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calculus

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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 - ,

    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 - ,

    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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