`int_(0)^(pi) sin 3x dx`

Answer

To solve the definite integral \( \int_{0}^{\pi} \sin(3x) \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = 3x \). Then, differentiate both sides:
\[ dt = 3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3} \]

### Step 2: Change the limits
Now we need to change the limits of integration according to our substitution:
- When \( x = 0 \), \( t = 3 \cdot 0 = 0 \)
- When \( x = \pi \), \( t = 3 \cdot \pi = 3\pi \)

### Step 3: Rewrite the integral
Substituting \( t \) and \( dx \) into the integral, we get:
\[ \int_{0}^{\pi} \sin(3x) \, dx = \int_{0}^{3\pi} \sin(t) \cdot \frac{dt}{3} = \frac{1}{3} \int_{0}^{3\pi} \sin(t) \, dt \]

### Step 4: Evaluate the integral
The integral of \( \sin(t) \) is \( -\cos(t) \). Thus, we can evaluate the integral:
\[ \frac{1}{3} \int_{0}^{3\pi} \sin(t) \, dt = \frac{1}{3} \left[ -\cos(t) \right]_{0}^{3\pi} \]
Calculating the limits:
\[ = \frac{1}{3} \left( -\cos(3\pi) + \cos(0) \right) \]

### Step 5: Calculate the cosine values
We know that:
\[ \cos(3\pi) = -1 \quad \text{and} \quad \cos(0) = 1 \]
Substituting these values back, we have:
\[ = \frac{1}{3} \left( -(-1) + 1 \right) = \frac{1}{3} \left( 1 + 1 \right) = \frac{1}{3} \cdot 2 = \frac{2}{3} \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{\pi} \sin(3x) \, dx \) is:
\[ \frac{2}{3} \]

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