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Posted by 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 - , 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 - , 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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