sin 3x cos 5x dx =....

integral 1/2 cos (a-ß) - cos (a+ß) dx 
integral 1/2 cos (-2x) - cos (8x) dx 
1/2 (1/2 sin 2x - 1/8 sin8x) 
1/4 sin 2x - 1/16 sin8x

[tex]\displaystyle \int\sin 3x\cos 5x\,dx=\int\frac{\sin(3x+5x)+\sin(3x-5x)}{2}\,dx\\\int\sin 3x\cos 5x\,dx=\int\frac{\sin8x-\sin2x}{2}\,dx\\\int\sin 3x\cos 5x\,dx=\frac{-8\cos8x+2\cos2x}{2}+C\\\boxed{\boxed{\int\sin 3x\cos 5x\,dx=-4\cos8x+\cos2x+C}}[/tex]