<s>[INST] Another way to evaluate $\int\frac{\cos5x+\cos4x}{1-2\cos3x}{dx}$? What I've done is this:$$\int\dfrac{\cos5x+\cos4x}{1-2\cos3x}{dx}$$
$$\int \dfrac{\sin 3x}{\sin 3x}\left[\dfrac{\cos5x+\cos4x}{1-2\cos3x}\right]{dx}$$
$$\dfrac {1}{2}\int\dfrac{\sin 8x -\sin 2x +\sin 7x -\sin x}{\sin 3x - \sin 6x}$$
$$-\dfrac {1}{2}\int\dfrac{ \sin \frac{7x}{2} +\sin \frac{5x}{2} } {\sin \frac{3x}{2} }$$
$$-\int\dfrac{ \sin {3x}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$
$$-\int\dfrac{2\sin \frac{3x}{2} \cos \frac{3x}{2}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$
$$-\int {2\cos \frac{3x}{2}\cos \frac{x}{2} }$$
$$ -\left(\frac{\sin 2x}{2} +\sin x \right) +c $$
Is there any other way to do so ? Is it possible to do it by substitution ?
 [/INST] 
Is there any other way to evaluate it ?

Hint. One may observe that
$$
\frac{\cos(5x)+\cos(4x)}{1-2\cos(3x)}=-\cos(2x)-\cos (x)
$$ then the evaluation is easier.

Edit. Here is a way to obtain such a simplification. One may set $u=e^{ix}$ then using De moivre's formula for $\cos(\cdot)$ one gets
$$
\begin{align}
\frac{\cos(5x)+\cos(4x)}{1-2\cos(3x)}&=\frac{\dfrac{u^5+\dfrac1{u^5}}2+\dfrac{u^4+\dfrac1{u^4}}2}{1-\big(u^3+\dfrac1{u^3}\big)}
\\\\&=-\frac12\:\frac{1+u+u^3+u^4}{u^2}
\\\\&=- \dfrac{u^2+\dfrac1{u^2}}2-\dfrac{u+\dfrac1{u}}2
\\\\&=-\cos(2x)-\cos (x).
\end{align}
$$
</s>