The value of the integral ` int_(0)^(pi) | sin 2x| dx ` is

Answer

To solve the integral \( \int_{0}^{\pi} |\sin 2x| \, dx \), we can follow these steps:

### Step 1: Analyze the function \( |\sin 2x| \)

The function \( \sin 2x \) has a period of \( \pi \). Within the interval \( [0, \pi] \), the function \( \sin 2x \) will cross the x-axis, which means it will have both positive and negative values. We need to find the points where \( \sin 2x = 0 \) to determine the intervals where the function changes sign.

### Step 2: Find the zeros of \( \sin 2x \)

Setting \( \sin 2x = 0 \), we have:
\[ 2x = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, the zeros in the interval \( [0, \pi] \) occur at:
\[ x = 0, \frac{\pi}{2}, \pi \]

### Step 3: Split the integral based on the zeros

We can split the integral into two parts:
\[ \int_{0}^{\pi} |\sin 2x| \, dx = \int_{0}^{\frac{\pi}{2}} \sin 2x \, dx + \int_{\frac{\pi}{2}}^{\pi} -\sin 2x \, dx \]

### Step 4: Evaluate the first integral

For the first integral:
\[ \int_{0}^{\frac{\pi}{2}} \sin 2x \, dx \]
Using the substitution \( u = 2x \), \( du = 2dx \) or \( dx = \frac{du}{2} \). The limits change as follows:
- When \( x = 0 \), \( u = 0 \)
- When \( x = \frac{\pi}{2} \), \( u = \pi \)

Thus, we have:
\[ \int_{0}^{\frac{\pi}{2}} \sin 2x \, dx = \frac{1}{2} \int_{0}^{\pi} \sin u \, du \]
The integral of \( \sin u \) is:
\[ -\cos u \Big|_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \]
So,
\[ \frac{1}{2} \cdot 2 = 1 \]

### Step 5: Evaluate the second integral

For the second integral:
\[ \int_{\frac{\pi}{2}}^{\pi} -\sin 2x \, dx \]
Using the same substitution \( u = 2x \):
- When \( x = \frac{\pi}{2} \), \( u = \pi \)
- When \( x = \pi \), \( u = 2\pi \)

Thus, we have:
\[ \int_{\frac{\pi}{2}}^{\pi} -\sin 2x \, dx = -\frac{1}{2} \int_{\pi}^{2\pi} \sin u \, du \]
Calculating this integral:
\[ -\frac{1}{2} \left(-\cos u \Big|_{\pi}^{2\pi}\right) = -\frac{1}{2} \left(-\cos(2\pi) + \cos(\pi)\right) = -\frac{1}{2} \left(-1 - 1\right) = -\frac{1}{2} \cdot (-2) = 1 \]

### Step 6: Combine the results

Now, we combine the two results:
\[ \int_{0}^{\pi} |\sin 2x| \, dx = 1 + 1 = 2 \]

### Final Answer

The value of the integral \( \int_{0}^{\pi} |\sin 2x| \, dx \) is \( 2 \).

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