TITLE: integral $\int_0^1\frac{1}{\sqrt{1+x^4}}\text{d}x$
QUESTION [6 upvotes]: How to prove that $$0.78<\int_0^1\frac{1}{\sqrt{1+x^4}}\text{d}x<0.93$$

approch: $x^2<\sqrt{1+x^4}<x^2+\frac{1}{2x^2}$
Any hint would be appreciated.

REPLY [2 votes]: Even though my previous answer solved the problem I hated it because there is no way when the problem was posed that it could have been intended to require $9$ terms of a slowly converging Taylor series. If you let $x^2=\tan(\frac{\theta}2)$ then
$$\int_0^1\frac1{\sqrt{1+x^4}}dx=\frac{\sqrt2}4\int_0^{\frac{\pi}2}\frac{d\theta}{\sqrt{\sin\theta}}=\frac{\sqrt2}8\text{B}\left(\frac14,\frac12\right)=\frac{\left(\Gamma\left(\frac14\right)\right)^2}{8\sqrt{\pi}}\approx0.927037$$
Since this is so close to the desired upper bound, a really good estimate must be made. Just recently I showed that
$$\begin{align}\int\frac{du}{u^4+1}= & -\frac1{4\sqrt2}\ln\left(u^2-\sqrt2u+1\right)+\frac1{2\sqrt2}\tan^{-1}\left(\sqrt2u+1\right)+\\
&\frac1{4\sqrt2}\ln\left(u^2+\sqrt2u+1\right)+\frac1{2\sqrt2}\tan^{-1}\left(\sqrt2u-1\right)+C\\
&=f(u)+C\end{align}$$
We can leverage this result to get estimates for the integral in question as follows:
$$\begin{align}\int_0^1\frac1{\sqrt{1+x^4}}dx & >\int_0^1\frac1{\sqrt{1+x^4+\frac14x^8}}dx=\int_0^1\frac1{1+\frac12x^4}dx\\
 & =2^{\frac14}\int_0^{\left(\frac12\right)^{\frac14}}\frac{du}{u^4+1}=2^{\frac14}\left(f\left(\left(\frac12\right)^{\frac14}\right)-f(0)\right)\\
 & \approx0.920788\end{align}$$
We have the good estimate
$$\frac1{\sqrt{1+x^4}}-1=\frac{1-\sqrt{1+x^4}}{\sqrt{1+x^4}}=\frac{-x^4}{\sqrt{1+x^4}+1+x^4}$$
So
$$\begin{align}\int_0^1\frac1{\sqrt{1+x^4}}dx & =\int_0^1\left(1-\frac{x^4}{\sqrt{1+x^4}+1+x^4}\right)dx\\
 & <1-\int_0^1\frac{x^4}{\sqrt{1+x^4+\frac14x^8}+1+x^4}dx\\
 & =1-\int_0^1\frac{x^4}{2+\frac32x^4}dx=1-\frac12\left(\frac43\right)^{\frac54}\int_0^{\left(\frac34\right)^{\frac14}}\frac{u^4}{1+u^4}du\\
 & =1-\frac12\left(\frac43\right)^{\frac54}\left(\left(\frac34\right)^{\frac14}-f\left(\left(\frac34\right)^{\frac14}\right)+f(0)\right)\\
 & \approx0.927798\end{align}$$
I think this approach is more consistent with the spirit of the problem.  
EDIT: There is also a way of estimating that integral in the upper bound without resorting to that intimidating interal used above. Since
$$\frac{x^4}{\sqrt{1+x^4}+1+x^4}\ge0$$
and
$$0\le\frac18(15x-10x^3+3x^5)\le1$$
for $0\le x\le1$, we can make the substitution $x^2=\sinh\theta$ to show that
$$\begin{align}\int_0^1\frac{x^4}{\sqrt{1+x^4}+1+x^4}dx & >\frac18\int_0^1(15x-10x^3+3x^5)\frac{x^4}{\sqrt{1+x^4}+1+x^4}dx\\
 & =\frac18\int_0^{\sinh^{-1}(1)}\frac{\left(15-10\sinh\theta+3\sinh^2\theta\right)\sinh^2\theta}{\cosh\theta+\cosh^2\theta}\cdot\frac12\cosh\theta\,d\theta\\
 & =\frac1{16}\int_0^{\sinh^{-1}(1)}\frac{\left(15-10\sinh\theta+3\sinh^2\theta\right)\left(\cosh^2\theta-1\right)}{1+\cosh\theta}\,d\theta\\
 & =\frac1{16}\int_0^{\sinh^{-1}(1)}\left(15-10\sinh\theta+3\sinh^2\theta\right)\left(\cosh\theta-1\right)\,d\theta\\
 & =\frac1{16}\left[15\sinh\theta-5\sinh^2\theta+\sinh^3\theta-15\theta+10\cosh\theta-\frac32\left(\frac12\sinh2\theta-\theta\right)\right]_0^{\sinh^{-1}(1)}\\
 & =\frac1{16}\left[15-5+1-15\sinh^{-1}(1)+10\left(\sqrt2-1\right)-\frac32\left(\sqrt2-\sinh^{-1}(1)\right)\right]\\
 & =\frac1{16}\left[1-\frac{27}2\ln\left(1+\sqrt2\right)+17/2\sqrt2\right]\\
 & \approx0.070142\end{align}$$
So then
$$\begin{align}\int_0^1\frac1{\sqrt{1+x^4}}dx & =1-\int_0^1\frac{x^4}{\sqrt{1+x^4}+1+x^4}dx\\
 & <1-\frac1{16}\left[1-\frac{27}2\ln\left(1+\sqrt2\right)+17/2\sqrt2\right]\approx0.929858\end{align}$$
Not as good an estimate, but (just barely) good enough, and a much easier integral.