Đáp án:
$\begin{array}{l}
\int {\frac{{1 + {{\sin }^2}x}}{{{{\sin }^2}2x}}dx} \\
= \int {\frac{1}{{{{\sin }^2}2x}} + \frac{{{{\sin }^2}x}}{{4{{\sin }^2}x.co{s^2}x}}dx} \\
= - \frac{{\cot 2x}}{2} + \int {\frac{1}{{4co{s^2}x}}dx} \\
= - \frac{{\cot 2x}}{2} + \frac{{\tan \,x}}{4} + C
\end{array}$