See this: $$\newcommand{\b}[1]{\left(#1\right)}\left\lfloor\frac{\displaystyle\int_0^{\pi/2}\cos^{\sqrt{13}-1}x{\rm d}x}{\displaystyle\int_0^{\pi/2}\cos^{\sqrt{13}+1}x{\rm d}x}\right\rfloor$$ Well I could only think of Cauchy-Schwarz, but it is also not fitting.
All calculations by hand, no Beta And that implicitly implies not using others like Gamma too.
As Daniel Fischer helped, using Integration by Parts: $$\int_0^{\pi/2}\cos^{\alpha+1}x{\rm d}x=\int_0^{\pi/2}\cos^{\alpha}x\cos x{\rm d}x=\cos^{\alpha}x\sin x\Bigg|_0^{\pi/2}+\alpha\int_0^{\pi/2}\cos^{\alpha-1}x\sin^2 x{\rm d}x=0+\alpha\int_0^{\pi/2}\cos^{\alpha-1}x{\rm d}x-\alpha\int_0^{\pi/2}\cos^{\alpha+1}x{\rm d}x$$ So: $$\frac{\int_0^{\pi/2}\cos^{\alpha-1}x{\rm d}x}{\int_0^{\pi/2}\cos^{\alpha+1}x{\rm d}x}=\frac{\alpha+1}{\alpha}$$
We needn't compute the integrals. Because $0\leq \cos x\leq1$ on $[0,\pi/2]$, the higher power of $\cos$ is less then the lower one, hence the floor is at least 1. And it is easy to estimate, that it cannot be 2 or more: $$ \int_0^{\pi/2}\cos^5 x\,dx=\frac8{15},\quad \int_0^{\pi/2}\cos^2 x\,dx=\frac\pi4 $$
Recall that $$\int_0^{\pi/2} \cos^a(x)dx = \dfrac12 \beta(1/2,(a+1)/2) = \dfrac12 \dfrac{\Gamma(1/2)\Gamma((a+1)/2)}{\Gamma((a+2)/2)}$$ where $\beta(x,y)$ is the beta function and $\Gamma(z)$ is the gamma function.
Hence, the value of your expression without the floor is $$\dfrac{\Gamma((a+1)/2)}{\Gamma((a+2)/2)} \cdot \dfrac{\Gamma((a+4)/2)}{\Gamma((a+3)/2)} = \dfrac{(a+4)/2-1}{(a+3)/2-1} = \dfrac{a+2}{a+1}=1+\dfrac1{\sqrt{13}}$$ where $a= \sqrt{13}-1$.
Another way is to write $$\cos^{a}(x) = \cos^{a+2}(x) + \sin^2(x)\cos^a(x)$$ Hence, we are after $$1+\dfrac{\int_0^{\pi/2}\sin^2(x)\cos^a(x)dx}{\int_0^{\pi/2}\cos^{a+2}(x)dx}$$ Clearly, $$\dfrac{\int_0^{\pi/2}\sin^2(x)\cos^a(x)dx}{\int_0^{\pi/2}\cos^{a+2}(x)dx} > 0$$ We will now show that $$\dfrac{\int_0^{\pi/2}\sin^2(x)\cos^a(x)dx}{\int_0^{\pi/2}\cos^{a+2}(x)dx} < 1$$ which on rearranging gives us that, it suffices to show $$ \text{ or }\int_0^{\pi/2}\cos(2x) \cos^{a}(x)dx > 0$$ We will show that the last statement is indeed true. \begin{align} \int_0^{\pi/2}\cos(2x) \cos^{a}(x)dx & = \int_0^{\pi/4}\cos(2x) \cos^{a}(x)dx + \int_{\pi/4}^{\pi/2}\cos(2x)\cos^a(x)dx\\ & = \int_0^{\pi/4}\cos(2x) \cos^{a}(x)dx + \int_{0}^{\pi/4}\cos(2(\pi/2-x))\cos^a(\pi/2-x)dx\\ & = \int_0^{\pi/4}\cos(2x) \cos^{a}(x)dx + \int_{0}^{\pi/4}\cos(\pi-2x)\sin^a(x)dx\\ & = \int_0^{\pi/4}\cos(2x) \cos^{a}(x)dx - \int_{0}^{\pi/4}\cos(2x)\sin^a(x)dx\\ & = \int_0^{\pi/4}\cos(2x) \left(\cos^{a}(x) - \sin^a(x) \right)dx > 0 \end{align} where the last inequality is true, since $\cos(x) > \sin(x)$ on the interval $[0,\pi/4)$. Hence, again we have $$1+\dfrac{\int_0^{\pi/2}\sin^2(x)\cos^a(x)dx}{\int_0^{\pi/2}\cos^{a+2}(x)dx} \in (1,2)$$
Integration by parts gives, for any $\eta>0$: $$\int_{0}^{\pi/2}\cos^{\eta+1}(x)\,dx = \frac{\eta}{\eta+1}\int_{0}^{\pi/2}\cos^{\eta-1}(x)\,dx \tag{1}$$ hence by taking $\eta=\sqrt{13}$ we just need to compute: $$\left\lfloor 1+\frac{1}{\sqrt{13}}\right\rfloor = \color{red}{1}.\tag{2}$$
