If the integral `I=∫(-(sinx)/(x)-ln x cosx)dx=f(x)+C` (where, C is the constant of integration) and `f(e )=-sine`, then the number of natural numbers less than `[f((pi)/(6))]` is equal to (where `[.]` is the greatest integer function)

Answer

To solve the integral \( I = \int \left( -\frac{\sin x}{x} - \ln x \cos x \right) dx = f(x) + C \), where \( C \) is the constant of integration and \( f(e) = -\sin e \), we need to find the value of \( f\left(\frac{\pi}{6}\right) \) and then determine the number of natural numbers less than \( [f\left(\frac{\pi}{6}\right)] \).

### Step 1: Find the expression for \( f(x) \)

From the integral, we can separate it into two parts:
\[ I = -\int \frac{\sin x}{x} \, dx - \int \ln x \cos x \, dx \]

Let’s denote:
\[ f(x) = -\int \frac{\sin x}{x} \, dx - \int \ln x \cos x \, dx + C \]

### Step 2: Evaluate \( f(e) \)

We know that \( f(e) = -\sin e \). This gives us a condition to find the constant \( C \) later.

### Step 3: Evaluate \( f\left(\frac{\pi}{6}\right) \)

Now we need to find \( f\left(\frac{\pi}{6}\right) \):
\[ f\left(\frac{\pi}{6}\right) = -\int \frac{\sin\left(\frac{\pi}{6}\right)}{\frac{\pi}{6}} \, dx - \int \ln\left(\frac{\pi}{6}\right) \cos\left(\frac{\pi}{6}\right) \, dx \]

Calculating \( \sin\left(\frac{\pi}{6}\right) \) and \( \cos\left(\frac{\pi}{6}\right) \):
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

Thus,
\[ f\left(\frac{\pi}{6}\right) = -\int \frac{\frac{1}{2}}{\frac{\pi}{6}} \, dx - \int \ln\left(\frac{\pi}{6}\right) \cdot \frac{\sqrt{3}}{2} \, dx \]

### Step 4: Simplify the expression

The first integral simplifies to:
\[ -\int \frac{3}{\pi} \, dx = -\frac{3}{\pi} x \]

The second integral becomes:
\[ -\frac{\sqrt{3}}{2} \int \ln\left(\frac{\pi}{6}\right) \, dx = -\frac{\sqrt{3}}{2} \cdot \ln\left(\frac{\pi}{6}\right) \cdot x \]

Combining these:
\[ f\left(\frac{\pi}{6}\right) = -\left(\frac{3}{\pi} \cdot \frac{\pi}{6} + \frac{\sqrt{3}}{2} \cdot \ln\left(\frac{\pi}{6}\right) \cdot \frac{\pi}{6}\right) \]

### Step 5: Evaluate \( f\left(\frac{\pi}{6}\right) \)

Calculating this gives:
\[ f\left(\frac{\pi}{6}\right) = -\left(\frac{3}{2} + \frac{\sqrt{3}}{12} \cdot \ln\left(\frac{\pi}{6}\right)\right) \]

### Step 6: Find the greatest integer function

Now we need to evaluate \( [f\left(\frac{\pi}{6}\right)] \). We need to find the numerical value of \( f\left(\frac{\pi}{6}\right) \) to determine how many natural numbers are less than this value.

Assuming \( \ln\left(\frac{\pi}{6}\right) \) is a small negative number, we can estimate:
\[ f\left(\frac{\pi}{6}\right) \approx -\left(\frac{3}{2} + \text{small positive value}\right) \]

This indicates that \( f\left(\frac{\pi}{6}\right) \) is slightly less than -1.

### Conclusion

Thus, the greatest integer less than \( f\left(\frac{\pi}{6}\right) \) is \( -2 \). Therefore, the number of natural numbers less than \( [f\left(\frac{\pi}{6}\right)] \) is:

\[ \text{Natural numbers less than } -2 = 0 \]

**Final Answer: 0**