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Verified answer

[tex]\sin x \sin (4x-7)=\frac{1}{2}(\cos((4x-7)-x)-\cos((4x-7)+x))= \frac{1}{2}(\cos(3x-7)-\cos(5x-7))\\\int {\sin x \sin (4x-7)} \, dx =\int {\frac{1}{2}(\cos(3x-7)-\cos(5x-7))} \, dx=\frac{1}{2}(\int {\cos (3x-7)} \, dx + \int {\cos (5x-7)} \, dx )=\frac{1}{2} (-\frac{\sin (3x-7)}{3}-\frac{\sin (5x-7)}{5} + C' )= -\frac{\sin (3x-7)}{6}-\frac{\sin (5x-7)}{10} + C[/tex]