Question

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

Answer 1

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

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

$⟹(-6\sin x+4\cos x)dx=dt$

$⟹(4\cos x-6\sin x)dx=dt$

$⟹(2\cos x-3\sin x)dx=\frac{1}{2}dt$

Now, $\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}dx=\int \frac{1}{2t}dt$

$=\frac{1}{2}\log t+c$

$=\frac{1}{2}\log (6\cos x+4\sin x)+c$