$\underset{0}{\overset{\pi /4}{\int }}\frac{{\sin }^{2}x.{\cos }^{2}x}{{({\sin }^{5}x+{\cos }^{3}x.{\sin }^{2}x+{\sin }^{3}x.{\cos }^{2}x+{\cos }^{5}x)}^2}dx$  is equal to K then 12k equal to

 $\underset{0}{\overset{\pi /4}{\int }}\frac{{\sin }^{2}x.{\cos }^{2}x}{{({\sin }^{3}x+{\cos }^{3}x)}^2}dx=\underset{0}{\overset{\pi /4}{\int }}\frac{{\tan }^{2}x.{\sec }^{2}x}{{(1+{\tan }^{3}x)}^2}dx$
Let  $1+{\tan }^{3}x=t$
$I=\frac{1}{3}\underset{1}{\overset{2}{\int }}\frac{dt}{t^2}dx={\left(\frac{-1}{3t}\right)}_1^2=-\frac{1}{3}[\frac{1}{2}-1]=\frac{1}{6}$