The value of `int_(0)^(pi) (2^(sin x)cos x)/(s^([sin x])).dx` when [.] denotes the greatest integer function is equal to

Options

A. 0
B. `(pi)/(log 2)`
C. `(2pi)/(log 2)`
D. none

Answer

To solve the integral

\[ I = \int_{0}^{\pi} \frac{2^{\sin x} \cos x}{s^{[\sin x]}} \, dx \]

where \([\cdot]\) denotes the greatest integer function, we can use the property of definite integrals and symmetry.

### Step 1: Define the function
Let

\[ f(x) = \frac{2^{\sin x} \cos x}{s^{[\sin x]}} \]

### Step 2: Evaluate \(f(\pi - x)\)
Now, we will evaluate \(f(\pi - x)\):

\[ f(\pi - x) = \frac{2^{\sin(\pi - x)} \cos(\pi - x)}{s^{[\sin(\pi - x)]}} \]

Using the properties of sine and cosine:

- \(\sin(\pi - x) = \sin x\)
- \(\cos(\pi - x) = -\cos x\)

Thus, we have:

\[ f(\pi - x) = \frac{2^{\sin x} (-\cos x)}{s^{[\sin x]}} = -\frac{2^{\sin x} \cos x}{s^{[\sin x]}} = -f(x) \]

### Step 3: Use the property of definite integrals
Since \(f(\pi - x) = -f(x)\), we can use the property of definite integrals:

\[ \int_{0}^{\pi} f(x) \, dx = \int_{0}^{\pi} f(\pi - x) \, dx \]

This gives us:

\[ \int_{0}^{\pi} f(x) \, dx = -\int_{0}^{\pi} f(x) \, dx \]

### Step 4: Set up the equation
Let \(I = \int_{0}^{\pi} f(x) \, dx\). Then we have:

\[ I = -I \]

### Step 5: Solve for \(I\)
Adding \(I\) to both sides gives:

\[ 2I = 0 \implies I = 0 \]

### Conclusion
Thus, the value of the integral is:

\[ \int_{0}^{\pi} \frac{2^{\sin x} \cos x}{s^{[\sin x]}} \, dx = 0 \]