Question
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
(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
Answers
drwls
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.
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.
Steve
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))
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))