<div>Evaluate $$\int \frac{2 \sin 2x - \cos x}{6 - \cos^2 x - 4 \sin x} \, dx$$</div>
Step-by-Step Solution
Key Concept: General
<div>$I = \int \frac{2 \sin 2x - \cos x}{6 - \cos^2 x - 4 \sin x} \, dx = \int \frac{(4 \sin x - 1) \cos x}{6 - (1 - \sin^2 x) - 4 \sin x} \, dx = \int \frac{(4 \sin x - 1) \cos x}{\sin^2 x - 4 \sin x + 5} \, dx$<br>Put $\sin x = t$, so that $\cos x \, dx = dt$.<br>$\therefore I = \int \frac{(4t - 1) \, dt}{(t^2 - 4t + 5)}$ ...(i)</div>
Correct Answer: A