What is the integration of [ \int {\frac{{{{\sin }^6}x + {{\cos }^6}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx]
~[NCERT Exemp. Q. 23,Page 164 ]

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


\[ = \int {\frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x – {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx\]


\[ = \int {\frac{{{{\sin }^4}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx + \int {\frac{{{{\cos }^4}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx – \int {\frac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx\]


\[ = \int {{{\tan }^2}} xdx + \int {{{\cot }^2}} xdx – \int 1 dx\]

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


\[ = \int {{{\sec }^2}} xdx + \int {{{{\mathop{\rm cosec}\nolimits} }^2}} xdx – 3\int d x\]

\[ I = \tan x – \cot x – 3x + C\]

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