Question

Integrate the function $\frac{2\cos x-3\sin x}{6\cos x+4\sin x}$

Answer 1

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

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

$\therefore (-3\sin x+2\cos x)dx=dt$

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

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

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

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