1. Key Concepts and Formulas

• Algebraic Identity: The expression under the square root resembles a perfect square trinomial: $a^2 - 2ab + b^2 = (a-b)^2$.
• Definition of Absolute Value: For any real number $A$, $\sqrt{A^2} = |A|$. This means we need to consider the sign of the expression inside the absolute value.
• Properties of Definite Integrals: If an integrand has different forms over different intervals, the integral can be split into the sum of integrals over those intervals. Specifically, $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$.
• Trigonometric Identities: We will use the half-angle identity for sine: $\sin^2(x/2) = \frac{1 - \cos x}{2}$.
• Integration of Basic Trigonometric Functions: $\int \cos x dx = \sin x + C$ and $\int \sin x dx = -\cos x + C$.

2. Step-by-Step Solution

Step 1: Simplify the integrand. The expression under the square root is $1 + 4\sin^2(x/2) - 4\sin(x/2)$. We can rewrite $1$ using the identity $\sin^2(x/2) + \cos^2(x/2) = 1$. So, $1 + 4\sin^2(x/2) - 4\sin(x/2) = (\sin^2(x/2) + \cos^2(x/2)) + 4\sin^2(x/2) - 4\sin(x/2)$. This does not immediately look like a perfect square. Let's try another approach.

Consider the expression $1 - 4\sin(x/2) + 4\sin^2(x/2)$. This is of the form $a^2 - 2ab + b^2$ if we let $a=1$ and $b=2\sin(x/2)$. Then, $a^2 = 1^2 = 1$, $b^2 = (2\sin(x/2))^2 = 4\sin^2(x/2)$, and $2ab = 2(1)(2\sin(x/2)) = 4\sin(x/2)$. Therefore, the expression $1 - 4\sin(x/2) + 4\sin^2(x/2)$ is a perfect square and can be written as $(1 - 2\sin(x/2))^2$.

The integral becomes: $I = \int\limits_0^\pi {\sqrt {(1 - 2\sin {x \over 2})^2} } dx$

Step 2: Apply the definition of the square root of a square. We know that $\sqrt{A^2} = |A|$. Thus, $I = \int\limits_0^\pi {|1 - 2\sin {x \over 2}|} dx$ To evaluate this integral, we need to determine when the expression $1 - 2\sin(x/2)$ is positive or negative within the interval $[0, \pi]$.

Step 3: Determine the sign of the expression $1 - 2\sin(x/2)$ in the interval $[0, \pi]$. We need to find the values of $x$ in $[0, \pi]$ for which $1 - 2\sin(x/2) = 0$. This implies $2\sin(x/2) = 1$, or $\sin(x/2) = 1/2$.

Let $u = x/2$. As $x$ varies from $0$ to $\pi$, $u$ varies from $0/2=0$ to $\pi/2$. So, we are looking for $u \in [0, \pi/2]$ such that $\sin u = 1/2$. The principal value for $u$ is $\pi/6$. So, $x/2 = \pi/6$, which means $x = 2(\pi/6) = \pi/3$.

Now we check the sign of $1 - 2\sin(x/2)$ in the intervals $[0, \pi/3)$ and $(\pi/3, \pi]$.

• For $x \in [0, \pi/3)$: This means $x/2 \in [0, \pi/6)$. In this interval, $\sin(x/2)$ is between $\sin(0)=0$ and $\sin(\pi/6)=1/2$. So, $0 \le \sin(x/2) < 1/2$. Multiplying by 2, we get $0 \le 2\sin(x/2) < 1$. Therefore, $1 - 2\sin(x/2) > 0$ for $x \in [0, \pi/3)$. So, $|1 - 2\sin(x/2)| = 1 - 2\sin(x/2)$ for $x \in [0, \pi/3]$.

• For $x \in (\pi/3, \pi]$: This means $x/2 \in (\pi/6, \pi/2]$. In this interval, $\sin(x/2)$ is between $\sin(\pi/6)=1/2$ and $\sin(\pi/2)=1$. So, $1/2 < \sin(x/2) \le 1$. Multiplying by 2, we get $1 < 2\sin(x/2) \le 2$. Therefore, $1 - 2\sin(x/2) < 0$ for $x \in (\pi/3, \pi]$. So, $|1 - 2\sin(x/2)| = -(1 - 2\sin(x/2)) = 2\sin(x/2) - 1$ for $x \in (\pi/3, \pi]$.

