Integrals involving reciprocal square root of a quartic For two integration below, what is the ratio of them? 
$$I_1 = \int_0^1\frac{dt}{\sqrt{1-t^4}}$$  $$I_2 = \int_0^1 \frac{dt}{\sqrt{1+t^4}}$$
What is the ration $\frac{I_1}{I_2}$?
I do not have any thoughts for solving this question so could anyone give me some hints?
Thank you!!
 A: The first one:let $t^2=\sin{x}$,then
$$I_{1}=\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{\sqrt{\sin{x}}}$$
The second one: let
$t^2=\tan{x}$
then
$$I_{2}=\dfrac{1}{2}\int_{0}^{\frac{\pi}{4}}\dfrac{dx}{\sqrt{\sin{x}\cos{x}}}=\dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{\sqrt{\sin{x}}}$$
so
$$\dfrac{I_{1}}{I_{2}}=\sqrt{2}$$
A: Hint:
$$I_1(x)=\int_x^1\frac{dt}{\sqrt{1-t^4}}=\frac{1}{\sqrt{2}}\int_{0}^{\arccos x}\frac{du}{\sqrt{1-\frac{1}{2}\sin^2u}}$$
$$I_2(x)=\int_0^x \frac{dt}{\sqrt{1+t^4}}=\frac{1}{2}\int_{0}^{\arccos\frac{1-x^2}{1+x^2}}\frac{du}{\sqrt{1-\frac{1}{2}\sin^2u}}$$
and since $$\arccos 0 = \arccos \frac{1-1^2}{1+1^2}$$
$$\frac{I_1(0)}{I_2(1)}=\sqrt{2}$$
From the hint the substitutions should be pretty straightforward.

Added due to the comment below:
The first one: let $t=\cos u$, $u=\arccos t$.
$$\frac{dt}{\sqrt{1-t^4}}=\frac{-\sin udu}{\sqrt{1-\cos^4 u}}=\frac{-\sin udu}{\sqrt{\sin^2 u(1+\cos^2 u)}}$$
and you get the RHS almost immediately.
The second one: let $t=\tan\frac{u}{2}$, $t^2=\frac{1}{\cos^2 \frac{u}{2}}-1=\frac{1-\cos u}{1+\cos u}$, $u=\arccos \frac{1-t^2}{1+t^2}$. Plug it in, and get the RHS also (almost) immediately.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\color{#00f}{\large I_{1}}&\equiv\int_{0}^{1}{\dd t \over \root{1 - t^{4}}}
=\int_{0}^{1}\pars{1 - t}^{-1/2}\,{1 \over 4}\,t^{-3/4}\,\dd t
={1 \over 4}\int_{0}^{1}t^{-3/4}\pars{1 - t}^{-1/2}\,\dd t
={1 \over 4}\,{\rm B}\pars{{1 \over 4},\half}
\\[3mm]&={1 \over 4}\,{\Gamma\pars{1/4}\Gamma\pars{1/2} \over \Gamma\pars{3/4}}
={1 \over 4}\,{\Gamma\pars{1/4}\root{\pi} \over \pi/\bracks{\Gamma\pars{1/4}\sin\pars{\pi/4}}}=
\color{#00f}{\large{1 \over 4\root{2\pi}}\,\Gamma^{\,2}\pars{1 \over 4}}
\approx 1.3110
\end{align}

${\rm B}\pars{x,y}$ and $\Gamma\pars{z}$ are the Beta and Gamma functions, respectively. We used well known properties of them.

\begin{align}
\color{#00f}{\large I_{2}}&\equiv\int_{0}^{1}{\dd t \over \root{1 + t^{4}}}
=\half\int_{0}^{\infty}{\dd t \over \root{1 + t^{4}}}
\end{align}
Lets $\ds{x \equiv {1 \over 1 + t^{4}}\quad\iff\quad t = \pars{{1 \over x} - 1}^{1/4}}$
\begin{align}
\color{#00f}{\large I_{2}}&=\half\int_{0}^{\infty}{\dd t \over \root{1 + t^{4}}}
=\half\int_{1}^{0}x^{1/2}\,{1 \over 4}\,\pars{1 - x \over x}^{-3/4}
\pars{-\,{\dd x \over x^{2}}}
\\[3mm]&={1 \over 8}\int_{0}^{1}x^{-3/4}\pars{1 - x}^{-3/4}\,\dd x
={1 \over 8}\,{\rm B}\pars{{1 \over 4},{1 \over 4}}
={1 \over 8}\,{\Gamma\pars{1/4}\Gamma\pars{1/4} \over \Gamma\pars{1/2}}
\\[3mm]&=\color{#00f}{\large{1 \over 8\root{\pi}}\,\Gamma^{\,2}\pars{1 \over 4}}
\approx 0.9270
\end{align}

$$
\color{#00f}{\large{I_{1} \over I_{2}}} = {1/\pars{4\root{2}} \over 1/8}
=\color{#00f}{\large\root{2}}
$$

A: As shown in this answer,
$$
\int_0^1t^{\alpha-1}\,(1-t)^{\beta-1}\,\mathrm{d}t
=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}
$$
and
$$
\int_0^\infty t^{\alpha-1}\,(1+t)^{-\beta}\,\mathrm{d}t
=\frac{\Gamma(\alpha)\Gamma(\beta-\alpha)}{\Gamma(\beta)}
$$
Substituting $t\mapsto1/t$ and then $t\mapsto t^{1/4}$, we get
$$
\begin{align}
\int_0^1\frac{\mathrm{d}t}{\sqrt{1+t^4}}
&=\int_1^\infty\frac{\mathrm{d}t}{\sqrt{1+t^4}}\\
&=\frac12\int_0^\infty\frac{\mathrm{d}t}{\sqrt{1+t^4}}\\
&=\frac18\int_0^\infty t^{-3/4}(1+t)^{-1/2}\,\mathrm{d}t\\
&=\frac18\frac{\Gamma(1/4)^2}{\Gamma(1/2)}
\end{align}
$$
Furthermore,
$$
\begin{align}
\int_0^1\frac{\mathrm{d}t}{\sqrt{1-t^4}}
&=\frac14\int_0^1t^{-3/4}(1-t)^{-1/2}\,\mathrm{d}t\\
&=\frac14\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)}
\end{align}
$$
As shown in this answer, $\Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)$. Therefore,
$$
\begin{align}
\frac{\displaystyle\int_0^1\frac{\mathrm{d}t}{\sqrt{1-t^4}}}{\displaystyle\int_0^1\frac{\mathrm{d}t}{\sqrt{1+t^4}}}
&=2\frac{\Gamma(1/2)^2}{\Gamma(1/4)\Gamma(3/4)}\\
&=2\frac{\pi\csc(\pi/2)}{\pi\csc(\pi/4)}\\[12pt]
&=\sqrt2
\end{align}
$$
