$\frac{1}{2}\int_{-\infty}^{\infty} \frac{dx}{(e^x+e^{-x})^n}=\frac{1}{2}\int_{-\infty}^{\infty} \frac{dx}{e^{nx}[1+\frac{e^{-x}}{e^x}]^n}=\frac{1}{2}\int_{-\infty}^{\infty} \frac{e^{-nx}dx}{(1+e^{-2x})^n}$

Put,

$e^{-2x}=t \hspace{0.3cm}=\gt\hspace{0.3cm} -2e^{-2x}dx=dt $

When,

$x=-\infty ,t=\infty,x=\infty,t=0 $

$\frac{1}{2}\int_{-\infty}^{0} \frac{t^{n/2}dt}{(1+t)^n2(-t)}=\frac{1}{4}\int_{0}^{\infty} \frac{t^{n/2}dt}{(1+t)^n} $

$= \frac{1}{4}\beta(\frac{n}{2},\frac{n}{2}) $

$ \int_{0}^{\infty} sech^8xdx=\int_{0}^{\infty}(\frac{2}{e^x+e^{-x}})^8dx$

$ =2^8* \frac{1}{4} \beta (\frac{8}{2},\frac{8}{2}) =2^6 \beta(4,4) $

$=2^6 \frac{\sqrt{4}\sqrt{4}}{\sqrt{8}} $

$ = \frac{16}{35} $