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 with the expression for \( I \):
\[ I = \int_{0}^{\frac{\pi}{2}} (2 \log \sin x - \log \sin 2x) \, dx \]

### Step 2: Substitute \( \sin 2x \)
Using the double angle formula, we know that:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite \( \log \sin 2x \):
\[ \log \sin 2x = \log(2 \sin x \cos x) = \log 2 + \log \sin x + \log \cos x \]
Substituting this back into the integral gives:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( 2 \log \sin x - (\log 2 + \log \sin x + \log \cos x) \right) \, dx \]

### Step 3: Simplify the integral
Now, simplify the expression inside 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:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( \log \sin x - \log 2 - \log \cos x \right) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\sin x}{2 \cos x} \right) \, dx \]

### Step 4: Split the integral
We can split the integral into two 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:
\[ \int_{0}^{\frac{\pi}{2}} \log 2 \, dx = \frac{\pi}{2} \log 2 \]

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

### Final Result
Therefore, the value of the integral is:
\[ \boxed{-\frac{\pi}{2} \log 2} \]