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The problem is:

$$\int_0^\pi\sqrt{\frac{1+\cos(2x)}{2}}dx$$

It is obvious that by using trig identities we can come up with:

$$\int_0^\pi\sqrt{\cos^2(x)} =$$ $$= \int_0^\pi|\cos(x)|dx$$

Here I have a slight problem. My professor offered a solution:

$$\int_0^\pi|\cos(x)|dx = \int_0^\frac{\pi}{2}\cos(x)dx + \int_\frac{\pi}{2}^\pi-\cos(x)dx$$

From here, it is simple to calculate, but I do not understand this separation into two integrals. I figure it's because of the absolute value of the cosine function, but I cannot quite grasp it... The minus before the cosine in the second integral is also confusing. If someone could elaborate, I would be very grateful. Thanks!

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    $\begingroup$ Use $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx$. Set $a=0, b=\pi, c=\pi /2$. And the rest follows by the definition of absolute values. $\endgroup$ – zxcvber Mar 28 '18 at 6:10
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    $\begingroup$ You're correct, $\cos x\ge 0$ for $x\in[0,\pi/2]$ and $\cos x\le 0$ for $x\in[\pi/2,\pi]$. So $|\cos x| = \cos x$ for $x\in[0,\pi/2]$ and $-\cos x$ for $x\in[\pi/2,\pi]$. $\endgroup$ – Quang Hoang Mar 28 '18 at 6:13
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$|\cos(x)|$ means absolute value of $\cos (x)$

\begin{align} |\cos(x) |= \begin{cases} \cos(x) &, \text{if } \cos(x) \ge 0\\ -\cos (x)& , \text{if } \cos(x) <0 \end{cases} \end{align}

$\cos(x) \ge 0 $ when $x$ is from $0$ to $\frac{\pi}2$

and $\cos(x) \le 0$ when $x$ is from $\frac{\pi}{2}$ to $\pi$.

\begin{align}\int_0^\pi|\cos(x)|dx&= \int_0^\frac{\pi}{2}|\cos(x)|dx + \int_\frac{\pi}{2}^\pi|\cos(x)|dx \\& = \int_0^\frac{\pi}{2}\cos(x)dx + \int_\frac{\pi}{2}^\pi-\cos(x)dx \end{align}

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