Let I = ∫ sin 3x dx  ...............1

Let 3x = t

=> 3 * dx = dt

=> dx = dt/3

From equation 1, we get

     I = ∫ sin t dt/3

=> I = (1/3) * ∫ sin t

=> I = -(1/3) * cos t + C

=> I = -(cos t)/3 + C

=> I = -(cos 3x)/3 + C       {since t = 3x}

So, ∫ sin 3x dx = -(cos 3x)/3 + C