If $\underset{0}{\overset{\frac{2}{\pi }}{\int }}{\sin }^{6}xdx=\frac{5\pi }{32},$ on solving $\underset{-\pi }{\overset{\pi }{\int }}({\sin }^{6}x+co{s}^{6}x)dx$ is equal to $\frac{5\pi }{2^{k-1}}$ , then the value of $\text{'}K\text{'}$ is  ____

GT $\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\sin }^{6}xdx=\frac{5\pi }{32}$

$\underset{-\pi }{\overset{\pi }{\int }}({\sin }^{6}x+{\cos }^{6}x)dx=2\underset{0}{\overset{\pi }{\int }}({\sin }^{6}x+{\cos }^{6}x)dx$

$=4\underset{0}{\overset{\frac{\pi }{2}}{\int }}({\sin }^{6}x+{\cos }^{6}x)dx$

$=4\left[\underset{0}{\overset{\frac{{\displaystyle \pi }}{{\displaystyle 2}}}{\int }}{\sin }^{6}xdx+\underset{2}{\overset{\frac{{\displaystyle \pi }}{{\displaystyle 2}}}{\int }}{\cos }^{6}xdx\right]$

$=8\left[\underset{0}{\overset{\frac{{\displaystyle \pi }}{{\displaystyle 2}}}{\int }}{\sin }^{6}xdx\right]$

$=8\times \frac{5\pi }{32}=\frac{5\pi }{4}=\frac{5\pi }{2^{3-1}}$

$\Rightarrow k=3$