Official solution
Given,
\[\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}}\,dx \]
\[ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left[ {{{\sin }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\cos }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]}^2}}}} \,dx\]
\[ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}^2}}}} dx\]
\[ = \int {{{{{\sin }^2}x\,\,{{\cos }^2}x} \over {{{\cos }^6}x{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,\,dx\]
\[ = \int {{{{{\sin }^2}x} \over {{{\cos }^4}x{{\left( {1 + {{\tan }^3}} \right)}^2}}}} \,\,dx\]
\[ = \int {{{{{\tan }^2}x.{{\sec }^2}x} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,dx\]
[ Let \[\,\,\,\,\,\,{\tan ^3}x = t\]
\[\,\,\,\,\,\,\] \[ \Rightarrow 3{\tan ^2}x\,{\sec ^2}x\,\,dx = dt\] ]
\[ = {1 \over 3}\int {{{3{{\tan }^2}x\,\,{{\sec }^2}x\,\,dx} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \]
\[ = {1 \over 3}\int {{{dt} \over {{{\left( {1 + t} \right)}^2}}}} \]
\[ = {1 \over 3} \times {{{{\left( {1 + t} \right)}^{ - 2 + 1}}} \over { - 2 + 1}} + c\]
\[ = {1 \over 3} \times {{ - 1} \over {\left( {1 + t} \right)}} + c\]
\[ = {{ - 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + c\]