calculus - Solve in terms of the Gamma function

Show:
\begin{align*}
\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm d x
&=\frac{\sqrt \pi}{4}\left(\frac{\Gamma \left(\frac14\right)}{\Gamma\left(\frac34\right)}-4\frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)
\end{align*}

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First attempt:
\begin{align*}
\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm d x
&=\int\limits_0^1{\left(1-x^2\right)}^{1/2}{\left(1+x^2\right)}^{-1/2}\,\mathrm d x\\
\qquad\text{let } x=u^{1/2}\\
\,\mathrm d x =\frac 1 2 u ^{-1/2}\,\mathrm d u\\
&=\frac 1 2 \int\limits_0^1{\left(1-u\right)}^{1/2}{\left(1+u\right)}^{-1/2}u^{-1/2}\,\mathrm d u\\\qquad \text{let }1-u= y\\ u= 1-y,\\
\,\mathrm du=-\,\mathrm dy\\

&=-\frac 1 2 \int_0^1y^{1/2}\left(2-y\right)^{-1/2}(1-y)^{-1/2}\,\mathrm d y\\
\\&=\quad\color{orange}{????}
\\
\end{align*}

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Second attempt:
\begin{align*}
\\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm dx\\

x^2=\cos(2\theta)\\
\mathrm dx=\frac{-\sin(2\theta)}x\,\mathrm d\theta\\
x=0\rightarrow\theta=\pi\\
x=1\rightarrow\theta=0\\
&=\int\limits_\pi^0\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}\cdot\frac{-\sin(2\theta)}{\sqrt{\cos(2\theta)}}\,\mathrm d\theta\\
2\sin^2(x)=1-\cos(2x)\\
2\cos^2(x)=1+\cos(2x)\\
\sin(2x)=2\sin(x)\cos(x)\\
&=\int\limits_\pi^0\sqrt{\frac{2\sin^2\theta }{2\cos^2\theta}}\cdot\frac{-2\sin(\theta)\cos(\theta)}{\sqrt{\cos(2\theta)}}\,\mathrm d\theta\\
&=2\int\limits^\pi_0\tan(\theta)\cdot\frac{\sin(\theta)\cos(\theta)}{\sqrt{\cos(2\theta)}}\,\mathrm d\theta\\

&=2\int\limits^\pi_0\frac{\sin^2(\theta)}{\sqrt{\cos(2\theta)}}\,\mathrm d\theta\\
\text{let } \theta=2\phi\\
\mathrm d\theta =2\mathrm d\phi\\
\theta=0\rightarrow\phi=0\\\theta=\pi\rightarrow\phi=\pi/2\\
&=2\int\limits_0^{\pi/2}\frac{\sin^2(2\phi)}{\sqrt{\cos(4\phi)}}2\,\mathrm d\phi\\
&=4\int\limits_0^{\pi/2}\frac{\sin^2(2\phi)}{\sqrt{\cos(4\phi)}}\,\mathrm d\phi\\
\\& =\quad\color{green}{????}\\
\end{align*}

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@doraemonpaul's Method:
\begin{align*}
\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm dx
&=\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\times\sqrt{\frac{1-x^2}{1-x^2}}\,\mathrm dx\\
&=\int\limits_0^1\frac{1-x^2}{\sqrt{1-x^4}}\,\mathrm dx\\
&=\int\limits_0^1(1-x^4)^{-1/2}\,\mathrm dx-\int\limits_0^1x^2(1-x^4)^{-1/2}\,\mathrm dx\\
\text{let }x=u^{1/4}\\
\mathrm dx=\frac14 u^{-3/4}\,\mathrm du\\
&=\int\limits_0^1(1-u)^{-1/2}\cdot\frac14 u^{-3/4}\,\mathrm du-\int\limits_0^1u^{1/2}(1-u)^{-1/2}\cdot\frac14 u^{-3/4}\,\mathrm du\\

