Show that $$\int_0^1\frac{x^2}{\sqrt{1-x^4}}\,dx \times \int_0^1\frac{1}{\sqrt{1+x^4}}\,dx=\frac{\pi}{4\sqrt{2}}$$
I'm trying to solve this problem but I can't prove it. In the end, I got $\Gamma(3/8)$, and $\Gamma(5/8)$ but I don't know how to solve their values.
Firstly I put $x^2=\sin(\theta)$ and solve the first part and $x^2=\tan(\theta)$. I don't know whether to combine both functions or solve them separately.I tried both but was not able to solve them. Please help guys
For the first one, let $x^2=\sin \theta$, then $$ \begin{aligned} I & =\frac{1}{2} \int_0^{\frac{\pi}{2}} \sin ^{2\left(\frac{3}{4}\right)-1} \theta \cos ^{2 \left( \frac{1}{2}\right)-1} \theta d \theta \\ & =\frac{1}{4} B\left(\frac{3}{4}, \frac{1}{2}\right)=\frac{\sqrt{\pi}}{4} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \end{aligned} $$ For the second, let $x^2=\tan \theta$, then
$$ \begin{aligned} J & =\int_0^{\frac{\pi}{4}} \frac{1}{\sec \theta} \cdot \frac{1}{2 \sqrt{\tan \theta}} \sec ^2 \theta d \theta \\ & =\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{\sin \theta \cos \theta}} d \theta \\ & =\frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{\sin 2 \theta}} d \theta \\ & =\frac{1}{2 \sqrt{2}} \int_0^{\frac{\pi}{2}} \sin ^{2\left(\frac{1}{4}\right)-1} \theta \cos ^{2\left(\frac{1}{2}\right)-1} \theta d \theta\\&=\frac{1}{4 \sqrt{2}} B\left(\frac{1}{4} ,\frac{1}{2}\right) \end{aligned} $$ Multiplying them yields$$ \begin{aligned} I \times J= & \frac{\sqrt{\pi}}{4} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \cdot \frac{\sqrt{\pi}}{4 \sqrt{2}} \cdot \frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)} \\ & =\frac{\pi}{16 \sqrt{2}} \cdot \frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \\ & =\frac{\pi}{16 \sqrt{2}} \cdot \frac{\Gamma\left(\frac{1}{4}\right)}{\frac{1}{4} \Gamma\left(\frac{1}{4}\right)} \\ & =\frac{\pi}{4 \sqrt{2}} \end{aligned} $$