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
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.
