$\int {\cfrac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}}}$equals

Option b is correct

Let $I = \int {\cfrac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}}} = \int {\cfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx$

$= \int {\left( {{{\sec }^2}x + \cos e{c^2}x} \right)} dx = \tan x - \cot x + C$

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