Evaluate $\int \frac{(\cos 5x + \cos 4x)}{1 - 2\cos 3x} \, dx$

$-\frac{1}{2} \sin 2x - \sin x + C$

The correct answer is Option (2) → $-\frac{1}{2} \sin 2x - \sin x + C$

Let $I = \int \frac{\cos 5x + \cos 4x}{1 - 2\cos 3x} \, dx = \int \frac{2 \cos \frac{9x}{2} \cdot \cos \frac{x}{2}}{1 - 2 \left( 2 \cos^2 \frac{3x}{2} - 1 \right)} \, dx$

$\left[ ∵\cos C + \cos D = 2 \cos \frac{C+D}{2} \cdot \cos \frac{C-D}{2} \text{ and } \cos 2x = 2 \cos^2 x - 1 \right]$

$∴I = \int \frac{2 \cos \frac{9x}{2} \cdot \cos \frac{x}{2}}{3 - 4 \cos^2 \frac{3x}{2}} \, dx = -\int \frac{2 \cos \frac{9x}{2} \cdot \cos \frac{x}{2}}{4 \cos^2 \frac{3x}{2} - 3} \, dx$

$= -\int \frac{2 \cos \frac{9x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3x}{2}}{4 \cos^3 \frac{3x}{2} - 3 \cos \frac{3x}{2}} \, dx \quad \left[ \text{multiply and divide by } \cos \frac{3x}{2} \right]$

$= -\int \frac{2 \cos \frac{9x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3x}{2}}{\cos 3 \cdot \frac{x}{2}} \, dx = -\int 2 \cos \frac{3x}{2} \cdot \cos \frac{x}{2} \, dx$  $\left[ ∵\cos 3x = 4 \cos^3 x - 3 \cos x \right]$

$= -\int \left\{ \cos \left( \frac{3x}{2} + \frac{x}{2} \right) + \cos \left( \frac{3x}{2} - \frac{x}{2} \right) \right\} \, dx$   $\left[ ∵2 \cos a \cos b = \cos(a+b) + \cos(a-b) \right]$

$= -\int (\cos 2x + \cos x) \, dx$

$= -\left[ \frac{\sin 2x}{2} + \sin x \right] + C$

$= -\frac{1}{2} \sin 2x - \sin x + C$