$\int \frac{\cos 5 x + \cos 4 x}{1 - 2 \cos 3 x} dx is equal to$

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$\frac{\sin 2 x}{2} - \cos 2 x + c$

$\frac{\sin 2 x}{2} + \cos x + c$

$- \frac{\sin 2 x}{2} - \sin x + c$

$\frac{\sin 2 x}{2} - \cos x + c$

answer is C.

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Detailed Solution

$\begin{array}{r} I = \int \frac{\cos 5 x + \cos 4 x}{1 - 2 \cos 3 x} d x = \int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2}}{1 - 2 (2 \cos^{2} \frac{3 x}{2} - 1)} d x \\ = \int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2}}{3 - 4 \cos^{2} \frac{3 x}{2}} d x \\ = \int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{3 \cos \frac{3 x}{2} - 4 \cos^{3} \frac{3 x}{2}} d x \\ = \int \frac{2 \cos \frac{9 x}{2} \cos \frac{3 x}{2} \cos \frac{x}{2}}{- \cos \frac{9 x}{2}} d x \end{array}$

$\begin{aligned} = \int 2 \cos \frac{3 x}{2} \cos \frac{x}{2} d x \\ = - \int (\cos 2 x + \cos x) d x \\ = - \frac{\sin 2 x}{2} - \sin x + c \end{aligned}$