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When you are solving an integral, and you have a square root of a square, for instance

$$ \int \sqrt{1+\cos x} = \int \sqrt{\frac{\sin^2x}{1-\cos x}}$$

Do I have to take the absolute value of $\sin x$, or can I just take the positive value?

$$= \int \frac{|\sin x|}{\sqrt{1-\cos x}}$$ OR $$= \int \frac{\sin x}{\sqrt{1-\cos x}}$$

If I have to take the absolute value, should I look for a better method when solving integrals than that one, because two solutions seem more complicated.

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    $\begingroup$ It depends on your interval. As what I know we usually just take the positive value for indefinite integral. $\endgroup$ – Crazy Jul 18 '17 at 10:15
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    $\begingroup$ $\sqrt{\sin^{2}(x)} = \vert \sin(x) \vert$. To remove the absolute values, you need to specify the range for $x$. For instance: $x \in [0,\pi]$. $\endgroup$ – jibounet Jul 18 '17 at 10:15
  • $\begingroup$ @jibounet So, what happens if it is indefinite integral? And what happens when it's definite integral? Could you please get into more detail, and write an answer? $\endgroup$ – Shocky2 Jul 18 '17 at 13:08
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For the sake of simplicity, let us consider

$$ \int \sqrt{\sin^{2}(x)} \, dx \tag{1}$$

It is not completely clear what $(1)$ means to me.

Consider:

$$ \forall s \in \mathbb{R}, \; F(s) = \int_{0}^{s} \sqrt{\sin^{2}(x)} \; dx. $$

$F$ is the anti-derivative of $x \mapsto \sqrt{\sin^{2}(x)}$ which vanishes at $0$. [Note that the choice of $0$ is arbitrary.]

As you know: for all $x \in \mathbb{R}, \; \sqrt{\sin^{2}(x)} = \vert \sin(x) \vert$. Therefore, $F$ is always given by:

$$ \forall s \in \mathbb{R}, \; F(s) = \int_{0}^{s} \vert \sin(x) \vert \; dx. \tag{2}$$

If you are interested in, say, $F(\pi)$, the sign of $\vert \sin(x) \vert$ is constant on $[0,\pi]$. Therefore:

$$ F(\pi) = \int_{0}^{\pi} \vert \sin(x) \vert \, dx = \int_{0}^{\pi} \sin(x) \, dx. $$

However, if $s$ is arbitrary, the sign of $\vert \sin(x) \vert$ might change as $x$ ranges from $0$ to $s$. Then, one cannot remove the absolute values. One possible workaround is to use Chasles relation for integrals and split $F$ into a sum of integrals for which the sign of $\vert \sin(x) \vert$ is constant. Example:

$$ \begin{align*} F(3\pi/2) & = {} \int_{0}^{3\pi/2} \vert \sin(x) \vert \; dx \\[2mm] & = \int_{0}^{\pi} \vert \sin(x) \vert \; dx + \int_{\pi}^{3\pi/2} \vert \sin(x) \vert \, dx \\[2mm] & = \int_{0}^{\pi} \sin(x) \, dx - \int_{\pi}^{3\pi/2} \sin(x) \, dx \end{align*}. $$

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In general the square root sign denotes the principal (positive) root, in which case you could omit the abs-brackets, but it always depends on the task.

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    $\begingroup$ You can't omit the brackets in general. It depends on the interval over which we're integrating (or, in the case of an indefinite integral, you can't drop them at all). This is because $\sin x$ can be both positive and negative. $\endgroup$ – Arthur Jul 18 '17 at 10:17

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