If `I_(1)=int_(0)^((pi)/(2))e^(sinx)(1+x cos x)dx and I_(2)=int_(0)^((pi)/(2))e^(cosx)(1-x sin x)dx, ` then `[(I_(1))/(I_(2))]` is equal to (where `[x]` denotes the greatest integer less than or equal to x)

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and then find the ratio \( \frac{I_1}{I_2} \).

### Step 1: Define the integrals
We have:
\[ I_1 = \int_{0}^{\frac{\pi}{2}} e^{\sin x} (1 + x \cos x) \, dx \]
\[ I_2 = \int_{0}^{\frac{\pi}{2}} e^{\cos x} (1 - x \sin x) \, dx \]

### Step 2: Use substitution for \( I_1 \)
Let \( t = x e^{\sin x} \). We differentiate this:
\[ \frac{dt}{dx} = e^{\sin x} (1 + x \cos x) \]
Thus, we can rewrite \( I_1 \) as:
\[ I_1 = \int_{0}^{\frac{\pi}{2}} e^{\sin x} (1 + x \cos x) \, dx = \left[ x e^{\sin x} \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} e^{\sin(\frac{\pi}{2})} - 0 = \frac{\pi}{2} e \]

### Step 3: Use substitution for \( I_2 \)
Let \( v = x e^{\cos x} \). We differentiate this:
\[ \frac{dv}{dx} = e^{\cos x} (1 - x \sin x) \]
Thus, we can rewrite \( I_2 \) as:
\[ I_2 = \int_{0}^{\frac{\pi}{2}} e^{\cos x} (1 - x \sin x) \, dx = \left[ x e^{\cos x} \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} e^{\cos(0)} - 0 = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2} \]

### Step 4: Calculate the ratio \( \frac{I_1}{I_2} \)
Now we can find the ratio:
\[ \frac{I_1}{I_2} = \frac{\frac{\pi}{2} e}{\frac{\pi}{2}} = e \]

### Step 5: Evaluate \( e \)
The value of \( e \) is approximately \( 2.718 \).

### Step 6: Find the greatest integer less than or equal to \( \frac{I_1}{I_2} \)
Thus, the greatest integer less than or equal to \( e \) is:
\[ \lfloor e \rfloor = 2 \]

### Final Answer
The answer is:
\[ \lfloor \frac{I_1}{I_2} \rfloor = 2 \]