The value of `int_(0)^(pi//2) (2log sin x-log sin 2x)dx`, is

Options

A. `(pi)/(2)log2`
B. `-(pi)/(2)log2`
C. `pi log2`
D. `-pi log 2`

Explanation

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

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} (2 \log \sin x - \log \sin 2x) \, dx \]

### Step 2: Use the Identity for \(\sin 2x\)
We know that:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can express \(\log \sin 2x\) as:
\[ \log \sin 2x = \log(2 \sin x \cos x) = \log 2 + \log \sin x + \log \cos x \]

### Step 3: Substitute into the Integral
Substituting this back into the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( 2 \log \sin x - (\log 2 + \log \sin x + \log \cos x) \right) \, dx \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( \log \sin x - \log 2 - \log \cos x \right) \, dx \]

### Step 4: Split the Integral
We can split the integral into three parts:
\[ I = \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx - \int_{0}^{\frac{\pi}{2}} \log 2 \, dx - \int_{0}^{\frac{\pi}{2}} \log \cos x \, dx \]
The second integral is straightforward:
\[ \int_{0}^{\frac{\pi}{2}} \log 2 \, dx = \log 2 \cdot \frac{\pi}{2} \]

### Step 5: Use Symmetry for \(\log \cos x\)
Using the property of definite integrals:
\[ \int_{0}^{\frac{\pi}{2}} \log \cos x \, dx = \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx \]
Let \( J = \int_{0}^{\frac{\pi}{2}} \log \sin x \, dx \). Then:
\[ I = J - \frac{\pi}{2} \log 2 - J = -\frac{\pi}{2} \log 2 \]

### Step 6: Conclusion
Thus, we find that:
\[ I = -\frac{\pi}{2} \log 2 \]

### Final Answer
The value of the integral is:
\[ \boxed{-\frac{\pi}{2} \log 2} \]