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

Options

A. `pi/2log 2`
B. `-pi/2 log 2`
C. `pi/4 log 2`
D. `(-pi)/4 log 2`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} (2 \log \sin x - \log \sin 2x) \, dx \), we will 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 Property of Definite Integrals
We can use the property of definite integrals that states:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \frac{\pi}{2} \), so we have:
\[ I = \int_{0}^{\frac{\pi}{2}} (2 \log \sin\left(\frac{\pi}{2} - x\right) - \log \sin(2(\frac{\pi}{2} - x))) \, dx \]

### Step 3: Simplify the Functions
Using the identities \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \sin(2(\frac{\pi}{2} - x)) = \sin(\pi - 2x) = \sin 2x \):
\[ I = \int_{0}^{\frac{\pi}{2}} (2 \log \cos x - \log \sin 2x) \, dx \]

### Step 4: Combine the Two Integrals
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{\frac{\pi}{2}} (2 \log \sin x - \log \sin 2x) \, dx \)
2. \( I = \int_{0}^{\frac{\pi}{2}} (2 \log \cos x - \log \sin 2x) \, dx \)

Adding these two equations gives:
\[ 2I = \int_{0}^{\frac{\pi}{2}} (2 \log \sin x + 2 \log \cos x - 2 \log \sin 2x) \, dx \]

### Step 5: Simplify the Integral
Using the logarithmic identity \( \log a + \log b = \log(ab) \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left(2 \log(\sin x \cos x) - 2 \log \sin 2x\right) \, dx \]
Since \( \sin 2x = 2 \sin x \cos x \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left(2 \log(\sin x \cos x) - 2 \log(2 \sin x \cos x)\right) \, dx \]

### Step 6: Factor Out the Common Terms
This can be simplified to:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left(2 \log(\sin x \cos x) - 2 \log 2 - 2 \log(\sin x \cos x)\right) \, dx \]
Thus,
\[ 2I = -2 \int_{0}^{\frac{\pi}{2}} \log 2 \, dx \]
\[ 2I = -2 \cdot \log 2 \cdot \frac{\pi}{2} \]

### Step 7: Solve for \( I \)
Dividing both sides by 2:
\[ I = -\frac{\pi}{2} \log 2 \]

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