Evaluate : `int_((pi)/(6))^((pi)/(3)) cos x dx `.

Answer

To evaluate the integral \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x \, dx \), we can follow these steps:

### Step 1: Identify the integral
We need to evaluate the integral of \( \cos x \) from \( \frac{\pi}{6} \) to \( \frac{\pi}{3} \).

### Step 2: Find the antiderivative
The antiderivative of \( \cos x \) is \( \sin x \). Therefore, we can write:
\[ \int \cos x \, dx = \sin x + C \]
where \( C \) is the constant of integration.

### Step 3: Apply the limits of integration
Now we will evaluate the definite integral using the limits:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x \, dx = \left[ \sin x \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \]
This means we will compute \( \sin\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{6}\right) \).

### Step 4: Calculate \( \sin\left(\frac{\pi}{3}\right) \) and \( \sin\left(\frac{\pi}{6}\right) \)
We know:
- \( \sin\left(\frac{\pi}{3}\right) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
- \( \sin\left(\frac{\pi}{6}\right) = \sin(30^\circ) = \frac{1}{2} \)

### Step 5: Substitute the values into the expression
Now we substitute these values into the expression:
\[ \sin\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2} \]

### Step 6: Simplify the expression
To simplify:
\[ \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x \, dx = \frac{\sqrt{3} - 1}{2} \]

---