Use integration to find I = ∫ xsin3x dx

Use integration by parts, let U = x, the derivative of U = 1, let the derivative of V = sin3x and intergrate the derivative of V to arrive at V = (-1/3)(cos3x). Substitute the value into the formula uv − ∫ vdu dx dx, arrive at I = (x)(-1/3)(cos3x) - ∫(1)(-1/3)(cos3x)dx which can be written us I = (-x/3)(cos3x) +∫(1/3)(cos3x)dx. ∫(1)(1/3)(cos3x)dx = (1/9)(sin3x). Now put that into the original equation giving the final answer I = (-x/3)(cos3x)+ (1/9)(sin3x) + c,