The value of the integral `int_(0)^(1) log sin ((pix)/(2))dx` is

Options

A. log 2
B. `-log 2`
C. `(pi)/(2)log 2`
D. `-(pi)/(2) log 2`

Answer

To solve the integral \( I = \int_{0}^{1} \log \sin\left(\frac{\pi x}{2}\right) \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Substitution**:
Let \( t = \frac{\pi x}{2} \). Then, we differentiate to find \( dx \):
\[ dx = \frac{2}{\pi} dt \]
When \( x = 0 \), \( t = 0 \) and when \( x = 1 \), \( t = \frac{\pi}{2} \).

2. **Change of Limits and Integral**:
Substitute \( x \) and \( dx \) in the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin t) \cdot \frac{2}{\pi} dt = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt \]

3. **Define a New Integral**:
Let \( J = \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt \). Then, we have:
\[ I = \frac{2}{\pi} J \]

4. **Using Symmetry**:
We can use the property of definite integrals:
\[ J = \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt = \int_{0}^{\frac{\pi}{2}} \log(\cos t) \, dt \]
Thus, we can write:
\[ 2J = \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt + \int_{0}^{\frac{\pi}{2}} \log(\cos t) \, dt = \int_{0}^{\frac{\pi}{2}} \log(\sin t \cos t) \, dt \]

5. **Simplifying the Integral**:
Using the identity \( \sin t \cos t = \frac{1}{2} \sin(2t) \):
\[ 2J = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2} \sin(2t)\right) \, dt \]
This can be split into two integrals:
\[ 2J = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right) \, dt + \int_{0}^{\frac{\pi}{2}} \log(\sin(2t)) \, dt \]

6. **Evaluating the First Integral**:
The first integral evaluates to:
\[ \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right) \, dt = \log\left(\frac{1}{2}\right) \cdot \frac{\pi}{2} = -\frac{\pi}{2} \log(2) \]

7. **Evaluating the Second Integral**:
For the second integral, use the substitution \( u = 2t \):
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin(2t)) \, dt = \frac{1}{2} \int_{0}^{\pi} \log(\sin u) \, du \]
By a known result, \( \int_{0}^{\pi} \log(\sin u) \, du = -\pi \log(2) \). Therefore:
\[ \frac{1}{2} \int_{0}^{\pi} \log(\sin u) \, du = -\frac{\pi}{2} \log(2) \]

8. **Combining Results**:
Thus, we have:
\[ 2J = -\frac{\pi}{2} \log(2) - \frac{\pi}{2} \log(2) = -\pi \log(2) \]
Therefore:
\[ J = -\frac{\pi}{2} \log(2) \]

9. **Final Calculation**:
Substitute \( J \) back into the expression for \( I \):
\[ I = \frac{2}{\pi} \left(-\frac{\pi}{2} \log(2)\right) = -\log(2) \]

### Final Answer:
\[ I = -\log(2) \]