$\int \frac{\cos 8x-\cos 7x}{1+2\cos 5x}dx$ is equal to 

$\begin{array}{rcl} & & \int \frac{\cos 8\mathrm{x}-\cos 7\mathrm{x}}{1+2\cos 5\mathrm{x}} \mathrm{dx}\\ & =& \int \frac{-2 \sin {\displaystyle \frac{15\mathrm{x}}{2}} \sin {\displaystyle \frac{\mathrm{x}}{2}}}{1+2 \cos 5\mathrm{x}}\mathrm{dx}\\ & =& (-2)\int \frac{\left(3 \sin {\displaystyle \frac{5\mathrm{x}}{2}}-4 {\sin }^3{\displaystyle \frac{5\mathrm{x}}{2}}\right)}{1+2(1-2 {\sin }^2{\displaystyle \frac{5\mathrm{x}}{2}})}\\ & =& (-2)\int \frac{\cancel{\left(3-4 {\sin }^2{\displaystyle \frac{5\mathrm{x}}{2}}\right)}\sin {\displaystyle \frac{5\mathrm{x}}{2}} \sin {\displaystyle \frac{\mathrm{x}}{2}}}{\cancel{3-4 {\sin }^2{\displaystyle \frac{5\mathrm{x}}{2}}}}\\ & =& \int \left(\cos 3\mathrm{x}-\cos 2\mathrm{x}\right)\mathrm{dx}\\ & =& \frac{\sin 3\mathrm{x}}{3}-\frac{\sin 2\mathrm{x}}{2}+\mathrm{c}\\ & & \end{array}$