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. 1

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 will use a symmetry property of definite integrals.

### Step 1: Define the integral
Let
\[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) dx \]

### Step 2: Use the property of definite integrals
We can use the property that
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
In our case, we will evaluate \( I \) by substituting \( x \) with \( \frac{\pi}{2} - x \):
\[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin\left(\frac{\pi}{2} - x\right)}{4 + 3 \cos\left(\frac{\pi}{2} - x\right)}\right) dx \]

### Step 3: Simplify using trigonometric identities
Using the identities \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \), we have:
\[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \cos x}{4 + 3 \sin x}\right) dx \]

### Step 4: Add the two expressions for \( I \)
Now we have two expressions for \( I \):
1. \( I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) dx \)
2. \( I = \int_0^{\frac{\pi}{2}} \log\left(\frac{4 + 3 \cos x}{4 + 3 \sin x}\right) dx \)

Adding these two equations:
\[ 2I = \int_0^{\frac{\pi}{2}} \left( \log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x}\right) + \log\left(\frac{4 + 3 \cos x}{4 + 3 \sin x}\right) \right) dx \]

### Step 5: Use properties of logarithms
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:
\[ 2I = \int_0^{\frac{\pi}{2}} \log(1) \, dx \]

### Step 6: Evaluate the integral
Since \( \log(1) = 0 \):
\[ 2I = \int_0^{\frac{\pi}{2}} 0 \, dx = 0 \]
Thus,
\[ I = 0 \]

### Final Answer
The value of the integral is:
\[ \boxed{0} \]