The value of `int_0^(pi//2) log ((4+3 sin x)/(4+3 cos x)) dx` is:

Options

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

Answer

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 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 states:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
In our case, we will substitute \( x \) with \( \frac{\pi}{2} - x \):
\[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin\left(\frac{\pi}{2} - x\right)}{4 + 3 \cos\left(\frac{\pi}{2} - x\right)}\right) \, dx \]
Since \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \), we can rewrite the integral:
\[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \cos x}{4 + 3 \sin x}\right) \, dx \]

3. **Add the Two Integrals**:
Now we have two expressions for \( I \):
\[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) \, dx \]
\[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \cos x}{4 + 3 \sin x}\right) \, dx \]
Adding these two equations:
\[ 2I = \int_0^{\frac{\pi}{2}} \left[ \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) + \log\left(\frac{4 + 3 \cos x}{4 + 3 \sin x}\right) \right] \, dx \]

4. **Simplify the Logarithm**:
Using the property of logarithms \( \log a + \log b = \log(ab) \):
\[ 2I = \int_0^{\frac{\pi}{2}} \log\left(1\right) \, dx \]
Since \( \log(1) = 0 \):
\[ 2I = 0 \]

5. **Solve for \( I \)**:
Therefore,
\[ I = 0 \]

### Final Answer:
\[ \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) \, dx = 0 \]