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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.

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  • $\begingroup$ How does ye olde “replace $\cos(x)$ by exponentials” tricke work in this case? $\endgroup$ – Circonflexe May 12 '15 at 13:54
  • $\begingroup$ @Circonflexe can't say, maybe that's overkill sinc eI only want to find the floor. $\endgroup$ – RE60K May 12 '15 at 13:55
  • $\begingroup$ Use the reduction formula for Wallis' integrals. $\endgroup$ – Lucian May 12 '15 at 21:13
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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: $$\frac{\int_0^{\pi/2}\cos^{\alpha-1}x{\rm d}x}{\int_0^{\pi/2}\cos^{\alpha+1}x{\rm d}x}=\frac{\alpha+1}{\alpha}$$

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    $\begingroup$ Could it be that the result is actually $\frac{\alpha + 1}{\alpha}$? $\endgroup$ – PhoemueX May 12 '15 at 21:30
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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: $$ \int_0^{\pi/2}\cos^5 x\,dx=\frac8{15},\quad \int_0^{\pi/2}\cos^2 x\,dx=\frac\pi4 $$

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  • $\begingroup$ is all this intutive? I like rigorousity better. $\endgroup$ – RE60K May 12 '15 at 13:58
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    $\begingroup$ @ADG It can be made into a fully rigorous argument quite directly. $\endgroup$ – GPerez May 12 '15 at 13:59
  • $\begingroup$ so would you help me eliminating 2 as a possibility, cuz I can't see it now. $\endgroup$ – RE60K May 12 '15 at 14:00
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    $\begingroup$ @ADG Sign error, it's $\frac{\alpha}{\alpha+1}$. So here, you have $\left\lfloor \frac{\sqrt{13}+1}{\sqrt{13}}\right\rfloor$. $\endgroup$ – Daniel Fischer May 12 '15 at 14:16
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    $\begingroup$ @ADG I reserve the right to be too lazy for that. You can write it up if you want. $\endgroup$ – Daniel Fischer May 12 '15 at 14:18
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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 $a= \sqrt{13}-1$.

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  • $\begingroup$ no beta plz, edited the post. $\endgroup$ – RE60K May 12 '15 at 13:56
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    $\begingroup$ @ADG What do you mean? $\endgroup$ – Adhvaitha May 12 '15 at 13:56
  • $\begingroup$ (+1)plz don't use beta or gamma function, that's not my course level yet, sorry for not mentioning it before, but you can kee this here, that'll help other visiters, if you're looking for acceptance, maybe post another answer [sorry times3] $\endgroup$ – RE60K May 12 '15 at 13:58
  • $\begingroup$ @ADG Done. Below is another answer, which uses nothing but basic trigonometry. $\endgroup$ – Adhvaitha May 12 '15 at 14:09
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    $\begingroup$ "Recall that..." $\endgroup$ – djechlin May 12 '15 at 20:42
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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 $$1+\dfrac{\int_0^{\pi/2}\sin^2(x)\cos^a(x)dx}{\int_0^{\pi/2}\cos^{a+2}(x)dx} \in (1,2)$$

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  • $\begingroup$ @ADG "Better" is a poor choice of word. My other answer gave you the exact value and this answer rigorously gives you a bound to work with. $\endgroup$ – Adhvaitha May 12 '15 at 14:17
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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: $$\left\lfloor 1+\frac{1}{\sqrt{13}}\right\rfloor = \color{red}{1}.\tag{2}$$

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