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

Options

A. `2pilog_(e)(1/2)`
B. `pi log_(e)2`
C. `(pi)/2log_(e)(1/2)`
D. `pilog_(e)(1/2)`

Answer

To solve the integral \( I = \int_0^{\pi} \log(\sin^2 x) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
Using the logarithmic property \( \log(a^b) = b \log(a) \), we can rewrite the integral:
\[ I = \int_0^{\pi} \log(\sin^2 x) \, dx = \int_0^{\pi} 2 \log(\sin x) \, dx = 2 \int_0^{\pi} \log(\sin x) \, dx \]

### Step 2: Use Symmetry Property
We can use the property of definite integrals:
\[ \int_0^{a} f(x) \, dx = \int_0^{a} f(a - x) \, dx \]
For our case, let \( f(x) = \log(\sin x) \). Then:
\[ \int_0^{\pi} \log(\sin x) \, dx = \int_0^{\pi} \log(\sin(\pi - x)) \, dx \]
Since \( \sin(\pi - x) = \sin x \), we have:
\[ \int_0^{\pi} \log(\sin(\pi - x)) \, dx = \int_0^{\pi} \log(\sin x) \, dx \]
Thus, we can conclude:
\[ \int_0^{\pi} \log(\sin x) \, dx = \int_0^{\pi} \log(\sin x) \, dx \]

### Step 3: Combine the Integrals
Now, we can combine the two integrals:
\[ 2I = \int_0^{\pi} \log(\sin x) \, dx + \int_0^{\pi} \log(\sin(\pi - x)) \, dx = \int_0^{\pi} \log(\sin x) \, dx + \int_0^{\pi} \log(\sin x) \, dx = 2 \int_0^{\pi} \log(\sin x) \, dx \]
Thus, we have:
\[ I = 2 \int_0^{\pi} \log(\sin x) \, dx \]

### Step 4: Change of Variable
Now, we can change the variable in the integral:
Let \( x = \frac{\pi}{2} - t \), then \( dx = -dt \). The limits change from \( x = 0 \) to \( x = \pi \) which corresponds to \( t = \frac{\pi}{2} \) to \( t = -\frac{\pi}{2} \). Thus:
\[ \int_0^{\pi} \log(\sin x) \, dx = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \log(\sin(\frac{\pi}{2} - t)) (-dt) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log(\cos t) \, dt \]

### Step 5: Combine the Results
Now we have:
\[ I = 2 \left( \int_0^{\pi} \log(\sin x) \, dx + \int_0^{\pi} \log(\cos x) \, dx \right) \]
Using the known result:
\[ \int_0^{\frac{\pi}{2}} \log(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \log(\cos x) \, dx = -\frac{\pi}{2} \log 2 \]
Thus:
\[ I = 2 \left( -\frac{\pi}{2} \log 2 + -\frac{\pi}{2} \log 2 \right) = 2 \left( -\pi \log 2 \right) = -2\pi \log 2 \]

### Final Result
Therefore, the final result is:
\[ I = -2\pi \log 2 \]