2}} + b^{2})\\
&= 2a^{3}\sqrt{a^{2}+b^{2}} + 2a^{2}(a^{2}+b^{2}) + ab^{2}\sqrt{a^{2}+b^{2}}\\
&= 2a^{3}\sqrt{a^{2}+b^{2}} + 2a^{4} + 2a^{2}b^{2} + ab^{2}\sqrt{a^{2}+b^{2}}\\
&= 2a^{4} + 2a^{2}b^{2} + 2a^{3}\sqrt{a^{2}+b^{2}} +ab^{2}\sqrt{a^{2}+b^{2}}. \qquad (5)
\end{align*}
So, now that we have verified that the cross-terms match we can use $(5)$ and thus have $(4)$ equal to
\begin{align*}
\pi\left(\frac{a\sqrt{a^{2}+b^{2}}(\sqrt{a^{2}+b^{2}}+a)^{2}}{(\sqrt{a^{2}+b^{2}}+a)^{2}} + b^{2}\ln(\phi)\right)
&= a\pi\sqrt{a^{2}+b^{2}} + \pi b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right)
\end{align*}
which was what was to be shown.