If ${\int }_0^{50\pi } \left({\sin }^{4}x+{\cos }^{4}x\right)dx$$=k{\int }_0^{\pi /2} \left(\frac{3}{4}+\frac{1}{4}\cos 4x\right)dx,$ then $k=$

$\begin{array}{r}{\sin }^{4}x+{\cos }^{4}x={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x\\ =1-\frac{1}{2}{\sin }^{2}2x=1-\frac{1}{4}(1-\cos 4x)=\frac{3}{4}+\frac{1}{4}\cos 4x\end{array}$

Hence the period of L.H.S. = Period of R.H.S.

Thus L.H.S. is a periodic function of period $\frac{\mathrm{\pi }}{2}$

$\begin{array}{l}I={\int }_0^{100\pi /2} \left(\frac{3}{4}+\frac{1}{4}\cos 4x\right)dx\\ =100{\int }_0^{\pi /2} \left(\frac{3}{4}+\frac{1}{4}\cos 4x\right)dx\\ \therefore k=100\end{array}$