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

Answer

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

### Step 1: Identify the integral
We need to evaluate the integral:
\[ \int_{0}^{\pi} \sin(3x) \, dx \]

### Step 2: Find the antiderivative
The antiderivative of \( \sin(3x) \) can be found using the formula for the integral of sine:
\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \]
In our case, \( k = 3 \). Therefore, the antiderivative is:
\[ -\frac{1}{3} \cos(3x) + C \]

### Step 3: Apply the limits
Now we will apply the limits from \( 0 \) to \( \pi \):
\[ \left[-\frac{1}{3} \cos(3x)\right]_{0}^{\pi} \]

### Step 4: Evaluate at the upper limit
First, we evaluate at the upper limit \( x = \pi \):
\[ -\frac{1}{3} \cos(3\pi) = -\frac{1}{3} \cdot (-1) = \frac{1}{3} \]

### Step 5: Evaluate at the lower limit
Next, we evaluate at the lower limit \( x = 0 \):
\[ -\frac{1}{3} \cos(3 \cdot 0) = -\frac{1}{3} \cdot 1 = -\frac{1}{3} \]

### Step 6: Subtract the lower limit from the upper limit
Now we subtract the value at the lower limit from the value at the upper limit:
\[ \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

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

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