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 analyze the integrals \( I_1 \) and \( I_2 \) given by:

\[ I_1 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{e^{\sin x} + 1}{e^{\cos x} + 1} \, dx \]

\[ I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{e^{\cos x} + 1}{e^{\sin x} + 1} \, dx \]

### Step 1: Analyze the interval
In the interval \( \left[\frac{\pi}{4}, \frac{\pi}{2}\right] \):

- \( \sin x \) is increasing and \( \cos x \) is decreasing.
- At \( x = \frac{\pi}{4} \), \( \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).
- At \( x = \frac{\pi}{2} \), \( \sin\left(\frac{\pi}{2}\right) = 1 \) and \( \cos\left(\frac{\pi}{2}\right) = 0 \).

Thus, in this interval, \( \sin x \) is greater than \( \cos x \).

### Step 2: Compare the integrands
Since \( \sin x > \cos x \) in the interval, we can analyze the expressions:

\[ \frac{e^{\sin x} + 1}{e^{\cos x} + 1} \quad \text{and} \quad \frac{e^{\cos x} + 1}{e^{\sin x} + 1} \]

From the fact that \( e^{\sin x} > e^{\cos x} \), we can conclude:

\[ \frac{e^{\sin x} + 1}{e^{\cos x} + 1} > 1 \quad \text{and} \quad \frac{e^{\cos x} + 1}{e^{\sin x} + 1} < 1 \]

### Step 3: Establish the inequality
This implies:

\[ I_1 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{e^{\sin x} + 1}{e^{\cos x} + 1} \, dx > \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 1 \, dx \]

And similarly:

\[ I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{e^{\cos x} + 1}{e^{\sin x} + 1} \, dx < \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 1 \, dx \]

### Step 4: Conclusion
Since \( I_1 > I_2 \), we conclude that:

\[ I_1 > I_2 \]

### Final Answer
Thus, the answer is that \( I_1 \) is greater than \( I_2 \).

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