Step 4: Split the integral based on the sign changes. We split the integral at $x = \pi/3$: $I = \int_0^{\pi/3} (1 - 2\sin(x/2)) dx + \int_{\pi/3}^\pi (2\sin(x/2) - 1) dx$

Step 5: Evaluate the first integral. $I_1 = \int_0^{\pi/3} (1 - 2\sin(x/2)) dx$ Let $u = x/2$, so $du = (1/2)dx$, which means $dx = 2du$. When $x=0$, $u=0$. When $x=\pi/3$, $u=\pi/6$. $I_1 = \int_0^{\pi/6} (1 - 2\sin u) (2du) = 2 \int_0^{\pi/6} (1 - 2\sin u) du$ $I_1 = 2 [u - 2(-\cos u)]_0^{\pi/6} = 2 [u + 2\cos u]_0^{\pi/6}$ $I_1 = 2 [(\pi/6 + 2\cos(\pi/6)) - (0 + 2\cos(0))]$ $I_1 = 2 [(\pi/6 + 2(\sqrt{3}/2)) - (0 + 2(1))]$ $I_1 = 2 [\pi/6 + \sqrt{3} - 2]$ $I_1 = \pi/3 + 2\sqrt{3} - 4$

Step 6: Evaluate the second integral. $I_2 = \int_{\pi/3}^\pi (2\sin(x/2) - 1) dx$ Again, let $u = x/2$, so $dx = 2du$. When $x=\pi/3$, $u=\pi/6$. When $x=\pi$, $u=\pi/2$. $I_2 = \int_{\pi/6}^{\pi/2} (2\sin u - 1) (2du) = 2 \int_{\pi/6}^{\pi/2} (2\sin u - 1) du$ $I_2 = 2 [-2\cos u - u]_{\pi/6}^{\pi/2}$ $I_2 = 2 [(-2\cos(\pi/2) - \pi/2) - (-2\cos(\pi/6) - \pi/6)]$ $I_2 = 2 [(-2(0) - \pi/2) - (-2(\sqrt{3}/2) - \pi/6)]$ $I_2 = 2 [-\pi/2 - (-\sqrt{3} - \pi/6)]$ $I_2 = 2 [-\pi/2 + \sqrt{3} + \pi/6]$ $I_2 = 2 [\sqrt{3} + \pi/6 - \pi/2]$ $I_2 = 2 [\sqrt{3} + (1/6 - 3/6)\pi]$ $I_2 = 2 [\sqrt{3} - 2\pi/6]$ $I_2 = 2\sqrt{3} - 2\pi/3$

Step 7: Add the results of the two integrals. $I = I_1 + I_2$ $I = (\pi/3 + 2\sqrt{3} - 4) + (2\sqrt{3} - 2\pi/3)$ $I = (\pi/3 - 2\pi/3) + (2\sqrt{3} + 2\sqrt{3}) - 4$ $I = -\pi/3 + 4\sqrt{3} - 4$ $I = 4\sqrt{3} - 4 - \pi/3$

3. Common Mistakes & Tips

• Forgetting the Absolute Value: The most common mistake is to treat $\sqrt{A^2}$ as $A$ instead of $|A|$. This leads to an incorrect sign for the integrand over parts of the interval.
• Incorrectly Determining the Sign: Carefully analyze the interval for $x$ and the corresponding interval for $x/2$ to correctly identify where the expression inside the absolute value is positive or negative.
• Errors in Trigonometric Values: Ensure accurate recall of trigonometric values for standard angles like $\pi/6$, $\pi/3$, and $\pi/2$.
• Substitution Errors: When using substitution, remember to change the limits of integration and to adjust the differential element ($dx$).

4. Summary

The problem required simplifying an integrand containing a square root of a perfect square. This led to an absolute value function. We identified the point where the expression inside the absolute value changed sign ($\sin(x/2) = 1/2$ at $x = \pi/3$) and split the integral accordingly. By evaluating the definite integrals for each part and summing the results, we obtained the final value of the integral. The simplification of the integrand to $(1 - 2\sin(x/2))^2$ and careful handling of the absolute value were the key steps.

The final answer is $4\sqrt 3 - 4 - {\pi \over 3}$.

The final answer is \boxed{4\sqrt 3 - 4 - {\pi \over 3}}.