Evaluate: `int_(0)^((pi)/(2)) log (sin x) dx`

Answer

To evaluate the integral \( I = \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \), we can use the property of definite integrals. Here’s a step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \]

### Step 2: Use the Property of Definite Integrals
Using the property of integrals, we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin(\frac{\pi}{2} - x)) \, dx \]
Since \( \sin(\frac{\pi}{2} - x) = \cos x \), we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx \]

### Step 3: Combine the Two Integrals
Now we can add both expressions for \( I \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx + \int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx \]
This can be combined using the logarithm property:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \log(\sin x) + \log(\cos x) \right) \, dx = \int_{0}^{\frac{\pi}{2}} \log(\sin x \cos x) \, dx \]

### Step 4: Simplify the Logarithm
Using the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \), we can rewrite the integral:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2} \sin(2x)\right) \, dx \]
This can be separated as:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right) \, dx + \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx \]

### Step 5: Evaluate the First Integral
The first integral evaluates to:
\[ \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right) \, dx = \log\left(\frac{1}{2}\right) \cdot \frac{\pi}{2} = -\frac{\pi}{2} \log(2) \]

### Step 6: Change of Variable for the Second Integral
For the second integral, we make the substitution \( u = 2x \) which gives \( du = 2dx \) or \( dx = \frac{du}{2} \). The limits change from \( 0 \) to \( \pi \):
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx = \frac{1}{2} \int_{0}^{\pi} \log(\sin u) \, du \]
Since \( \int_{0}^{\pi} \log(\sin u) \, du = 2 \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = 2I \), we have:
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx = \frac{1}{2} \cdot 2I = I \]

### Step 7: Combine Everything
Now substituting back into the equation for \( 2I \):
\[ 2I = -\frac{\pi}{2} \log(2) + I \]
Rearranging gives:
\[ 2I - I = -\frac{\pi}{2} \log(2) \implies I = -\frac{\pi}{2} \log(2) \]

### Final Result
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = -\frac{\pi}{2} \log(2) \]