$\int {\frac{{{{\sin }^6}x + {{\cos }^6}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx$

Let $I = \int {\frac{{{{\sin }^6}x + {{\cos }^6}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx = \int {\frac{{{{\left( {{{\sin }^2}x} \right)}^3} + {{\left( {{{\cos }^2}x} \right)}^3}}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx$

$= \int {\frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx$

$= \int {\frac{{{{\sin }^4}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx + \int {\frac{{{{\cos }^4}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx - \int {\frac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx$

$= \int {{{\tan }^2}} xdx + \int {{{\cot }^2}} xdx - \int 1 dx$

$= \int {\left( {{{\sec }^2}x - 1} \right)} dx + \int {\left( {{{{\mathop{\rm cosec}\nolimits} }^2}x - 1} \right)} dx - \int 1 dx$

$= \int {{{\sec }^2}} xdx + \int {{{{\mathop{\rm cosec}\nolimits} }^2}} xdx - 3\int d x$

$I = \tan x - \cot x - 3x + C$