Answer A or D for Calc Integration Question

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The discussion centers around a calculus integration question, specifically whether \(\frac{\sin^2(x)}{2}\) equals \(\frac{-\cos(2x)}{4}\). Participants analyze the problem by substituting values and differentiating the options. The consensus is that the correct answer is "D," as demonstrated through simplification and differentiation of the functions involved. One user highlights the importance of recognizing that any antiderivative can include a constant. The conversation concludes with a clarification on how to evaluate the integral correctly.
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question and work is http://home.earthlink.net/~urban-xrisis/clip_image002.jpg

I got the answer down to letter A and D. Now I feel like it's letter A but not sure...
 
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Youre asking if \frac{sin^2(x)}{2} = \frac{-cos(2x)}{4}

Try x = 0

Or notice that f(x) = sin(x)cos(x) = sin(2x)/2, then the integral is real easy.
 
The easiest method is to differentiate each of the 3 results...

"D" is the correct answer.

Daniel.
 
Take the derivative of each of the choices.

*Haha, too late
 
whozum said:
Youre asking if \frac{\sin^2{x}}{2}=\frac{-\cos 2x}{4}

Not at all. If F(x) is an antiderivative of f(x) then so is F(x)+C for any C. Here,

\frac{\sin^2 x}{2} = \left(-\frac{\cos 2x}{4}\right) + \frac{1}{4}
 
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dextercioby said:
The easiest method is to differentiate each of the 3 results...

"D" is the correct answer.

Daniel.

how did you get III to be true?
 
look at my last post.
 
Convert f(x) to what I recommended and the integral evaluates to D directly.
 
Data said:
Not at all. If F(x) is an antiderivative of f(x) then so is F(x)+C for any C. Here,

\frac{\sin^2 x}{2} = \left(-\frac{\cos 2x}{4}\right) + \frac{1}{4}

Good call, I didnt see that.
 
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