`int_(pi//6)^(pi//3) sin(3x)dx=`

Options

A. `(1)/(3)`
B. `-(1)/(3)`
C. `1`
D. `3`

Explanation

To solve the integral \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin(3x) \, dx \), we will follow these steps:

### Step 1: Find the antiderivative of \( \sin(3x) \)

The integral of \( \sin(kx) \) is given by:
\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \]
In our case, \( k = 3 \), so:
\[ \int \sin(3x) \, dx = -\frac{1}{3} \cos(3x) + C \]

### Step 2: Evaluate the definite integral

Now we need to evaluate the definite integral from \( \frac{\pi}{6} \) to \( \frac{\pi}{3} \):
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin(3x) \, dx = \left[-\frac{1}{3} \cos(3x)\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \]

### Step 3: Calculate the limits

Now we will calculate the values at the limits:
1. For \( x = \frac{\pi}{3} \):
\[ -\frac{1}{3} \cos\left(3 \cdot \frac{\pi}{3}\right) = -\frac{1}{3} \cos(\pi) = -\frac{1}{3} \cdot (-1) = \frac{1}{3} \]

2. For \( x = \frac{\pi}{6} \):
\[ -\frac{1}{3} \cos\left(3 \cdot \frac{\pi}{6}\right) = -\frac{1}{3} \cos\left(\frac{\pi}{2}\right) = -\frac{1}{3} \cdot 0 = 0 \]

### Step 4: Subtract the two results

Now we subtract the two results:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin(3x) \, dx = \frac{1}{3} - 0 = \frac{1}{3} \]

### Final Answer

Thus, the value of the integral \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin(3x) \, dx \) is:
\[ \frac{1}{3} \]

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