The value of the integral $\int \sin 3x \cos 4x \mathrm{dx}$ is

Let $\begin{array}{l}\mathrm{I}=\int \sin 3\mathrm{xcos} 4\mathrm{xdx}=\frac{1}{2}\int 2\sin 3\mathrm{xcos} 4\mathrm{xdx}\\ =\frac{1}{2}\int \{\sin (3\mathrm{x}+4\mathrm{x})+\sin (3\mathrm{x}-4\mathrm{x}\left)\right\}\mathrm{dx}\\ =\frac{1}{2}\int (\sin 7x-\sin x) dx (\because 2\sin \mathrm{Acos} \mathrm{B}=\sin (\mathrm{A}+\mathrm{B})+\sin (\mathrm{A}-\mathrm{B}\left)\right)\\ =\frac{1}{2}\left(\frac{-\cos 7\mathrm{x}}{7}+\cos \mathrm{x}\right)+\mathrm{C}\left[\because \int \sin \mathrm{a\theta dx}=\frac{-\cos \mathrm{a\theta }}{\mathrm{a}}\right]\\ =\frac{-\cos 7\mathrm{x}}{14}+\frac{\cos \mathrm{x}}{2}+\mathrm{C}\end{array}$