&=\frac14\int\limits_0^1u^{-3/4}(1-u)^{-1/2}\cdot \,\mathrm du-\frac14\int\limits_0^1u^{-1/4}(1-u)^{-1/2}\,\mathrm du\\
\small{\text B(m,n)=\int\limits_0^1t^{m-1}(1-t)^{n-1} \,\mathrm d t}\\
\small{m_1-1=-3/4\quad n_1-1}&\small{=-1/2\quad m_2-1=-1/4\quad n_2-1=-1/2}\\
\small{m_1=1/4\quad \quad n_1=1/2\quad}&\small{\quad m_2=3/4\quad\quad n_2=1/2}\\
\\
&=\frac14\text B\left(\tfrac 14,\tfrac12\right)-\frac14\text B\left(\tfrac34,\tfrac12\right)\\
\small{\text B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}}\\
&=\frac 14\left(\frac{\Gamma(\frac 14)\Gamma(\frac 12)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)\Gamma(\frac 12)}{\Gamma(\frac 54)}\right)\\
&=\frac {\sqrt\pi}4\left(\frac{\Gamma(\frac 14)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)}{\Gamma(\frac 54)}\right)\\
&=\frac {\sqrt\pi}4\left(\frac{\Gamma(\tfrac 14)}{\Gamma(\tfrac 34)}-\frac{\Gamma(\tfrac34)}{\tfrac 14\Gamma(\tfrac 14)}\right)\\

&=\frac{\sqrt\pi}4\left(\frac{\Gamma(\tfrac 14)}{\Gamma(\tfrac 34)}-4\frac{\Gamma(\tfrac34)}{\Gamma(\tfrac 14)}\right)\\
\end{align*}

NB: this question has been edited post comments ,

Answer

I think you treat this question too complicated.

First it should note that integral of the form $\int_0^1(1-x^2)^a(1+x^2)^b~dx$ is impossible to have easy-look substitution to transform it to a integral of the form $\int_0^1x^p(1-x)^q~dx$ (i.e. beta function).

Second it should note that trigonometric substitution also often bring this type of question too complicated.

But I admit that this question has a level of lucky or tricky.

$\int_0^1\sqrt{\dfrac{1-x^2}{1+x^2}}~dx$

$=\int_0^1\dfrac{1-x^2}{\sqrt{1-x^2}\sqrt{1+x^2}}~dx$

$=\int_0^1\dfrac{1-x^2}{\sqrt{1-x^4}}~dx$

$=\int_0^1(1-x^4)^{-\frac{1}{2}}~dx-\int_0^1x^2(1-x^4)^{-\frac{1}{2}}~dx$

$=\int_0^1(1-x)^{-\frac{1}{2}}~d(x^{\frac{1}{4}})-\int_0^1x^{\frac{1}{2}}(1-x)^{-\frac{1}{2}}~d(x^{\frac{1}{4}})$

$=\dfrac{1}{4}\int_0^1x^{-\frac{3}{4}}(1-x)^{-\frac{1}{2}}~dx-\dfrac{1}{4}\int_0^1x^{-\frac{1}{4}}(1-x)^{-\frac{1}{2}}~dx$

$=\dfrac{B\left(\dfrac{1}{4},\dfrac{1}{2}\right)}{4}-\dfrac{B\left(\dfrac{3}{4},\dfrac{1}{2}\right)}{4}$

$=\dfrac{\Gamma\left(\dfrac{1}{4}\right)\Gamma\left(\dfrac{1}{2}\right)}{4\Gamma\left(\dfrac{3}{4}\right)}-\dfrac{\Gamma\left(\dfrac{3}{4}\right)\Gamma\left(\dfrac{1}{2}\right)}{4\Gamma\left(\dfrac{5}{4}\right)}$

$=\dfrac{\sqrt{\pi}~\Gamma\left(\dfrac{1}{4}\right)}{4\Gamma\left(\dfrac{3}{4}\right)}-\dfrac{\sqrt{\pi}~\Gamma\left(\dfrac{3}{4}\right)}{\Gamma\left(\dfrac{1}{4}\right)}$

$=\dfrac{\sqrt{\pi}~\Gamma\left(\dfrac{1}{4}\right)}{4\times\dfrac{\sqrt{2}\pi}{\Gamma\left(\dfrac{1}{4}\right)}}-\dfrac{\sqrt{\pi}~\Gamma\left(\dfrac{3}{4}\right)}{\dfrac{\sqrt{2}\pi}{\Gamma\left(\dfrac{3}{4}\right)}}$

$=\dfrac{1}{4\sqrt{2\pi}}\Gamma^2\left(\dfrac{1}{4}\right)-\dfrac{1}{\sqrt{2\pi}}\Gamma^2\left(\dfrac{3}{4}\right)$