Question

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

Answer 1

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

Let 3cos x+ 2sin x=t

∴ (-3sin x + 2cos 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

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