Question

2cosx - 3cosx /6cosx +4sinx​

Answer 1

Answer:

∫2cosx−3sinx6cosx+9sinxdx

let 6cosx+9sinx=t

(2cosx-3sinx)dx=(dt)/2

=∫1tdt2

=12log|t|+C

=12log|6cosx+4sinx|+C

Step-by-step explanation:

Answer 2

Answer:

$\dfrac{2\cos x-3\cos x}{6\cos x+4\sin x}=\dfrac{-\cos x}{6\cos x+4\sin x}=-\dfrac{\cos x}{6\cos x+4\sin x}$