`I_(1)=int_(0)^((pi)/2)Ln (sinx)dx, I_(2)=int_(-pi//4)^(pi//4)Ln(sinx+cosx)dx`. Then

Options

A. `I_(1)=2I_(2)`
B. `I_(2)=2I_(1)`
C. `I_(1)=4I_(2)`
D. `I_(2)=4I_(1)`

Explanation

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and establish a relationship between them.

### Step 1: Define the integrals
Let:
\[ I_1 = \int_0^{\frac{\pi}{2}} \ln(\sin x) \, dx \]
\[ I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) \, dx \]

### Step 2: Transform \( I_2 \)
To analyze \( I_2 \), we can use the property of definite integrals. We replace \( x \) with \( -x \):
\[ I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin(-x) + \cos(-x)) \, dx \]
Using the properties of sine and cosine:
\[ \sin(-x) = -\sin x \quad \text{and} \quad \cos(-x) = \cos x \]
Thus, we have:
\[ I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(-\sin x + \cos x) \, dx \]
This can be rewritten as:
\[ I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos x - \sin x) \, dx \]

### Step 3: Add the two forms of \( I_2 \)
Now we can add the two expressions for \( I_2 \):
\[ 2I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) \, dx + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos x - \sin x) \, dx \]
Using the property of logarithms:
\[ \ln(a) + \ln(b) = \ln(ab) \]
This gives us:
\[ 2I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln((\sin x + \cos x)(\cos x - \sin x)) \, dx \]
Now, simplifying the product:
\[ (\sin x + \cos x)(\cos x - \sin x) = \cos^2 x - \sin^2 x = \cos(2x) \]
Thus:
\[ 2I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos(2x)) \, dx \]

### Step 4: Use symmetry of the integral
Since the function \( \ln(\cos(2x)) \) is even, we can write:
\[ 2I_2 = 2 \int_0^{\frac{\pi}{4}} \ln(\cos(2x)) \, dx \]
So we have:
\[ I_2 = \int_0^{\frac{\pi}{4}} \ln(\cos(2x)) \, dx \]

### Step 5: Change of variables
Now, let \( t = 2x \), then \( dt = 2dx \) or \( dx = \frac{dt}{2} \). Changing the limits accordingly:
- When \( x = 0 \), \( t = 0 \)
- When \( x = \frac{\pi}{4} \), \( t = \frac{\pi}{2} \)

Thus:
\[ I_2 = \int_0^{\frac{\pi}{2}} \ln(\cos(t)) \frac{dt}{2} = \frac{1}{2} \int_0^{\frac{\pi}{2}} \ln(\cos(t)) \, dt \]

### Step 6: Relate \( I_1 \) and \( I_2 \)
We know from the properties of integrals that:
\[ \int_0^{\frac{\pi}{2}} \ln(\sin(t)) \, dt = \int_0^{\frac{\pi}{2}} \ln(\cos(t)) \, dt \]
Thus:
\[ I_1 = \int_0^{\frac{\pi}{2}} \ln(\sin(t)) \, dt = \int_0^{\frac{\pi}{2}} \ln(\cos(t)) \, dt \]
This gives us:
\[ I_1 = -\frac{\pi}{2} \ln(2) \]
And since:
\[ I_2 = \frac{1}{2} \int_0^{\frac{\pi}{2}} \ln(\cos(t)) \, dt = \frac{1}{2} \cdot I_1 \]
Thus:
\[ I_1 = 2I_2 \]

### Conclusion
The relationship between \( I_1 \) and \( I_2 \) is:
\[ I_1 = 2I_2 \]