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

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.
