Đáp án:

• Đặt \( I = \int \frac{\sin x}{2 \cos x + 3 \sin x} \, dx \quad \text{và} \quad J = \int \frac{\cos x}{2 \cos x + 3 \sin x} \, dx \)

• Ta có:  
\(
\begin{cases}
3I + 2J = \int dx = x + c \\
-2I + 3J = \int \frac{3 \cos x - 2 \sin x}{3 \sin x + 2 \cos x} \, dx = \ln |3 \sin x + 2 \cos x| + c
\end{cases}
\)

\( \Rightarrow I = \frac{1}{13} \left[ 3x - 2 \ln |3 \sin x + 2 \cos x| \right] + c \)

• \( J = \frac{1}{13} \left[ 3 \ln |3 \sin x + 2 \cos x| + 2x \right] + c \)