Proving $\frac12 < \int_0^1 \frac1{\sqrt{4-x+x^3}} dx < \frac{ \pi }6$ I have a question where I have to show 
$$\frac12 < \int_0^1 \frac1{\sqrt{4-x+x^3}} dx < \frac{ \pi }6 \approx 0.52359$$
using the result 
$$\frac12 < \int_0^{1/2} \frac{1}{\sqrt{1-x^{2n}}} dx \leq \frac{ \pi }6 ~, \quad \forall~n \in \mathbb{N}$$ which I have already proved, where the right equality can be attained only at $n = 1$.
I assume that I need to find a way to reduce the first expression to the second, but I cannot work out how to do this. What step can I do to help me show this?
Thank you!
 A: Noting that
$$
\int_0^1\frac{\mathrm{d}x}2\lt\int_0^1\frac{\mathrm{d}x}{\sqrt{4-x+x^3}}\lt\int_0^1\frac{\mathrm{d}x}{\sqrt{4-2x+2x^2}}
$$
we get
$$
\frac12\lt\int_0^1\frac{\mathrm{d}x}{\sqrt{4-x+x^3}}\lt\sqrt2\sinh^{-1}\left(\vcenter{\tfrac1{\sqrt7}}\right)
$$
and while it is true that $\sqrt2\sinh^{-1}\left(\vcenter{\frac1{\sqrt7}}\right)\lt\frac\pi6$, that is not immediately apparent.
However, using the first $3$ terms of the series for $\sinh^{-1}(x)$, we get that $\sqrt2\sinh^{-1}\left(\vcenter{\tfrac1{\sqrt7}}\right)\lt\sqrt{\frac27}\frac{5749}{5880}$ and $\frac{111}{212}\lt\frac\pi6$ follows from the pretty well known underestimate $\frac{333}{106}\lt\pi$. Therefore,
$$
\sqrt2\sinh^{-1}\left(\vcenter{\tfrac1{\sqrt7}}\right)\lt\sqrt{\frac27}\frac{5749}{5880}\lt\frac{111}{212}\lt\frac\pi6
$$

Noting that
$$
\int_0^1\frac{\mathrm{d}x}2\lt\int_0^1\frac{\mathrm{d}x}{\sqrt{4-x+x^3}}\lt\int_0^1\frac{\mathrm{d}x}{2-x/2+x^2/2}
$$
we get
$$
\frac12\lt\int_0^1\frac{\mathrm{d}x}{\sqrt{4-x+x^3}}\lt\frac8{\sqrt{15}}\tan^{-1}\left(\vcenter{\tfrac1{\sqrt{15}}}\right)
$$
and while it is true that $\frac8{\sqrt{15}}\tan^{-1}\left(\vcenter{\tfrac1{\sqrt{15}}}\right)\lt\frac\pi6$, that is not immediately apparent.
However, using the first $3$ terms of the series for $\tan^{-1}(x)$, we get that $\frac8{\sqrt{15}}\tan^{-1}\left(\vcenter{\tfrac1{\sqrt{15}}}\right)\lt\frac{2936}{5625}$ and $\frac{111}{212}\lt\frac\pi6$ follows from the pretty well known underestimate $\frac{333}{106}\lt\pi$. Therefore,
$$
\frac8{\sqrt{15}}\tan^{-1}\left(\vcenter{\tfrac1{\sqrt{15}}}\right)\lt\frac{2936}{5625}\lt\frac{111}{212}\lt\frac\pi6
$$
