How find integral $\int_{[0,\frac{\pi}{4}]^n}\frac{dx_{1}dx_{2}\cdots dx_{n}}{\sqrt{1+\sec^2{x_{1}}+\sec^2{x_{2}}+\cdots+\sec^2{x_{n}}}}$ 
show that
  $$\int_{[0,\dfrac{\pi}{4}]^n}\dfrac{dx_{1}dx_{2}\cdots dx_{n}}{\sqrt{1+\sec^2{x_{1}}+\sec^2{x_{2}}+\cdots+\sec^2{x_{n}}}}=\dfrac{\pi^n(n!)^2}{(2n+1)!}$$

first I see this problem I want use
$\sec{x_{i}}=y_{i}$,But follow is very ugly,maybe this idea works.
this problem is my  friends ask me this not key,I hope this problem is  solve  is key
 A: We'll do this by induction. For $n=1$ we have:
$$\int_0^{\pi/4}\frac{\cos x}{\sqrt{1+\cos^2x}} \:dx=\frac{\pi}{6}$$
I'm sure you can handle this so I'll omit the details. Now suppose it's true for $n$ and let's prove it for $n+1.$
We have:
$$\int_{[0,\frac{\pi}{4}]^{n+1}} \frac{dx_1...dx_{n+1}}{\sqrt{1+\sum_{i=1}^{n+1}\sec^2x_i}}=\int_{[0,\frac{\pi}{4}]^{n}} dx_1...dx_n\int_0^{\pi/4}\frac{dx_{n+1}}{\sqrt{(1+C)+\frac{1}{\cos^2 x_{n+1}}}{}}$$
Where $C=\sum_{i=1}^{n} \sec^2{x_i}$.Rewriting it:
$$\int_{[0,\frac{\pi}{4}]^{n}} dx_1...dx_n\int_0^{\pi/4}\frac{\cos x_{n+1}\:dx_{n+1}}{\sqrt{(1+C)\cos^2 x_{n+1}+1}}$$
Using $\cos^2x_{n+1}=1-\sin^2 x_{n+1}$ we have:
$$\int_{[0,\frac{\pi}{4}]^{n}} dx_1...dx_n\int_0^{\pi/4}\frac{\cos x_{n+1}\:dx_{n+1}}{\sqrt{(2+C)-(1+C)\sin^2 x_{n+1}}}$$
Now using the substitution $\sin x_{n+1}=u \implies \cos x_{n+1} \:dx_{n+1}=du$ the integrals become:
$$\int_{[0,\frac{\pi}{4}]^{n}} \frac{dx_1...dx_n}{\sqrt{1+C}}\int_0^{\sqrt 2/2} \frac{du}{\sqrt{\frac{2+C}{1+C}-u^2}}$$
This equals:
$$\int_{[0,\frac{\pi}{4}]^{n}}\arcsin\bigg(\sqrt{\frac{1+C}{2+C}}\frac{\sqrt 2}{2}\bigg) \frac{dx_1...dx_n}{\sqrt{1+C}}$$
I'm confident this can be reduced to the form desired, but now I've got to go.
