The integral of the absolute value of $\sin(x)$, denoted by $\int |\sin(x)| \, dx$, can be approached by breaking the function into intervals where $\sin(x)$ is positive or negative. This is because the absolute value function affects the behavior of $\sin(x)$ differently depending on the interval.

Step-by-Step Solution

1. Break the integral into intervals:

   $\sin(x)$ is positive on the interval $[2n\pi, (2n+1)\pi]$, and negative on the interval $[(2n+1)\pi, (2n+2)\pi]$ for integer $n$. Thus, the absolute value of $\sin(x)$ can be handled by splitting the integral into two parts.

2. Positive intervals (where $\sin(x) \geq 0$): On $[2n\pi, (2n+1)\pi]$, $|\sin(x)| = \sin(x)$. The integral in this region is: $\int \sin(x) \, dx = -\cos(x) + C$

3. Negative intervals (where $\sin(x) < 0$): On $[(2n+1)\pi, (2n+2)\pi]$, $|\sin(x)| = -\sin(x)$. The integral in this region is: $\int -\sin(x) \, dx = \cos(x) + C$

4. General formula: The integral of $|\sin(x)|$ over a full period, from $0$ to $2\pi$, is the sum of the integrals over the intervals $[0, \pi]$ and $[\pi, 2\pi]$: $\int_0^\pi \sin(x) \, dx + \int_\pi^{2\pi} -\sin(x) \, dx$

   Evaluating these: $\int_0^\pi \sin(x) \, dx = [-\cos(x)]_0^\pi = 2$ $\int_\pi^{2\pi} -\sin(x) \, dx = [\cos(x)]_\pi^{2\pi} = 2$

   Therefore, over a full period: $\int_0^{2\pi} |\sin(x)| \, dx = 2 + 2 = 4$

Final Answer:

The indefinite integral of $|\sin(x)|$ is a piecewise function, depending on the intervals where $\sin(x)$ is positive or negative. Over one period $[0, 2\pi]$, the integral evaluates to $4$.

Would you like more details on any step or further clarifications?

Here are some related questions:

1. What is the integral of $\cos(x)$?
2. How can you calculate the area under the curve of $\sin(x)$?
3. How do you handle integrals involving absolute values of more complex functions?
4. How does the integral of $\sin(x)$ change over different intervals?
5. What is the Fourier series of $|\sin(x)|$?

Tip: Always check the periodicity of trigonometric functions when solving integrals, as they can help simplify the calculations.