$\text{ If }{\mathrm{I}}_{\mathrm{m},\mathrm{n}}=\int {\cos }^{\mathrm{m}}\mathrm{xsin}\mathrm{nxdx},\text{ then }7{\mathrm{I}}_{4,3}-4{\mathrm{I}}_{3,2}\text{ is equal to}$

$I_{4,3}=\int {\cos }^{4}x\sin 3xdx$

Integrating by parts, we have
$\begin{array}{l}{\mathrm{I}}_{4,3}=-\frac{\cos 3{\mathrm{xcos}}^4\mathrm{x}}{3}-\frac{4}{3}\int {\cos }^3\mathrm{xsin}\mathrm{xcos}3\mathrm{xdx}\\ \text{ But }\sin \mathrm{xcos}3\mathrm{x}=-\sin 2\mathrm{x}+\sin 3\mathrm{xcos}\mathrm{x}.\\ \text{ So, }{\mathrm{I}}_{4,3}=-\frac{\cos {\mathrm{xcos}}^4\mathrm{x}}{2}+\frac{4}{2}\int {\cos }^3\mathrm{xsin}2\mathrm{xdx}-\frac{4}{3}\int {\cos }^4\mathrm{xsin}3\mathrm{xdx}+\mathrm{C}\end{array}$

$=-\frac{\cos 3x{\cos }^{4}x}{3}+\frac{4}{3}{I}_{3,2}-\frac{4}{3}{I}_{4,3}+C$

$\text{ Therefore, }\frac{7}{3}{I}_{4,3}-\frac{4}{3}{I}_{3,2}=-\frac{\cos 3x{\cos }^{4}x}{3}+C$

$\text{ or }7I_{4,3}-4I_{3,2}=-\cos 3x{\cos }^{4}x+C$