Key Concepts and Formulas:

1. Symmetry Property of Definite Integrals: For a function $f(x)$, if $f(2a-x) = f(x)$, then $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$.
2. Absolute Value Definition: $|u| = u$ if $u \ge 0$, and $|u| = -u$ if $u < 0$.
3. Trigonometric Identity for $\cos^3 x$: $\cos^3 x = \frac{1}{4}(3\cos x + \cos 3x)$.
4. Standard Integration Formulas: $\int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C$ and $\int \cos x \, dx = \sin x + C$.

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Step-by-step Solution:

Let the given integral be $I$. $I = \int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$

Step 1: Utilize Symmetry Property Let $f(x) = |\cos x|^3$. We check for symmetry around $x = \pi/2$. Consider $f(\pi - x)$: $f(\pi - x) = |\cos(\pi - x)|^3$ Since $\cos(\pi - x) = -\cos x$, we have: $f(\pi - x) = |-\cos x|^3 = |\cos x|^3 = f(x)$ The condition $f(2a-x) = f(x)$ is satisfied with $2a = \pi$, so $a = \pi/2$. Therefore, we can use the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$: $I = 2 \int\limits_0^{\pi/2} {{{\left| {\cos x} \right|}^3}} \,dx$

Step 2: Remove Absolute Value In the interval $[0, \pi/2]$, the cosine function, $\cos x$, is non-negative. Specifically, $\cos x \ge 0$ for $x \in [0, \pi/2]$. Thus, $|\cos x| = \cos x$ for $x \in [0, \pi/2]$. Substituting this into the integral: $I = 2 \int\limits_0^{\pi/2} {{{\cos^3} x}} \,dx$

Step 3: Simplify the Integrand using Trigonometric Identity We use the identity for $\cos^3 x$: $\cos^3 x = \frac{3\cos x + \cos 3x}{4}$ Substituting this into the integral: $I = 2 \int\limits_0^{\pi/2} {\frac{3\cos x + \cos 3x}{4}} \,dx$ $I = \frac{2}{4} \int\limits_0^{\pi/2} {(3\cos x + \cos 3x)} \,dx$ $I = \frac{1}{2} \int\limits_0^{\pi/2} {(3\cos x + \cos 3x)} \,dx$

Step 4: Integrate Term by Term Now we integrate the expression with respect to $x$: $I = \frac{1}{2} \left[ 3\sin x + \frac{\sin 3x}{3} \right]_0^{\pi/2}$

Step 5: Apply the Limits of Integration We evaluate the expression at the upper limit ($x = \pi/2$) and the lower limit ($x = 0$).

At the upper limit $x = \pi/2$: $3\sin(\pi/2) + \frac{\sin(3 \cdot \pi/2)}{3} = 3(1) + \frac{\sin(3\pi/2)}{3}$ Since $\sin(3\pi/2) = -1$: $3 + \frac{-1}{3} = 3 - \frac{1}{3} = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}$

At the lower limit $x = 0$: $3\sin(0) + \frac{\sin(3 \cdot 0)}{3} = 3(0) + \frac{\sin(0)}{3}$ Since $\sin(0) = 0$: $0 + \frac{0}{3} = 0$

Now, subtract the value at the lower limit from the value at the upper limit: $I = \frac{1}{2} \left[ \frac{8}{3} - 0 \right]$ $I = \frac{1}{2} \cdot \frac{8}{3}$ $I = \frac{4}{3}$

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Common Mistakes & Tips:

• Incorrectly Handling Absolute Value: The most common mistake is to either ignore the absolute value or to incorrectly determine the sign of $\cos x$ over the integration interval.
• Mistaking Periodicity for Symmetry: While $|\cos x|$ has a period of $\pi$, the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ relies on the specific symmetry $f(2a-x) = f(x)$, not just the general periodicity of the function.
• Trigonometric Identity Errors: Ensure the correct trigonometric identity for $\cos^3 x$ is used and that it is correctly manipulated. Alternatively, one could use $\cos^3 x = \cos^2 x \cdot \cos x = (1-\sin^2 x)\cos x$ and a substitution $u = \sin x$.

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Summary:

The problem requires careful handling of the absolute value function and utilizes a symmetry property of definite integrals. By first reducing the integration interval from $[0, \pi]$ to $[0, \pi/2]$ using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ for $f(\pi-x)=f(x)$, we simplify the integrand. In the interval $[0, \pi/2]$, $\cos x \ge 0$, so $|\cos x| = \cos x$. The integral then becomes $\int_0^{\pi/2} \cos^3 x \, dx$. Using the identity $\cos^3 x = \frac{3\cos x + \cos 3x}{4}$, we can integrate the expression and evaluate it using the limits of integration. The final result is $\frac{4}{3}$.

The final answer is $\boxed{4/3}$.