Inductively prove that this sequence of integrals is bounded. EDIT: I have an attempted solution to this in a post below, it is very long, but still incomplete.
EDIT:Alright, I've pretty much almost finished my solution, but my biggest problem is the 2nd integral, rightmost inequality. I cannot find a way to evaluate it and compare it to the n=1 case for the upper integral, which should give me what I need.
EDIT: still stuck!
EDIT: 10 days later, I still need help with this..
Can someone please help me with this? I'm not even sure where to go from here.
Here is the problem:

Prove that
$$\frac{1}{2}<\int_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^{2n}}}\le0.52359,$$
for any integer $n\ge1$,
And that 
$$\frac{1}{2}<\int_{0}^{1}\frac{dx}{\sqrt{4-x+x^{3}}}\le0.52359$$

Now I can do this for the $n=1$ case, as obviously this simply integrates to $\arcsin{\frac{1}{2}}$ giving $\frac{\pi}{6}$, but I'm completely stumped on any doing any further for this question.
All help very appreciated, thank you for your time.
 A: In the interest of completeness, and at the risk of duplication, I will include my solution of the first part of the question.
$$
\frac12\lt\int_0^{1/2}\frac{\mathrm{d}x}{\sqrt{1-x^{2n}}}
$$
since $1\lt\dfrac1{\sqrt{1-x^{2n}}}$ on $\left(0,\frac12\right)$.
Furthermore, since $\dfrac1{\sqrt{1-x^{2n}}}\le\dfrac1{\sqrt{1-x^2}}$ on $\left(0,\frac12\right)$ for $n\ge1$, we have
$$
\begin{align}
\int_0^{1/2}\frac{\mathrm{d}x}{\sqrt{1-x^{2n}}}
&\le\int_0^{1/2}\frac{\mathrm{d}x}{\sqrt{1-x^2}}\\
&=\arcsin\left(\tfrac12\right)-\arcsin(0)\\
&=\frac\pi6\\[6pt]
&\doteq0.5235987755983
\end{align}
$$

Second Part
$$
\frac12<\int_0^1\frac{\mathrm{d}x}{\sqrt{4-x+x^3}}
$$
since $\frac12\lt\dfrac1{\sqrt{4-x+x^3}}$ on $(0,1)$.
Furthermore, $4-x+x^3$ is convex on $[0,1]$. Looking at its tangents at $0$ and $1$, and its minimum on $[0,1]$,
$\hspace{3cm}$
we get
$$
\frac1{\sqrt{4-x+x^3}}\le\min\left(\frac1{\sqrt{4-x}},\frac1{\sqrt{4-2/\sqrt{27}}},\frac1{\sqrt{2+2x}}\right)
$$
Therefore,
$$
\begin{align}
&\int_0^1\frac{\mathrm{d}x}{\sqrt{4-x+x^3}}\\
&\le\int_0^{2/\sqrt{27}}\frac{\mathrm{d}x}{\sqrt{4-x}}
+\int_{2/\sqrt{27}}^{1-1/\sqrt{27}}\frac{\mathrm{d}x}{\sqrt{4-2/\sqrt{27}}}
+\int_{1-1/\sqrt{27}}^1\frac{\mathrm{d}x}{\sqrt{2+2x}}\\
&=2\left(2-\sqrt{4-2/\sqrt{27}}\right)
+\frac{1-3/\sqrt{27}}{\sqrt{4-2/\sqrt{27}}}
+\left(2-\sqrt{4-2/\sqrt{27}}\right)\\
&=6-\frac{11-3/\sqrt{27}}{\sqrt{4-2/\sqrt{27}}}\\[6pt]
&\doteq0.5182655150624\\[18pt]
&\lt0.5235987755983
\end{align}
$$

On Strict Inequality
Suppose we know that $f(x)\le g(x)$ on $[a,b]$ and that $f(x)\lt g(x)$ on $(a,b)$. If $f$ and $g$ are continuous and $a\lt b$, then we can pick any $a'$ and $b'$ so that $a\lt a'\lt b'\lt b$. $g-f$ is continuous on the compact set $[a',b']$ so it attains its minimum on $[a',b']$ that is
$$
g(x)-f(x)\ge m\gt0\text{ for }x\in[a',b']\subset(a,b)
$$
Therefore,
$$
\begin{align}
\int_a^b(g(x)-f(x))\,\mathrm{d}x
&\ge\int_{a'}^{b'}(g(x)-f(x))\,\mathrm{d}x\\
&\ge\int_{a'}^{b'}m\,\mathrm{d}x\\[6pt]
&=m(b'-a')\\[12pt]
&\gt0
\end{align}
$$
so that we have the strict inequality
$$
\int_a^bf(x)\,\mathrm{d}x\lt\int_a^bg(x)\,\mathrm{d}x
$$
A: First note that the leftmost inequality is trivial, since $\frac{1}{\sqrt{1-x^{2n}}}\geq 1$.
Now note that for $x<1$, $x^{2(n+1)} = x^2x^{2n} \leq x^{2n}$. Hence $\frac{1}{\sqrt{1-x^{2(n+1)}}}\leq\frac{1}{\sqrt{1-x^{2n}}}$ for all $x<1$. Thus $\displaystyle\int_0^\frac{1}{2}\frac{\mathrm{d}x}{\sqrt{1-x^{2(n+1)}}} \leq \displaystyle\int_0^\frac{1}{2}\frac{\mathrm{d}x}{\sqrt{1-x^{2n}}}$. The rightmost inequality now follows by induction.
