Choose the correct answer. The value of `int_0^(pi/2) log((4+3sinx)/(4+3cosx))dx`

Options

A. `2`
B. `3/4`
C. `0`
D. `-2`

Explanation

To solve the integral \( I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) dx \), we can use a symmetry property of definite integrals.

### Step-by-step Solution:

1. **Define the Integral:**
Let \( I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) dx \).

2. **Use the Property of Definite Integrals:**
We can use the property that \( \int_0^{a} f(x) dx = \int_0^{a} f(a - x) dx \). Here, we set \( a = \frac{\pi}{2} \):
\[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin\left(\frac{\pi}{2} - x\right)}{4 + 3 \cos\left(\frac{\pi}{2} - x\right)}\right) dx \]

3. **Simplify the Expression:**
Recall that \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \). Thus, we have:
\[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \cos x}{4 + 3 \sin x}\right) dx \]

4. **Combine the Two Integrals:**
Now we can add the two expressions for \( I \):
\[ 2I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) dx + \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \cos x}{4 + 3 \sin x}\right) dx \]

5. **Use Logarithmic Properties:**
Using the property \( \log a + \log b = \log(ab) \):
\[ 2I = \int_0^{\frac{\pi}{2}} \log\left(\frac{(4 + 3 \sin x)(4 + 3 \cos x)}{(4 + 3 \cos x)(4 + 3 \sin x)}\right) dx \]
This simplifies to:
\[ 2I = \int_0^{\frac{\pi}{2}} \log(1) dx \]

6. **Evaluate the Integral:**
Since \( \log(1) = 0 \):
\[ 2I = \int_0^{\frac{\pi}{2}} 0 \, dx = 0 \]

7. **Conclusion:**
Thus, we have:
\[ I = 0 \]

### Final Answer:
The value of the integral \( \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) dx \) is \( 0 \).