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.

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 $${1\over 4}\leq {1\over 3+\cos x}< {2\over 7}$$ 
$${1\over 6}\leq \int_0^1 {1-x^2\over 3+\cos x}dx<{4\over 21}<{2\over 9}$$
