If ${\mathrm{I}}_{\mathrm{m},\mathrm{n}}=\int {\cos }^{\mathrm{m}}\mathrm{xsin}\mathrm{nx}$ dx and $7{\mathrm{I}}_{4,3}-4{\mathrm{I}}_{3,2}=-\cos \mathrm{Axco}{s}^{\mathrm{B}}\mathrm{x}+\mathrm{c}$ then A + B =

$$\begin{gathered} I_{4,3}=\int {\cos }^{4}x\sin 3xdx \\ =-\frac{\cos 3x{\cos }^{4}x}{3}-\frac{4}{3}\int {\cos }^{3}x\sin x\cos 3xdx \\ \text{ But }\sin x\cos 3x=-\sin 2x+\sin 3x\cos x\text{. So, } \\ \begin{array}{l}{\mathrm{I}}_{4,3}=-\frac{\cos 3x{\cos }^4\mathrm{x}}{3}+\frac{4}{3}\int {\cos }^3\mathrm{xsin}2\mathrm{xdx}-\frac{4}{3}\int {\cos }^4\mathrm{xsin}3\mathrm{xdx}+\mathrm{K}\\ {\mathrm{I}}_{4,3}=-\frac{\cos 3x{\cos }^4\mathrm{x}}{3}+\frac{4}{3}{\mathrm{I}}_{3,2}-\frac{4}{3}{\mathrm{I}}_{4,3}\\ \frac{7}{3}{\mathrm{I}}_{4,3}=-\frac{\cos 3x{\cos }^4\mathrm{x}}{3}+\frac{4}{3}{\mathrm{I}}_{3,2}\\ \frac{7}{3}{\mathrm{I}}_{4,3}-\frac{4}{3}{\mathrm{I}}_{3,2}=-\frac{\cos 3x{\cos }^3\mathrm{x}}{3}+\mathrm{C}\\ \text{ or }7{\mathrm{I}}_{4,3}-4{\mathrm{I}}_{3,2}=-\cos 3{\mathrm{xcos}}^4\mathrm{x}+\mathrm{K}\end{array} \end{gathered}$$