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))
To evaluate the integral of (x)cos(3x)dx, we can use integration by parts. The formula for integration by parts is as follows:
∫(u dv) = uv - ∫(v du)
In this case, we can choose:
u = x --> Derivative of u: du = dx
dv = cos(3x) --> Integral of dv: v = (1/3)sin(3x)
Now, using the integration by parts formula, we have:
∫(x)cos(3x)dx = (x)((1/3)sin(3x)) - ∫((1/3)sin(3x))(dx)
The first part, (x)((1/3)sin(3x)), can be simplified as:
(x)((1/3)sin(3x)) = (1/3)(x)(sin(3x))
For the second part, we need to integrate sin(3x) with respect to x. The integral of sin(3x) is:
∫sin(3x)dx = -(1/3)cos(3x) + C
Now, substituting back into the original equation:
(1/3)(x)(sin(3x)) - ∫((1/3)sin(3x))(dx) = (1/3)(x)(sin(3x)) + (1/3)cos(3x) + C
Therefore, the correct answer is C. (1/3)(x)(sin(3x)) + (1/9)(cos(3x)) + C.