Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$ Integrate 

$$I=\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$$

Substitution $x=\sqrt{\tan(u)}\rightarrow dx={\sec^2(u)\over 2\sqrt{\tan(u)}}du$
$x=1\rightarrow u={\pi\over 4}$
$x=0\rightarrow u=0$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-\tan^2(u)}\over 1+\tan^2(u)}\cdot {\sec^2(u)\over \sqrt{\tan(u)}}du$$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-\tan^2(u)\over \tan(u)}}du$$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{\cot(u)-\tan(u)}}du$$
Recall
$$\cot(u)-\tan(u)={\cos^2(u)-\sin^2(u)\over \sin(u)\cos(u)}=2\cot(2u)$$
Substitute back into I
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{2\cot(2u)}}du$$
$$I={\sqrt2\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{\cot(2u)}}du$$
Well I know that $$\int{\cos(2u)\over\sin(2u)}du={1\over 2}\ln(\sin(2u))+C$$ but
$$\int\sqrt{{\cos(2u)\over\sin(2u)}}du$$ I have not idea, so can anyone please give a hand? Thank.
 A: Let $y = x^2$ and $z = \sqrt{y^{-1} - y}$, we have
$$\begin{align}
\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4}dx
= & \int_0^1 \frac{x\sqrt{x^{-2} - x^2}}{x^2(x^{-2}+x^2)}dx
= \int_0^1 \frac{\sqrt{x^{-2} - x^2}}{(x^{-2}+x^2)}\frac{dx}{x}\\
= & \frac12 \int_0^1 \frac{\sqrt{y^{-1}-y}}{y^{-1}+y}\frac{dy}{y}
= -\frac12 \int_{y=0}^1 \frac{\sqrt{y^{-1}-y}}{y^{-1}+y}\frac{d(y^{-1} - y)}{y^{-1}+y}\\
= & \int_0^\infty \frac{z^2 dz}{z^4+4}
= \int_0^\infty \frac{z^2 dz}{(z^2+2)^2 - (2z)^2}
= \int_0^\infty \frac{z^2 dz}{(z^2-2z+2)(z^2+2z+2)}\\
= & \frac14 \int_0^\infty\left(\frac{z}{z^2-2z+2} - \frac{z}{z^2+2z+2}\right)dz\\
= & \frac14 \int_0^\infty \left[\frac12\log\left(\frac{z^2-2z+2}{z^2+2z+2}\right)'
+ \left(\frac{1}{z^2-2z+2} + \frac{1}{z^2+2z+2}\right)\right] dz
\end{align}
$$
The first piece contributes
$\displaystyle\;\frac18 \left[\log\left(\frac{z^2-2z+2}{z^2+2z+2}\right)\right]_0^\infty\;$ which clearly vanishes.
For the second piece, substitute $z$ by $-z$ in its second term, we get:
$$\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4}dx 
= \frac14 \int_{-\infty}^\infty\frac{dz}{z^2-2z+2} 
= \frac14 \int_{-\infty}^\infty\frac{dz}{(z-1)^2+1} 
= \frac{\pi}{4}$$
A: You were so close!
