Question

Solve by integration by part
Integrate x²cos3xdx
Show your calculations

Answer 1

Let u = x² and dv = cos3x dx

du = 2x dx and v = -1/3 sin3x

Integrate x²cos3xdx = x²(-1/3 sin3x) - ∫(-1/3 sin3x)2x dx

Integrate x²cos3xdx = x²(-1/3 sin3x) + (2/3)∫sin3x dx

Integrate x²cos3xdx = x²(-1/3 sin3x) + (2/3)(-1/3 cos3x) + C

Integrate x²cos3xdx = x²(-1/3 sin3x) - (2/9)cos3x + C

Answer 2

Ella, if you did look at the video I suggested, this one comes out as:

(1/3)x^2 sin 3x - 2/9 x cos 3x + 2/27 sin 3x + c

clearly the bot is wrong