If ${\displaystyle \int \frac{\cos 8x-\cos 7x}{1+2\cos 5x}}dx=K$ $\sin 3x+M\sin 2x+C,$ then

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

$I={\displaystyle \int \frac{-2\sin \left(\frac{15x}{2}\right)\sin \left(\frac{x}{2}\right)}{1+2\left(1-2{\sin }^2\frac{5x}{2}\right)}}dx$       $\left[\begin{array}{c}\cos C-\cos D=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}\\ \cos 2A=2{\cos }^{2}A-1=1-2{\sin }^{2}A\end{array}\right]$

$=-{\displaystyle \int \frac{2\sin \frac{15x}{2}.\sin \frac{x}{2}}{3-4{\sin }^2\frac{5x}{2}}}dx.\frac{\sin \frac{5x}{2}}{\sin \frac{5x}{2}}$

$=-{\displaystyle \int \frac{2\sin \frac{15x}{2}.\sin \frac{x}{2}.\sin \frac{5x}{2}dx}{3\sin \frac{5x}{2}-4{\sin }^3\frac{5x}{3}}}$

$=-{\displaystyle \int \frac{2\sin \frac{15}{x}\sin \frac{x}{2}.\sin \frac{5x}{2}dx}{\sin \frac{15x}{2}}}$

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

$=\frac{\sin 3x}{3}-\frac{\sin 2x}{2}+C$

$K=\frac{1}{3},M=-\frac{1}{2}$ .