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

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

and $\left. {\cos 2x = 2{{\cos }^2}x - 1} \right]$

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

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

$= - \int {\frac{{2\cos \frac{{9x}}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{{3x}}{2}}}{{4{{\cos }^3}\frac{{3x}}{2} - 3\cos \frac{{3x}}{2}}}} dx\quad \left[ {} \right.$ multiplying and dividing by $\left. {\cos \frac{{3x}}{2}} \right]$

$= - \int {\frac{{2\cos \frac{{9x}}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{{3x}}{2}}}{{\cos 3 \cdot \frac{{3x}}{2}}}} dx = - \int 2 \cos \frac{{3x}}{2} \cdot \cos \frac{x}{2}dx$

$= - \int {\left\{ {\cos \left( {\frac{{3x}}{2} + \frac{x}{2}} \right) + \cos \left( {\frac{{3x}}{2} - \frac{x}{2}} \right)} \right\}} dx$

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

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

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