calculus
posted by Tracy .
Evaluate the integral of
(x)cos(3x)dx
A. (1/6)(x^2)(sin)(3x)+C
B. (1/3)(x)(sin)(3x)(1/3)(sin)(3x)+C
C. (1/3)(x)(sin)(3x)+(1/9)(cos)(3x)+C
D. (1/2)(x^2)+(1/18)(sin)^2(3x)+C

An easy way to do this is to take the derivatives of the four choices, and see what gives you back the original function.
The derivative of function C. is
(1/3)sin(3x) + (x/3)*3*cos(3x) (3/9)sin(3x) = x cos(3x)
Looks like a winner to me. 
Just as easy as taking all those derivatives (to me anyway) is to use integration by parts:
u = x
du = dx
dv = cos 3x dx
v = 1/3 sin 3x
Int(u dv) = uv  Int(v du)
= x/3 sin 3x  Int(1/3 sin 3x dx)
= x/3 sin 3x + 1/9 cos 3x
= (C)
PS (sin)(3x) is bogus notation. (sin) is not a value like (1/3) or (x). It is a function.
Try (sin(3x))