Evaluate :
`int_(0)^(pi//2)Log ((5+3cosx)/(5+3sin x))dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{5 + 3 \cos x}{5 + 3 \sin x} \right) dx, \]

we can use a symmetry property of definite integrals.

### Step 1: Define the integral
Let
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{5 + 3 \cos x}{5 + 3 \sin x} \right) dx. \]

### Step 2: Use the property of integrals
Using the property of integrals, we can express \( I \) in terms of \( \pi/2 - x \):
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{5 + 3 \cos\left(\frac{\pi}{2} - x\right)}{5 + 3 \sin\left(\frac{\pi}{2} - x\right)} \right) dx. \]

### Step 3: Simplify the expression
Using the trigonometric identities \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \) and \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \), we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{5 + 3 \sin x}{5 + 3 \cos x} \right) dx. \]

### Step 4: Add the two expressions for \( I \)
Now, we have two expressions for \( I \):
1. \( I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{5 + 3 \cos x}{5 + 3 \sin x} \right) dx \)
2. \( I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{5 + 3 \sin x}{5 + 3 \cos x} \right) dx \)

Adding these two equations gives:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \log \left( \frac{5 + 3 \cos x}{5 + 3 \sin x} \right) + \log \left( \frac{5 + 3 \sin x}{5 + 3 \cos x} \right) \right) dx. \]

### Step 5: Combine the logarithms
Using the property of logarithms \( \log a + \log b = \log(ab) \), we can combine the logarithms:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{(5 + 3 \cos x)(5 + 3 \sin x)}{(5 + 3 \sin x)(5 + 3 \cos x)} \right) dx. \]

### Step 6: Simplify the integral
The expression simplifies to:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(1) \, dx = \int_{0}^{\frac{\pi}{2}} 0 \, dx = 0. \]

### Step 7: Solve for \( I \)
Thus, we have:
\[ 2I = 0 \implies I = 0. \]

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