Asked 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

Answers

Answered by 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.
Answered by 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))
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