Posted by Tracy on Saturday, November 19, 2011 at 10:10pm.
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

calculus  drwls, Saturday, November 19, 2011 at 10:50pm
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.

calculus  Steve, Sunday, November 20, 2011 at 7:05am
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))
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