text: You will need to use integration by parts.\nLet u = x^2 and dv = sin(ax)dx, then\ndu = 2x dx and v = (-1/a)cos(ax)\nuv - integral(v du)\n= (-1/a)x^2 cos(ax) + integral [ (2/a)x cos(ax) dx]\nNow to integrate (2/a) cos(ax) dx, we need to use integration by parts one more time.\nLet u = (2/a)x and dv = cos(ax) dx, then\ndu = (2/a)x and v = (1/a)sin(ax)\nuv - integral(v du)\n= (2/a^2)x sin(ax) - integral [ (2/a^2)sin(ax) dx]\n= (2/a^2)x sin(ax) - (2/a^2)(-1/a)cos(ax) + C\n= (2/a^2)x sin(ax) + (2/a^3)cos(ax) + C\nNow you need to put this with the first part to get your final answer.\nAnswer: (-1/a)x^2 cos(ax) + (2/a^2)x sin(ax) + (2/a^3)cos(ax) + C