Evaluate: ∫ (cos x)/(3cos x - 5) dx

$\frac{\cos x}{3\cos x-5}=\frac13+\frac{5}{3}\cdot\frac{1}{3\cos x-5}$. Put $t=\tan\frac x2$: $\cos x=\frac{1-t^2}{1+t^2}$, $dx=\frac{2\,dt}{1+t^2}$, $3\cos x-5=-\frac{2+8t^2}{1+t^2}$. $\int\frac{dx}{3\cos x-5}=-\int\frac{dt}{1+4t^2}=-\frac12\tan^{-1}(2t)$.