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The original question was " Integrate $\sqrt{1+\cos x}$ from $0$ to $200\pi$ " , so what I am in doubt is that in area we don't cancel positive and negative areas, but in definite integrals it should come out to be $0$ but in the solution, the $\sqrt{1+\cos x}$ is evaluated as $\left|\cos\frac{x}{2}\right|$, I don't understand here why the modulus of $\cos\frac{x}{2}$ is there, should not it be simply $\cos\frac{x}{2}$ as the square root of any number can be both positive and negative?

Jack D'Aurizio
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satyatech
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  • $\sqrt{1+\cos x}$ is a non-negative function, which cancellation are you referring to? – Jack D'Aurizio Mar 27 '17 at 17:17
  • Can't the sqrt have both +ve and - ve values @JackD'Aurizio – satyatech Mar 27 '17 at 17:18
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    $\sqrt{\cdot}$ is by definition nonnegative. To indicate the negative root, simply prefix a minus sign. So, "$\sqrt{4}$" is $2$, and not $-2$; likewise "$-\sqrt{4}$" is $-2$. It's quite unambiguous. You should distinguish carefully in your language between "THE square root" and "A square root". The former is nonnegative, while the latter includes both the nonnegative and its opposite. – MPW Mar 27 '17 at 17:18
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    This is not your first question on MSE, you should be aware that formatting matters. If you do not care about writing a proper question, why should we care about giving a proper answer? Anyway, $\sqrt{x^2}$ is $|x|$, not $x$. The square root function, where defined, is non-negative. – Jack D'Aurizio Mar 27 '17 at 17:22
  • Why is it not -2 as (-2*-2=4) – satyatech Mar 27 '17 at 17:23
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    Because $\sqrt{(-2)^2}=2$ by the definition of $\sqrt{\cdot}$. – Jack D'Aurizio Mar 27 '17 at 17:23
  • @JackD'Aurizio,thanks bro but I am really in hurry now as my exams are just tomorrow.Anyways thanks for answering this simple doubt. So finally can I say that in all questions wherever sqrt is given I will always take as as $$\ |x|$$ – satyatech Mar 27 '17 at 17:28
  • @satyatech: please be careful, I have only said that $\sqrt{x^2}=|x|$. Since $1-\cos(x) = 2\sin^2\frac{x}{2}$, the square root of $1-\cos(x)$ is $\sqrt{2}$ times the absolute value of $\sin\frac{x}{2}$. – Jack D'Aurizio Mar 27 '17 at 17:31

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