Đáp án:
$\begin{array}{l}
{I_9} = \int {\frac{{dx}}{{\sin 3x + \sin \,x}}} \\
= \int {\frac{1}{{2.\sin \frac{{3x + x}}{2}.\cos \frac{{3x - x}}{2}}}dx} \\
= \int {\frac{1}{{2.\sin 2x.cosx}}dx} \\
= \int {\frac{1}{{4\sin x.{{\cos }^2}x}}dx} \\
= \frac{1}{4}\int {\frac{{\sin x}}{{\left( {1 - {{\cos }^2}x} \right).{{\cos }^2}x}}} dx\\
= \frac{1}{4}.\int {\frac{1}{{{{\cos }^2}x\left( {{{\cos }^2}x - 1} \right)}}d\cos x} \\
= \frac{1}{4}.\int {\frac{{{{\cos }^2}x - \left( {{{\cos }^2}x - 1} \right)}}{{{{\cos }^2}x\left( {{{\cos }^2}x - 1} \right)}}} d\cos x\\
= \frac{1}{4}.\int {\frac{1}{{{{\cos }^2} - 1}} - \frac{1}{{{{\cos }^2}x}}} d\cos x\\
= \frac{1}{4}.\left( {\ln \frac{{\cos x - 1}}{{\cos x + 1}} + \frac{1}{{\cos x}}} \right) + C
\end{array}$
