$\underset{0}{\overset{\pi /4}{\int }}\frac{{\sin }^{2}x{\cos }^{2}x}{{\left({\sin }^{3}x+{\cos }^{3}x\right)}^2}dx=$

$\begin{array}{l}\underset{0}{\overset{\pi /4}{\int }}\frac{{\sin }^{2}x{\cos }^{2}x}{{({\sin }^{3}x+{\cos }^{3}x)}^2}dx\\ =\underset{0}{\overset{\pi /4}{\int }}\frac{{\sin }^{2}x{\cos }^{2}x}{{\cos }^{6}x{({\tan }^{3}x+1)}^2}dx\\ =\underset{0}{\overset{\pi /4}{\int }}\frac{{\tan }^{2}x{\sec }^{2}xdx}{{(1+{\tan }^{3}x)}^2}\\ put\tan x=t\\ {\sec }^{2}xdx=dt\\ x=0\Rightarrow t=0\\ x=\frac{\pi }{4}\Rightarrow t=1\\ =\underset{0}{\overset{1}{\int }}\frac{t^2}{{(1+t^3)}^2}\\ =\frac{1}{3}\underset{0}{\overset{1}{\int }}\frac{dy}{y^2}\\ Put1+t^3=y\\ 3t^{2}dt=dy\\ t^{2}dt=\frac{dy}{3}\\ t=0\Rightarrow y=0\\ t=1\Rightarrow y=1+1^3=2\\ =\frac{1}{3}{\left(\frac{-1}{y}\right)}_1^2\\ =\frac{1}{3}(\frac{1}{1}-\frac{1}{2})\\ =\frac{1}{6}\end{array}$