Are there any other methods to integrate the following integral than the ones cited on the Wikipedia page? Maybe with a more elementary tool set?
$$\int_0^\infty e^{-x^2}\,dx$$
Addendum: I am not looking for proofs that involve double or triple integration.
Maybe this is what you are looking? It requires mostly Calculus I and II knowledge. Although, I am no really sure what you mean by "elementary tool set".
Start by evaluating: $$(1)\ \int_0^1\left(1-x^2\right)^n\,dx\qquad (2)\ \int_0^{+\infty}\frac1{\left(1+x^2\right)^n}\,dx$$ For $(1)$ let $x=\cos(u)$ and $dx=-\sin(u)\,du$, and by applying the reduction formula on $\int_0^{\pi/2}\sin^{2n+1}(u)\,du$, we obtain: $$\int_0^1\left(1-x^2\right)^n\,dx=\frac23\cdot\frac45\cdot\frac67\cdots\frac{2n}{2n+1}=\prod_{k=1}^{n}\frac{2k}{2k+1}$$ Similarly, for $(2)$ let $x=\cot(u)$ and $dx=-\csc^2(u)\,du$, and by applying the reduction formula on $\int_0^{\frac\pi2}\sin^{2n-2}(u)\,du$, we obtain: $$\int_0^{+\infty}\frac1{\left(1+x^2\right)^n}dx=\frac\pi2\cdot\frac12\cdot\frac34\cdot\frac56\cdots\frac{2n-3}{2n-2}=\frac\pi2\cdot\prod_{k=2}^{n}\frac{2k-3}{2k-2}$$ We can also show that: $$(3)\ 1-x^2\le e^{-x^2}, \quad \text{for $0\le x\le1$} $$ $$(4)\ e^{-x^2}\le \frac1{1+x^2}, \quad \text{for $x\geq 0$}$$ For $(3)$, let: $$f(x)= e^{-x^{2}}+x^{2} -1$$ $$f'(x)= 2x(1-e^{-x^2})$$ We can clearly see that: $$f'(x)>0 \quad \text{for $x>0$}$$ Using the result just obtained, and noticing that $f(0)=0$, we have: $$f(x)\geq 0=e^{-x^2}\geq1-x^2 \quad \text{for $x\geq 0$}$$ A similar argument can be made for $(4)$ using $g(x)=e^{x^2}-x^2-1$.
Thus, integrating the relations just proved: $$\int_0^1\left(1-x^2\right)^n\,dx \le \int_0^1 e^{-n{x}^2}\,dx \le \int_0^{+\infty} e^{-n{x}^2}\,dx \le \int_0^{+\infty}\frac1{\left(1+x^2\right)^n}\,dx \Rightarrow$$ $$\prod_{k=1}^{n}\frac{2k}{2k+1} \le \int_0^1 e^{-n{x}^2}\,dx \le \int_0^{+\infty} e^{-n{x}^2}\,dx \le \frac\pi2\cdot\prod_{k=2}^{n}\frac{2k-3}{2k-2}$$
Letting $t=x\sqrt{n}$ and $(1/\sqrt{n})\,dt=dx$, we obtain: $$\sqrt{n}\cdot\prod_{k=1}^{n}\frac{2k}{2k+1} \le \int_0^{\sqrt{n}} e^{-t^2}\,dt \le \int_0^{+\infty} e^{-t^2}\,dt \le \frac{\pi\sqrt{n}}2\cdot\prod_{k=2}^{n}\frac{2k-3}{2k-2}$$ Using the Squeeze Lemma: $$\lim_{n\to+\infty}\left(\sqrt{n}\cdot\prod_{k=1}^{n}\frac{2k}{2k+1}\right) \le \int_0^\infty e^{-t^2}\,dt \le \lim_{n\to+\infty}\left(\frac{\pi\sqrt{n}}2\cdot\prod_{k=2}^{n}\frac{2k-3}{2k-2}\right)$$ We can evaluate both limits by recalling the Wallis Product: $$\lim_{n\to+\infty}\prod_{k=1}^n \left(\frac{(2k)^2}{(2k+1)(2k-1)}\right)=\frac{\pi}{2}$$ For the first limit notice that: $$\begin{aligned} \prod_{k=1}^n\frac{2k}{2k+1}&=\frac{2\cdot 1}{2\cdot 1+1}\frac{2\cdot 2}{2\cdot 2+1}\frac{2\cdot 3}{2\cdot 3+1}\cdots \frac{2\cdot n}{2\cdot n+1}\\ &=2\cdot 1\frac{2\cdot 2}{2\cdot 1+1}\frac{2\cdot 3}{2\cdot 2+1}\cdots \frac{2\cdot n}{2\cdot (n-1)+1}\frac{1}{2n+1}\\ &=\frac{2}{2n+1}\prod_{k=2}^n\frac{2k}{2k-1}=\frac{1}{2n+1}\prod_{k=1}^n\frac{2k}{2k-1} \end{aligned}$$ By applying the relation just found to the Wallis product: $$\begin{aligned} \frac{\pi}{2}&=\lim_{n\to+\infty}\prod_{k=1}^n \left(\frac{(2k)^2}{(2k+1)(2k-1)}\right)\\ &=\lim_{n\to+\infty}\left(n\cdot\frac{2n+1}{n}\left(\prod_{k=1}^n\frac{2k}{2k+1}\right)^2\right)\\ \end{aligned}$$ And taking the square root to both sides: $$\sqrt{\frac{\pi}{2}}=\lim_{n\to+\infty}\sqrt{\frac{2n+1}{n}}\lim_{n\to+\infty}\left(\sqrt{n}\prod_{k=1}^n\frac{2k}{2k+1}\right)\Rightarrow$$ $$\lim_{n\to+\infty}\left(\sqrt{n}\prod_{k=1}^n\frac{2k}{2k+1}\right)=\frac{\sqrt{\pi}}{2}$$ A similar approach can be used for the second limit, leading to: $$\lim_{n\to+\infty}\left(\frac{\pi\sqrt{n}}2\cdot\prod_{k=2}^{n}\frac{2k-3}{2k-2}\right)=\frac\pi2\cdot\frac1{\sqrt\pi}=\frac{\sqrt{\pi}}{2}$$ Therefore: $$\int_0^{+\infty} e^{-t^2}\,dt=\boxed{\frac{\sqrt{\pi}}{2}}$$
