How to prove $\int_0^{3/\pi}\frac{\cos^2x}{\sqrt{9-\cos^{3}x}}dx\geq \frac{\pi}{36}$ 
Prove that $\displaystyle\int_0^{3/\pi}\frac{\cos^2x}{\sqrt{9-\cos^{3}x}}dx\geq \frac{\pi}{36}$.

I'm calculating its lowest and greatest bound and multiplying it by $\frac{\pi}{6}$ but I'm getting completely different answers than $\dfrac{\pi}{36}$.
What am I exactly doing wrong? I'm beginning to think the question has a typo in it.
 A: Here's one way you can prove it.  First, convince yourself that the integrand is positive over the whole domain of integration and that over this domain, $0 \leq \sin(x) \leq 1$.  Next, show that
$$\int_{0}^{3/\pi}\frac{\cos^2(x)}{\sqrt{9-\cos^3(x)}}dx \geq \int_{0}^{3/\pi}\frac{\cos^2(x)}{\sqrt{9-\cos^3(x)}}\sin(x)dx.$$
For the integral on the RHS, show that a $u$ substitution with $u = \cos(x)$ gives you
$$\int_{0}^{3/\pi}\frac{\cos^2(x)}{\sqrt{9-\cos^3(x)}}\sin(x)dx = \int_{1}^{\cos(3/\pi)}\frac{-u^2}{\sqrt{9-u^3}}du.$$
Show that the integral on the RHS is bigger than $\pi/36$ (this integral is easier than the original: do another substitution with $v = u^3$).
Combining these steps proves the desired result.
A: What you could do is to write
$$\frac{\cos^2x}{\sqrt{9-\cos^{3}x}}=\frac 1{\sqrt \pi}\sum_{n=0}^\infty  \frac{ \Gamma \left(n+\frac{1}{2}\right)}{3^{2 n+1}\,\Gamma(n+1)}\big[\cos (x)\big]^{(3 n+2)}$$ Over this range $\left\{0,\frac{3}{\pi }\right\}$ or $\left\{0,\frac{\pi }{3}\right\}$, all
$$\int_0^{a \leq \frac \pi 2} \big[\cos (x)\big]^{(3 n+2)}\,dx \quad > 0$$ Now, computing (for the worst case)
$$I_n=\frac 1{\sqrt \pi} \int_0^{\frac{\pi }{3}}\frac{ \Gamma \left(n+\frac{1}{2}\right)}{3^{2 n+1}\,\Gamma(n+1)}\big[\cos (x)\big]^{(3 n+2)}$$
$$I_0=\frac{\pi }{18}+\frac{1}{8 \sqrt{3}}$$ is already much larger than $\frac{\pi }{36}$
