Consider is `int_(0)^(pi//2) ln (sinx)dx` equal to ?
What is `int_(0)^(pi//2)ln (sinx) dx` equal to ?

Options

A. `4I`
B. `2I`
C. `I`
D. `I//2`

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx \), we will use a symmetry property of definite integrals.

### Step-by-step Solution:

1. **Define the Integral**:
Let \( I = \int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx \).

2. **Use the Property of the Integral**:
We can use the property of definite integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
For our case, we let \( a = \frac{\pi}{2} \). Thus,
\[ I = \int_{0}^{\frac{\pi}{2}} \ln(\sin(\frac{\pi}{2} - x)) \, dx \]

3. **Simplify the Integral**:
We know that \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \). Therefore, we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \ln(\cos x) \, dx \]

4. **Combine the Two Integrals**:
Now we have two expressions for \( I \):
\[ I = \int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \ln(\cos x) \, dx \]
Adding these two equations gives:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx + \int_{0}^{\frac{\pi}{2}} \ln(\cos x) \, dx \]

5. **Use Logarithmic Properties**:
Using the property of logarithms, we can combine the integrals:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \ln(\sin x \cos x) \, dx \]
We know that \( \sin x \cos x = \frac{1}{2} \sin(2x) \), so:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \ln\left(\frac{1}{2} \sin(2x)\right) \, dx \]

6. **Split the Integral**:
This can be split into two parts:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \ln\left(\frac{1}{2}\right) \, dx + \int_{0}^{\frac{\pi}{2}} \ln(\sin(2x)) \, dx \]
The first integral evaluates to:
\[ \int_{0}^{\frac{\pi}{2}} \ln\left(\frac{1}{2}\right) \, dx = \ln\left(\frac{1}{2}\right) \cdot \frac{\pi}{2} = -\frac{\pi}{2} \ln(2) \]

7. **Evaluate the Second Integral**:
For the second integral, we can change the variable \( u = 2x \), which gives \( du = 2dx \) or \( dx = \frac{du}{2} \). The limits change from \( 0 \) to \( \pi \):
\[ \int_{0}^{\frac{\pi}{2}} \ln(\sin(2x)) \, dx = \frac{1}{2} \int_{0}^{\pi} \ln(\sin u) \, du \]
We know that \( \int_{0}^{\pi} \ln(\sin u) \, du = -\pi \ln(2) \) (a known result).

8. **Combine Everything**:
Thus,
\[ 2I = -\frac{\pi}{2} \ln(2) + \frac{1}{2} \left(-\pi \ln(2)\right) \]
Simplifying gives:
\[ 2I = -\frac{\pi}{2} \ln(2) - \frac{\pi}{2} \ln(2) = -\pi \ln(2) \]
Therefore,
\[ I = -\frac{\pi}{2} \ln(2) \]

### Final Result:
\[ \int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx = -\frac{\pi}{2} \ln(2) \]