Đáp án:
\[B\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
F\left( x \right) = \int {f\left( x \right)} \\
= \int {\frac{{2\cos 4x}}{{{{\sin }^2}2x.\left( {1 - {{\sin }^2}2x} \right)}}dx} \\
= \int {\frac{{2\cos 4x}}{{{{\sin }^2}2x.{{\cos }^2}2x}}dx} \\
= \int {\frac{{8\cos 4x}}{{{{\left( {2\sin 2x.\cos 2x} \right)}^2}}}dx} \\
= \int {\frac{{2.\left( {4\cos 4xdx} \right)}}{{{{\sin }^2}4x}}} \\
= \int {\frac{{2d\left( {\sin 4x} \right)}}{{{{\sin }^2}4x}}} \\
= \frac{{ - 2}}{{\sin 4x}} + C\\
= \frac{{ - 2\left( {{{\sin }^2}2x + {{\cos }^2}2x} \right)}}{{2.\sin 2x.\cos 2x}} + C\,\,\,\,\,\,\,\,\left( {{{\sin }^2}2x + {{\cos }^2}2x = 1} \right)\\
= - \left( {\frac{{\sin 2x}}{{\cos 2x}} + \frac{{\cos 2x}}{{\sin 2x}}} \right) + C\\
= - \tan 2x - \cot 2x + C
\end{array}\)
