`int_(0)^(pi//2)log ((1+ sinx)/ (1+ cos x )) dx=`

Options

A. 0
B. `pi`
C. `pi/2`
D. `pi/4`

Answer

To solve the definite integral

\[ I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1 + \sin x}{1 + \cos x}\right) \, dx, \]

we will use a property of definite integrals.

### Step 1: Set up the integral
Let

\[ I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1 + \sin x}{1 + \cos x}\right) \, dx. \]

### Step 2: Use the property of definite integrals
We can use the property that

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In our case, \( a = 0 \) and \( b = \frac{\pi}{2} \). Therefore, we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1 + \sin\left(\frac{\pi}{2} - x\right)}{1 + \cos\left(\frac{\pi}{2} - x\right)}\right) \, dx. \]

### Step 3: Simplify the expression
Using the identities \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \), we have:

\[ I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1 + \cos x}{1 + \sin x}\right) \, dx. \]

### Step 4: Combine the two integrals
Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1 + \sin x}{1 + \cos x}\right) \, dx \)
2. \( I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1 + \cos x}{1 + \sin x}\right) \, dx \)

Adding these two equations, we get:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \log\left(\frac{1 + \sin x}{1 + \cos x}\right) + \log\left(\frac{1 + \cos x}{1 + \sin x}\right) \right) \, dx. \]

### Step 5: Use logarithmic properties
Using the property of logarithms \( \log a + \log b = \log(ab) \):

\[ 2I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{(1 + \sin x)(1 + \cos x)}{(1 + \cos x)(1 + \sin x)}\right) \, dx = \int_{0}^{\frac{\pi}{2}} \log(1) \, dx. \]

Since \( \log(1) = 0 \):

\[ 2I = \int_{0}^{\frac{\pi}{2}} 0 \, dx = 0. \]

### Step 6: Solve for \( I \)
Thus, we have:

\[ 2I = 0 \implies I = 0. \]

### Final Answer
Therefore,

\[ \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1 + \sin x}{1 + \cos x}\right) \, dx = 0. \]