`int_(0)^(pi//2) [2log sin x-log sin 2x] dx`=

Options

A. `pi log 2`
B. `-pi log 2`
C. `(pi)/(2) "log"(1)/(2)`
D. `-(pi)/(2) "log" (1)/(2)`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \left( 2 \log \sin x - \log \sin 2x \right) dx \), we will follow these steps:

### Step 1: Rewrite the Integral
Using the properties of logarithms, we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( 2 \log \sin x - \log \sin 2x \right) dx = \int_{0}^{\frac{\pi}{2}} \left( \log \sin^2 x - \log \sin 2x \right) dx \]
Using the property \( \log a - \log b = \log \frac{a}{b} \), we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\sin^2 x}{\sin 2x} \right) dx \]

### Step 2: Substitute for \(\sin 2x\)
Recall that \( \sin 2x = 2 \sin x \cos x \). Therefore, we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\sin^2 x}{2 \sin x \cos x} \right) dx \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\sin x}{2} \right) dx - \int_{0}^{\frac{\pi}{2}} \log (\cos x) dx \]

### Step 3: Split the Integral
Now we can split the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx - \int_{0}^{\frac{\pi}{2}} \log 2 \, dx - \int_{0}^{\frac{\pi}{2}} \log (\cos x) \, dx \]
The second integral evaluates to:
\[ \int_{0}^{\frac{\pi}{2}} \log 2 \, dx = \frac{\pi}{2} \log 2 \]
Thus, we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx - \frac{\pi}{2} \log 2 - \int_{0}^{\frac{\pi}{2}} \log \cos x \, dx \]

### Step 4: Use Symmetry
Using the property of integrals, we know:
\[ \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx = \int_{0}^{\frac{\pi}{2}} \log \cos x \, dx \]
Let \( J = \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx \). Therefore:
\[ I = J - \frac{\pi}{2} \log 2 - J = -\frac{\pi}{2} \log 2 \]

### Step 5: Final Result
Thus, we conclude that:
\[ I = -\frac{\pi}{2} \log 2 \]

### Final Answer
The value of the integral is:
\[ I = \frac{\pi}{2} \log \left( \frac{1}{2} \right) \]