Prove

$\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}=\frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}$

Let $3 \cos x+2 \sin x=t$

$\therefore(-3 \sin x+2 \cos x) d x=d t$

$\int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x=\int \frac{d t}{2 t}$

$=\frac{1}{2} \int \frac{1}{t} d t$

$=\frac{1}{2} \log |t|+\mathrm{C}$

$=\frac{1}{2} \log |2 \sin x+3 \cos x|+\mathrm{C}$