Đáp án:
$\begin{array}{l}
\int {\frac{{dx}}{{4{{\sin }^2}x + 9{{\cos }^2}x}}} \\
= \int {\frac{{\frac{1}{{{{\cos }^2}x}}}}{{\frac{{4{{\sin }^2}x}}{{{{\cos }^2}x}} + 9}}} dx\\
= \int {\frac{1}{{4{{\tan }^2}x + 9}}.\frac{1}{{{{\cos }^2}x}}dx} \\
= \int {\frac{1}{{4{{\tan }^2}x + 9}}d\left( {{\mathop{\rm tanx}\nolimits} } \right)} \left( {do:\left( {\tan x} \right)' = \frac{1}{{{{\cos }^2}x}}} \right)\\
Đặt\,\tan x = \frac{3}{2}\tan u\\
\Rightarrow u = \arctan \left( {\frac{2}{3}\tan x} \right)\\
\Rightarrow d\left( {{\mathop{\rm tanx}\nolimits} } \right) = \frac{3}{2}d\left( {\tan u} \right)\\
\Rightarrow \int {\frac{1}{{4{{\tan }^2}x + 9}}d\left( {{\mathop{\rm tanx}\nolimits} } \right)} \\
= \int {\frac{1}{{9{{\tan }^2}u + 9}}} .\frac{3}{2}d\tan u\\
= \int {\frac{1}{{6\left( {{{\tan }^2}u + 1} \right)}}d\tan u} \\
= \int {\frac{1}{6}.{{\cos }^2}u.\frac{1}{{{{\cos }^2}u}}du} \\
= \int {\frac{1}{6}du} \\
= \frac{1}{6}u + C\\
= \frac{1}{6}\arctan \left( {\frac{2}{3}\tan x} \right) + C
\end{array}$
