integrate xsin3xdx?

Answer 1

-Cos[3*d*x^2]/(6*d)

Answer 2

This is exactly the same as the other one, except that you have to play with it a little to get it in the right form.

x sin3x = 1/3 (3x sin3x)

u = 3x
du = 3dx

dv = sin3x dx
v = -1/3cos3x dx

∫ u dv = uv - ∫ v du

1/3∫ 3x sin3x dx = 1/3[3x(-1/3cos3x) - 1/3∫ -cos3x 3dx]
= 1/3[-xcos3x + 1/3 sin3x + C]
= -1/3 x cos3x + 1/9 sin3x + C

A little less sure about this one...going to check it by taking the derivative:

-1/3[ x (-3sin3x) + cos3x] + 1/9[ 3 cos3x]
x sin3x - 1/3 cos3x + 1/3 cos3x
x sin3x

I get what I wanted, so it must be right.