$$\begin{align}\int_0^{\frac{\pi}4}\sqrt{\cot2u}\,du&=\frac12\int_0^{\frac{\pi}2}\sqrt{\cot v}\,dv\tag{1}\\
&=\frac14\cdot2\int_0^{\frac{\pi}2}\cos^{1/2}v\sin^{-1/2}v\,dv\tag{2}\\
&=\frac14\text{B}\left(\frac34,\frac14\right)\tag{3}\\
&=\frac14\frac{\Gamma\left(\frac34\right)\Gamma\left(\frac14\right)}{\Gamma(1)}\tag{4}\\
&=\frac14\frac{\pi}{\sin\frac{\pi}4}\tag{5}\\
&=\frac{\pi}{2\sqrt2}\tag{6}\end{align}$$
EDIT: Some annotations:
$(1)$ Let $2u=v$
$(2)$ $\cot\theta=\frac{\cos\theta}{\sin\theta}$
$(3)$ Recognizing the trigonometric form of the Beta function
$(4)$ Using the relationship between the Beta and Gamma functions
$(5)$ The reflection formula for the Gamma function. We might equally have used the duplication formula
$$\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)=\sqrt{\pi}2^{1-2\left(\frac14\right)}\Gamma\left(2\left(\frac14\right)\right)=\sqrt{\pi}\sqrt2\sqrt{\pi}=\pi\sqrt2$$
$(6)$ Of course $\sin\frac{\pi}4=\frac1{\sqrt2}$. At this point we're home free and we can substitute the value of the integral into where you were stuck and arrive at
$$\int_0^1\frac{\sqrt{1-x^4}}{1+x^4}dx=\frac{\sqrt2}2\frac{\pi}{2\sqrt2}=\frac{\pi}4$$
A: Substitute $t^2=\frac{1-x^4}{2x^2}$. Then,
$dx= -\frac{2tx^3}{1+x^4}dt$ and $x^2=\sqrt{1+t^4}-t^2$
\begin{align}
\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx
=& \int_{0}^{\infty}{2\sqrt2 \ t^2 \over (\frac1{x^2}+x^2)^2}dt 
=\frac1{\sqrt2} \int_{0}^{\infty}{t^2 \over 1+t^4}dt
={\pi\over 4}
\end{align}
A: On the path of Achille Hui,
$\displaystyle J=\int_0^{\infty} \dfrac{x^2}{x^4+4}dx$
Perform the change of variable $y=\dfrac{x}{\sqrt{2}}$,
$\displaystyle J=\dfrac{\sqrt{2}}{2}\int_0^{\infty} \dfrac{x^2}{x^4+1}dx$
1)
Perform the change of variable $y=x^4$,
$\begin{align} J&=\dfrac{\sqrt{2}}{8}\int_0^{\infty} \dfrac{x^{-\tfrac{1}{4}}}{1+x}dx\\
&=\dfrac{\sqrt{2}}{8}\text{B}\left(\dfrac{3}{4},\dfrac{1}{4}\right)\\
&=\dfrac{\sqrt{2}}{8}\dfrac{\Gamma\left(\dfrac{3}{4}\right)\Gamma\left(\dfrac{1}{4}\right)}{\Gamma\left(1\right)}\\
&=\dfrac{\sqrt{2}}{8}\dfrac{\pi}{\sin\left(\tfrac{\pi}{4}\right)}\\
&=\dfrac{\sqrt{2}}{8}\dfrac{\pi}{\tfrac{\sqrt{2}}{2}}\\
&=\boxed{\dfrac{\pi}{4}}
\end{align}$
2) alternatively
$\displaystyle J=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty} \dfrac{x^2}{1+x^4}dx$
Perform the change of variable $y=\dfrac{1}{x}$,
$\displaystyle J=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty} \dfrac{1}{1+x^4}dx$
therefore,
$\displaystyle 2J=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty}\dfrac{1+\tfrac{1}{x^2}}{x^2+\tfrac{1}{x^2}}dx$
Observe that,
$x^2 +\dfrac{1}{x^2}=\left(x-\dfrac{1}{x}\right)^2+2$
Perform the change of variable $y=x-\dfrac{1}{x}$,
$\begin{align} 2J&=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty} \dfrac{1}{x^2+2}dx\\
&=\dfrac{\sqrt{2}}{4}\left[\frac{\arctan\left(\frac{x}{\sqrt{2}}\right)}{\sqrt{2}}\right]_{-\infty}^{+\infty}\\
&=\dfrac{\pi}{2}
\end{align}$
therefore,
$\boxed{J=\dfrac{\pi}{4}}$
A: Partial answer:
$$\begin{align*}
\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4} \, dx &= \sum_{k=0}^\infty (-1)^k \int_0^1 x^{4k} \sqrt{1-x^4} \, dx \\[1ex]
&= \sum_{k=0}^\infty \frac{(-1)^k}4 \int_0^1 x^{k-\frac34} (1-x)^{\frac12} \, dx & x\mapsto x^{\frac14} \\[1ex]
&= \sum_{k=0}^\infty \frac{(-1)^k}4 \operatorname{B}\left(k+\frac14, \frac32\right) \\[1ex]
&= \frac{\sqrt\pi}8 \sum_{k=0}^\infty (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)}
\end{align*}$$
The remaining sum has a value of $2\sqrt\pi$, I'm just not sure how to show it yet...
