Question:

Step 1: Identify the periodicity of the integrand. The integrand is $f(x) = |\cos x|$. The function $\cos x$ has a period of $2\pi$. However, due to the absolute value, $|\cos x|$ repeats its pattern more frequently. For $x \in [0, \pi/2]$, $\cos x \ge 0$. For $x \in [\pi/2, \pi]$, $\cos x \le 0$. For $x \in [\pi, 3\pi/2]$, $\cos x \le 0$. For $x \in [3\pi/2, 2\pi]$, $\cos x \ge 0$. Thus, $|\cos x|$ over $[0, \pi]$ is identical to its shape over $[\pi, 2\pi]$, and so on. Therefore, $|\cos x|$ is a periodic function with a period of $\pi$. Step 2: Apply the periodicity property to simplify the integral. For a periodic function $f(x)$ with period $T$, we have $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$. In this case, $f(x) = |\cos x|$, $T = \pi$, and the upper limit of integration is $4\pi$. So $n=4$. We can rewrite the given integral as: $$ I = \int_{0}^{4\pi} |\cos x| dx = 4 \int_{0}^{\pi} |\cos x| dx $$ Step 3: Split the integral over $[0, \pi]$ based on the sign of $\cos x$. To evaluate $\int_{0}^{\pi} |\cos x| dx$, we need to consider the intervals where $\cos x$ is positive and negative within $[0, \pi]$. $\cos x \ge 0$ for $x \in [0, \pi/2]$, so $|\cos x| = \cos x$. $\cos x \le 0$ for $x \in [\pi/2, \pi]$, so $|\cos x| = -\cos x$. Therefore, we split the integral: $$ \int_{0}^{\pi} |\cos x| dx = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{\pi} (-\cos x) dx $$ $$ \int_{0}^{\pi} |\cos x| dx = \int_{0}^{\pi/2} \cos x dx - \int_{\pi/2}^{\pi} \cos x dx $$ Step 4: Evaluate the definite integrals. Now, we evaluate each part of the integral. The antiderivative of $\cos x$ is $\sin x$. For the first part: $$ \int_{0}^{\pi/2} \cos x dx = [\sin x]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1 $$ For the second part: $$ \int_{\pi/2}^{\pi} \cos x dx = [\sin x]_{\pi/2}^{\pi} = \sin(\pi) - \sin(\pi/2) = 0 - 1 = -1 $$ Step 5: Substitute the evaluated integrals back to find the final result. Substitute the values from Step 4 into the expression from Step 3, and then multiply by 4 as determined in Step 2: $$ I = 4 \left[ \left( \int_{0}^{\pi/2} \cos x dx \right) - \left( \int_{\pi/2}^{\pi} \cos x dx \right) \right] $$ $$ I = 4 [1 - (-1)] $$ $$ I = 4 [1 + 1] $$ $$ I = 4 [2] $$ $$ I = 8 $$ The final answer is $\boxed{8}$.