Integral comparison question

Question

The question wants me to prove that $$\frac { \pi }{ 6 } <\int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { 4-{ x }^{ 2 }{ -x }^{ 3 } } } < } \frac { \pi }{ 4\sqrt { 2 } } $$

My first attemp was to show that the integrand is on the interval $(\frac { \pi }{ 6 },\frac { \pi }{ 4\sqrt {2} })$. Unfortunately, I couldn't prove that.

Thanks in advance.

Answer 1

Hint: The integrand is $\ge \frac{1}{\sqrt{4-x^2}}$ and $\le \frac{1}{\sqrt{4-2x^2}}$. Now integrate these, making the obvious substitutions.