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Prove that $$\frac{1}{6}<\int_0^1 \frac{1-x^2}{3+\cos(x)}dx < \frac{2}{9}. $$

I tried using known integral inequalities (Cauchy-Schwarz, Chebyshev) but I did not arrive at anything. Then I also tried considering functions of the form $$f(x) = \int_0^x \frac{1-t^2}{3+\cos(t)}dt - \frac{1}{6}$$ and then arrive at something using monotony, but still no answer.

I even tried to compute the integral using the substitution $\displaystyle t = \tan \left(\frac{x}{2} \right),$ but then I arrive at an integral of the form $ \displaystyle \int \frac{(\arctan(x))^2}{x^2+a}dx, a \in \mathbb{R}, $ which I do not know to compute.

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  • $\begingroup$ Please show your attempt through Cauchy-Schwarz since I am pretty sure it can prove a way sharper inequality than $\frac{3}{18}<I<\frac{4}{18}$. $\endgroup$ Commented Oct 7, 2018 at 11:06
  • $\begingroup$ @JackD'Aurizio My attempt using Cauchy-Schwarz was merely $\int_0^1 \frac{1-x^2}{3+\cos(x)} dx \leq \sqrt{\int_0^1{(1-x^2)^2}dx \cdot \int_0^1{\frac{1}{(3+\cos(x))^2}}dx}$ but I did not derive the upper bound from this. $\endgroup$
    – C_M
    Commented Oct 7, 2018 at 11:14
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    $\begingroup$ An efficient application of Cauchy-Schwarz to $\int_{0}^{1}f(x)\,dx$ comes from writing $f(x)$ as $g(x)\cdot h(x)$ where $g(x)$ and $h(x)$ have an approximately constant ratio on $(0,1)$. It is not the case for $1-x^2$ and $\frac{1}{3+\cos x}$. $\endgroup$ Commented Oct 7, 2018 at 11:20

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For any $x\in(0,1)$

$$ \frac{1-x^2}{3+\cos x}-\frac{1-x^2}{4} = \frac{(1-x^2)\sin^2\frac{x}{2}}{2(3+\cos x)}\in\left[0,\frac{x^2(1-x)}{12}\right] $$ hence the given integral is bounded between $\frac{1}{6}=\int_{0}^{1}\frac{1-x^2}{4}\,dx$ and $\frac{1}{6}+\frac{1}{144}$.

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  • $\begingroup$ That's a nice approach! Thank you. $\endgroup$
    – C_M
    Commented Oct 7, 2018 at 11:21
  • $\begingroup$ I can see $${(1-x^2)\sin^2(x/2)\over2(3+\cos x)}\le{(1-x^2)x^2\over24}$$ (which gives ${1\over6}+{1\over180}$ as an upper bound), but I don't see how you got to $x^2(1-x)/12$. Am I missing something easy and obvious? $\endgroup$ Commented Oct 7, 2018 at 13:52
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Hint: Since ${\pi\over 3}>1\geq x$ we have $\cos x > {1\over 2}$ so $$3+{1\over 2}<3+\cos x\leq 4$$ so $${1\over 4}\leq {1\over 3+\cos x}< {2\over 7}$$

$${1\over 6}\leq \int_0^1 {1-x^2\over 3+\cos x}dx<{4\over 21}<{2\over 9}$$

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This is just a minor variant on greedoid's answer. It suffices to note that $1\lt\pi/2$, so that $0\le\cos x\le1$ for $0\le x\le1$, and thus

$${1\over4}\le{1\over3+\cos x}\le{1\over3}$$

Consequently, since $\int_0^1(1-x^2)dx={2\over3}$, we have

$${1\over6}=\int_0^1{1-x^2\over4}dx\le\int_0^1{1-x^2\over3+\cos x}dx\le\int_0^1{1-x^2\over3}dx={2\over9}$$

The stronger inequality, $1\lt\pi/3$, in greedoid's answer gives, of course, a stronger upper bound (and Jack D'Aurizio's answer gives an even stronger upper bound). But if all you need is $2/9$, the inequality $1\lt\pi/2$ is good enough.

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