Consider `I_(1)=int_((pi)/(4))^((pi)/(2))(e^(sinx)+1)/(e^(cosx)+1)dx and I_(2)=int_((pi)/(4))^((pi)/(2))(e^(cosx)+1)/(e^(sinx)+1)dx`, then

Options

A. `I_(1)gtI_(2)`
B. `I_(1)ltI_(2)`
C. `I_(1)=I_(2)`
D. `I_(1)+I_(2)=0`

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and compare their values.

### Step 1: Define the integrals
We have:
\[ I_1 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{e^{\sin x} + 1}{e^{\cos x} + 1} \, dx \]
\[ I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{e^{\cos x} + 1}{e^{\sin x} + 1} \, dx \]

### Step 2: Analyze the functions involved
We need to analyze the behavior of the functions \( e^{\sin x} \) and \( e^{\cos x} \) over the interval \( \left[\frac{\pi}{4}, \frac{\pi}{2}\right] \).

- At \( x = \frac{\pi}{4} \):
\[ \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \]
Thus,
\[ e^{\sin\left(\frac{\pi}{4}\right)} = e^{\cos\left(\frac{\pi}{4}\right)} \]

- At \( x = \frac{\pi}{2} \):
\[ \sin\left(\frac{\pi}{2}\right) = 1, \quad \cos\left(\frac{\pi}{2}\right) = 0 \]
Thus,
\[ e^{\sin\left(\frac{\pi}{2}\right)} = e^1, \quad e^{\cos\left(\frac{\pi}{2}\right)} = e^0 = 1 \]

### Step 3: Compare \( \sin x \) and \( \cos x \)
In the interval \( \left[\frac{\pi}{4}, \frac{\pi}{2}\right] \):
- \( \sin x \) is increasing and \( \cos x \) is decreasing.
- At \( x = \frac{\pi}{4} \), both are equal.
- For \( x > \frac{\pi}{4} \), \( \sin x > \cos x \).

This implies:
\[ e^{\sin x} > e^{\cos x} \quad \text{for } x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \]

### Step 4: Establish inequalities
Since \( e^{\sin x} > e^{\cos x} \), we can add 1 to both sides:
\[ e^{\sin x} + 1 > e^{\cos x} + 1 \]
Thus,
\[ \frac{e^{\sin x} + 1}{e^{\cos x} + 1} > 1 \]
for \( x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \).

### Step 5: Analyze \( I_2 \)
For \( I_2 \):
\[ \frac{e^{\cos x} + 1}{e^{\sin x} + 1} < 1 \]
for \( x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \).

### Step 6: Conclusion
Since \( I_1 \) integrates a function that is greater than 1 and \( I_2 \) integrates a function that is less than 1 over the same interval, we conclude:
\[ I_1 > I_2 \]

### Final Result
Thus, the final answer is:
\[ I_1 > I_2 \]