<s>[INST] 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!!
 [/INST] 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.
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