Question

$I=\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}dx$. Find out I

Answer 1

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

$6\cos x+4\sin x=t$

$[-6\sin x+4\cos x]dx=dt$

or $(2\cos x-3\sin x)dx=\frac{dt}{2}$

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

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

$=\frac{1}{2}\ln t=\frac{1}{2}\ln |2\cos x-3\sin x|+c$