`int_(0)^(1) lnsin(pi/2x) dx`

Answer

To solve the integral \( I = \int_0^1 \ln(\sin(\frac{\pi}{2} x)) \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Define the Integral**:
\[ I = \int_0^1 \ln(\sin(\frac{\pi}{2} x)) \, dx \]

2. **Use the Symmetry Property of Integrals**:
We can use the property of integrals that states:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
Here, \( a = 0 \) and \( b = 1 \), so we have:
\[ I = \int_0^1 \ln(\sin(\frac{\pi}{2} (1 - x))) \, dx \]

3. **Simplify the Argument**:
Using the identity \( \sin(\frac{\pi}{2} (1 - x)) = \cos(\frac{\pi}{2} x) \):
\[ I = \int_0^1 \ln(\cos(\frac{\pi}{2} x)) \, dx \]

4. **Add the Two Integrals**:
Now, we have two expressions for \( I \):
\[ I = \int_0^1 \ln(\sin(\frac{\pi}{2} x)) \, dx \]
\[ I = \int_0^1 \ln(\cos(\frac{\pi}{2} x)) \, dx \]
Adding these two equations:
\[ 2I = \int_0^1 \left( \ln(\sin(\frac{\pi}{2} x)) + \ln(\cos(\frac{\pi}{2} x)) \right) \, dx \]

5. **Combine the Logarithms**:
Using the property \( \ln(a) + \ln(b) = \ln(ab) \):
\[ 2I = \int_0^1 \ln(\sin(\frac{\pi}{2} x) \cos(\frac{\pi}{2} x)) \, dx \]

6. **Use the Double Angle Identity**:
Recall that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ \sin(\frac{\pi}{2} x) \cos(\frac{\pi}{2} x) = \frac{1}{2} \sin(\pi x) \]
Thus:
\[ 2I = \int_0^1 \ln\left(\frac{1}{2} \sin(\pi x)\right) \, dx \]

7. **Separate the Integral**:
This can be separated into two integrals:
\[ 2I = \int_0^1 \ln(\sin(\pi x)) \, dx - \int_0^1 \ln(2) \, dx \]
The second integral simplifies to:
\[ \int_0^1 \ln(2) \, dx = \ln(2) \]
Therefore:
\[ 2I = \int_0^1 \ln(\sin(\pi x)) \, dx - \ln(2) \]

8. **Evaluate the Integral**:
The integral \( \int_0^1 \ln(\sin(\pi x)) \, dx \) is known and equals \( -\ln(2) \):
\[ 2I = -\ln(2) - \ln(2) \]
\[ 2I = -2\ln(2) \]

9. **Solve for \( I \)**:
Dividing both sides by 2:
\[ I = -\ln(2) \]

### Final Answer:
\[ \int_0^1 \ln(\sin(\frac{\pi}{2} x)) \, dx = -\ln(2) \]