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December 18, 2014

December 18, 2014

Posted by **Tracy** on Saturday, November 19, 2011 at 10:10pm.

(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 -
**drwls**, Saturday, November 19, 2011 at 10:50pmAn 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 -
**Steve**, Sunday, November 20, 2011 at 7:05amJust 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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